Even Odd

Show algebraically whether the function is even: Solution $f(x) = -3(x + 2)(3x + 1)^2(x - 1)^3$

1. Even
2. Not even

All functions are even when ƒ(x) = ƒ(-x) Substitute and see if above ^ is true... $\begin{align} f(x) & = -3(x + 2)(3x + 1)^2(x - 1)^3 \\ \\ f(-x) & = -3((-x) + 2)(3(-x) + 1)^2((-x) - 1)^3 \\ \\ f(-x) & = -3(-x + 2)(-3x + 1)^2(-x - 1)^3 \\ \\ \end{align}$ Comparing the factored forms, the two functions are not equivalent. Since ƒ(x) ≠ ƒ(-x), then the function is NOT even.

Show algebraically whether the function is odd: Solution $f(x) = -3x^2 - 4$

1. Odd
2. Not odd

All functions are odd when ƒ(-x) = -ƒ(x) Substitute and see if above ^ is true... $\begin{align} f(x) & = -3x^2 - 4 \\ \\ f(-x) & = -3(-x)^2 - 4 \\ \\ f(-x) & = -3x^2 - 4 \\ \\ \end{align}$ $\begin{align} f(x) & = -3x^2 - 4 \\ \\ -f(x) & = (-1)[-3x^2 - 4] \\ \\ -f(x) & = 3x^2 + 4 \\ \\ \end{align}$ Since ƒ(-x) ≠ -ƒ(x), then the function is NOT odd.

Set and Interval Notation

Write each of the following as the other form... in either set notation or interval notation. Remember that interval notation uses parentheses ( ) for less than (<) or greater than (>), and square brackets [ ] is used for greater than or equal to [≥], and less than or equal to [≤]. ±∞ ... uses ... ( )
>, < ... uses ... ( )
≥, ≤ ... uses ... [ ] You will need some of the symbols below, provided for your copy-paste convenience. ∞     ≤     ≥     <     >

Write {XER | x ≥ -4} in interval notation. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & \{XER \ \big| \ x ≥ -4\} \\ \\ & = [\,-4, \, ∞\,) \\ \\ \end{align}$

Write {YER | 23 < y ≤ 67} in interval notation. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & \{YER \ \big| \ 23 < y ≤ 67\} \\ \\ & = (\,23, 67\,] \\ \\ \end{align}$

Write ( 0, 4 ] in set notation in the form {XER | ______ }. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & (\,0, 4\,] \\ \\ & = \{XER \ \big| \ 0 < x ≤ 4 \} \\ \\ \end{align}$

Parent Function Features

Sort the parent functions into their defining features: Solution

$y = x^2$

$y = \sqrt{x}$

$y = 2^x$

$y = \dfrac{1}{x}$

y-intercept at (0,1)

Domain: {XER | x ≥ 0}

End behaviour: x → -∞, y → +∞

Has both vertical and horizontal asymptotes

• $y = x^2$: End behaviour: x → -∞, y → +∞
• $y = \sqrt{x}$: Domain: {XER | x ≥ 0}
• $y = 2^x$: y-intercept at (0,1)
• $y = \dfrac{1}{x}$: Has both vertical and horizontal asymptotes

Factoring polynomials

Factor the general quadratic equation. Solution abx² + (an + bc)x + cn

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Factoring: Difference of Squares and Fractions

Factor fully.

$x^4 - y^2$ Solution

⁰

¹

²

³

⁴

⁵

⁶

⁷

⁸

⁹

⁻

⁺

⁽

⁾

₀

₁

₂

₃

₄

₅

₆

₇

₈

₉

₋

₊

₍

₎

Incorrect Attempts:

CHECK

Hint Unavailable

y = x⁴ - y²

y = (x²)² - (y)²

y = (x² + y)(x² - y)

$25x^4 - 9y^2$ Solution

⁰

¹

²

³

⁴

⁵

⁶

⁷

⁸

⁹

⁻

⁺

⁽

⁾

₀

₁

₂

₃

₄

₅

₆

₇

₈

₉

₋

₊

₍

₎

Incorrect Attempts:

CHECK

Hint Unavailable

y = 25x⁴ - 9y²

y = (5x²)² - (3y)²

y = (5x² + 3y)(5x² - 3y)

Complete the Square

Converting to vertex form by completing the square would yield which of the following? Solution Video y = ax² + bx + c

1. $a\left(x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a} + c$
2. $a\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{b^2}{4a} + c$
3. $a\left(x - \dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a} + c$
4. $a\left(x + \dfrac{b}{a}\right)^2 - \dfrac{b^2}{2a} - c$

$\begin{align} y & = ax^2 + bx + c \\ \\ & = a\left(x^2 + \dfrac{b}{a}x\right) + c \\ \\ & = a\left(x^2 + \dfrac{b}{a}x + \dfrac{b^2}{(2a)^2} - \dfrac{b^2}{(2a)^2}\right) + c \\ \\ & = a\left(x^2 + \dfrac{b}{a}x + \dfrac{b^2}{4a^2}\right) - \dfrac{b^2}{4\cancelto{1}{a^2}}(\cancel{a}) + c \\ \\ & = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \\ \\ \end{align}$

Based on this work, determine the equation for the axis of symmetry. Solution

Hint Clear Info

$-\,\dfrac{\quad\quad^{^{^2}}}{\quad\quad} \ +$

Incorrect Attempts:

CHECK

Hint Unavailable

The equation of the axis of symmetry is the x-value of the vertex... $\begin{align} x = -\frac{b}{2a} \\ \\ \end{align}$ Interestingly (as an aside) this is the equation of the max or min value... $\begin{align} y = - \frac{b^2}{4a} + c \\ \\ \end{align}$

Transformations with Function Notation

Which function is vertically compressed by a factor of ½, and translated 2 units right? Solution

1. y = 2ƒ (x + 2)
2. y = 2ƒ (x + 2)
3. y = ½ƒ (x + 2)
4. y = ½ƒ (x - 2)

For the function:

y = a ƒ (k (x - d) ) + c

When |a| > 0, the function is vertically compressed by a factor of 'a'.

When d > 0, the function is translated 'd' units right.

Real and Unreal (Imaginary, Complex Number) Roots of Polynomial Functions

The degree of a polynomial equals the number of real plus unreal roots. This rule does not work well for linear polynomials. Anyways, determine the number of roots in each...

For quadratic (U-shaped) polynomials. Solution

2 real roots

2 real, equal roots

2 unreal roots

Drag each, to reveal correct answer.

For cubic (S-shaped) polynomials. Solution

3 real/equal roots

1 real root, and 2 real/equal roots

3 real roots

1 real root, and 2 unreal roots

Drag each, to reveal correct answer.

For quartic (U or W-shaped) polynomials. (Some categories will apply to more than one graph) Solution

4 real/equal roots

4 real roots

4 unreal roots

4 real roots (2 equal sets)

Drag each, to reveal correct answer.

For more quartic (U or W-shaped) polynomials. (Some categories will apply to more than one graph) Solution

2 real, and 2 unreal roots

2 real/equal, and 2 unreal roots

2 equal/real roots, and 2 other real roots

3 equal/real roots, and 1 real root

Drag each, to reveal correct answer.

Finite Differences of a Polynomial

Use finite differences to determine the nature of the function. Solution

x	y
-3	-88
-2	-29
-1	-6
0	-1
1	4
2	27
3	86

1. Linear
2. Quadratic
3. Cubic
4. Quartic

The same number for 3^rd differences is a cubic function, like x³...

Degree and End Behaviour

Which of the following is the correct end behaviour of the function $f(x) = -3x^3 - 2x^2 - 3x + 1$ Solution

1. As x → -∞, y → -∞
2. As x → +∞, y → -∞
3. As x → -∞, y → 0
4. As x → +∞, y → +∞

This is just one end behaviour, of the right-hand side. The end behaviour for the left-hand side is as x → -∞, y → +∞.

Which statement is incorrect regarding end behaviours of polynomial functions? Solution

1. If the degree is even, the end behaviours are on the same side.
2. If the degree is odd, the end behaviours are on opposite sides.
3. If the degree is odd and the leading coefficient is positive, the ends are in the first and third quadrants.
4. If the degree is odd and the leading coefficient is negative, the ends are in the second and fourth quadrants.
5. If the degree is even and the leading coefficient is positive, the ends are in the first and second quadrants.
6. If the degree is even and the leading coefficient is negative, the ends are in the first and third quadrants.

If the degree is even and the leading coefficient is negative, the ends are in the third and fourth quadrants.

• The even degree means the ends are on the same side of the x-axis
• A negative leading coefficient means the function is reflected vertically

Determine the end behaviours of the function below as x → +∞ and as x → -∞. Solution $f(x) = -4x^5 - 3x^4 + x^3 + 7x^2 - 3$

1. As x → +∞... y → -∞
   As x → -∞... y → +∞
2. As x → +∞... y → +∞
   As x → -∞... y → +∞
3. As x → +∞... y → -∞
   As x → +∞... y → +∞
4. As x → +∞... y → 0
   As x → +∞... y → +∞

The degree of the function is odd, 5. Therefore the function has end behaviours on opposite sides of the x-axis.

Since the degree is odd and the leading coefficient is negative, the ends are in the second and fourth quadrants.

As x → +∞... y → -∞ (in the fourth quadrant).

As x → -∞... y → +∞ (in the second quadrant).

Sketching Polynomial Functions

In your notebook, sketch a graph of the function $f(x) = \dfrac{1}{2}x^3(x + 5)^2(x^2 - 9)$ Solution

$\begin{align} f(0) & = \dfrac{1}{2}(0)^3((0) + 5)^2((0)^2 - 9) \\ \\ & = 0 \\ \\ & \text{The y-intercept is located at (0, 0)} \\ \\ \\ \\ \\ \\ 0 & = \dfrac{1}{2}x^3(x + 5)^2(x^2 - 9) \\ \\ 0 & = \dfrac{1}{2}x^3(x + 5)^2(x + 3)(x - 3) \\ \\ & \text{The roots are located at x = -5, -3, 0, 3} \\ \\ \\ \\ \\ \\ f(-6) & = -2916 \\ \\ f(-4) & = -224 \\ \\ f(-2) & = +180 \\ \\ f(1) & = -144 \\ \\ f(4) & = +18,144 \\ \\ & \text{Use points on either side of the roots as a guide when sketching} \\ \\ \end{align}$ This gives all the information needed to make a proper sketch. (Exact turning points not required.)

Transformations of Polynomial Functions

State the transformations on the following polynomial. Solution $f(x) = -\dfrac{2}{3}(5x - 5)^3 + 10$

Hint Clear Info

· Vertical compression by a factor of $\dfrac{\quad\quad}{}$
· Reflection across the
-axis · Horizontal compression by a factor of $\dfrac{\quad\quad}{}$
· Translation
unit(s) right · Translation unit(s) up

Incorrect Attempts:

CHECK

Hint Unavailable

Factor out the 'k' value first. $f(x) = -\dfrac{2}{3}(5(x - 1))^3 + 10$ Compare to: $f(x) = a(k(x - d))^n + c$

The transformations:
Vertical compression by a factor of $\dfrac{2}{3}$
Reflection accross the x-axis.
Horizontal compression by a factor of $\dfrac{1}{5}$
Translation 1 unit right
Translation 10 units up

The Remainder Theorem Part II

Determine the remainder, without the quotient. Solution Dividend: ƒ(x) = 3x³ - 2x² + 10x - 5
Divisor: (x + 4)(x + 3)

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

The remainder is always linear regardless of the degree of the dividend and divisor... It is either a constant, or has a degree of 1. Unless indicated otherwise, assume the remainder has a degree of 1, and represent as nx + m $\begin{align} Dividend & = (Divisor)(Quotient) + \color{SkyBlue}{Remainder} \\ \\ 3x^3 - 2x^2 + 10x - 5 & = (x + 4)(x + 3)(Quotient) + \color{SkyBlue}{nx + m} \end{align}$ From the root (x + 4), determine and substitute ƒ(-4) = -269... $\begin{align} 3(-4)^3 - 2(-4)^2 + 10(-4) - 5 & = \cancelto{0}{((-4) + 4)((-4) + 3)(Quotient)} + n(-4) + m \\ \\ -269 & = -4n + m \end{align}$ From the root (x + 3), determine and substitute ƒ(-3) = -134... $\begin{align} 3(-3)^3 - 2(-3)^2 + 10(-3) - 5 & = \cancelto{0}{((-3) + 4)((-3) + 3)(Quotient)} + n(-3) + m \\ \\ -134 & = -3n + m \end{align}$ Now solve the system of equations (using elimination here)... $\begin{align} -269 & = -4n + m \\ \\ \underline{(-) \quad -134} & = \underline{-3n + m} \\ \\ -135 & = -n \\ \\ n & = 135 \end{align}$ Determine 'm'.. $\begin{align} -269 & = -4(135) + m \\ \\ m & = 271 \\ \\ \end{align}$ Therefore the remainder is: 135x + 271

Given the conditions below, determine the remainder, without the quotient or dividend, when the same divisor is divided into 3·ƒ(x) - 2. Solution Dividend: ƒ(x)
Divisor: (x - 1)
Remainder: -2

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

The question indicates that the remainder is a constant, so use one variable, 'R' to represent the remainder as a constant. $\begin{align} Dividend & = (Divisor)(Quotient) + \color{SkyBlue}{Remainder} \\ \\ f(x) & = (x - 1)(Quotient) + \color{SkyBlue}{R} \end{align}$ Substitute... $\begin{align} f(x) & = (x - 1)(Quotient) + \color{SkyBlue}{R} \\ \\ f(1) & = \cancelto{0}{((1) - 1)(Quotient)} + (-2) \\ \\ f(1) & = -2 \end{align}$ Now use this value for 3·ƒ(x) - 2... $\begin{align} 3·f(x) - 2 & = (x - 1)(Quotient) + \color{SkyBlue}{R} \\ \\ 3·f(1) - 2 & = \cancelto{0}{((1) - 1)(Quotient)} + \color{SkyBlue}{R} \\ \\ 3(-2) - 2 & = \color{SkyBlue}{R} \\ \\ R & = -8 \end{align}$

Difference of Cubes, Sum of Cubes

Factor, use proper form.

64x³ - 27 Solution Video

⁰

¹

²

³

⁴

⁵

⁶

⁷

⁸

⁹

⁻

⁺

⁽

⁾

₀

₁

₂

₃

₄

₅

₆

₇

₈

₉

₋

₊

₍

₎

Incorrect Attempts:

CHECK

Hint Unavailable

125x³ + 216y³ Solution

⁰

¹

²

³

⁴

⁵

⁶

⁷

⁸

⁹

⁻

⁺

⁽

⁾

₀

₁

₂

₃

₄

₅

₆

₇

₈

₉

₋

₊

₍

₎

Incorrect Attempts:

CHECK

Hint Unavailable

Terms of Division

Classify the terms for the following. Solution $\dfrac{12x}{3x} = 4$

12x

3x

4

0

Quotient

Dividend

Remainder

Divisor

The divisor (3x) is what you divide the dividend (12x) by to give you the quotient (4).

There is no remainder, or a remainder of zero.

Solving Polynomial Equations

Determine the x-intercepts of $0 = (x^3 + 27)(2x - 3)$, giving exact values. Solution

1. $x = \dfrac{3}{2}, 3$
2. $x = \dfrac{3}{2}, -3$
3. $x = \dfrac{3}{2}, 3$
4. $x = \dfrac{2}{3}, -3$

$\begin{align} 2x - 3 & = 0 \\ \\ 2x & = 3 \\ \\ x & = \dfrac{3}{2} \\ \\ \\ \\ \\ \\ x^3 + 27 & = 0 \\ \\ x^3 & = -27 \\ \\ \sqrt[3]{x^3} & = \sqrt[3]{-27} \\ \\ x & = -3 \\ \\ \end{align}$

Solve $0 = -5(x^2 + 25)(x^3 + 125)$, giving exact values. Solution

1. 5
2. -5
3. 5, -5
4. None of the above

$\begin{align} x^2 + 25 & = 0 \\ \\ x^2 & = -25 \\ \\ \sqrt{x^2} & = \cancel{\sqrt{-25}} \\ \\ & \text{No real roots here, only 2 non-real roots} \\ \\ \\ \\ \\ \\ x^3 + 125 & = 0 \\ \\ x^3 & = -125 \\ \\ \sqrt[3]{x^3} & = \sqrt[3]{-125} \\ \\ x & = -5 \\ \\ \end{align}$

Exponents Review

Express each as a power with a base of 3

$81^{12}$ Solution

⁰

¹

²

³

⁴

⁵

⁶

⁷

⁸

⁹

⁻

⁺

⁽

⁾

₀

₁

₂

₃

₄

₅

₆

₇

₈

₉

₋

₊

₍

₎

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = 81^{12} \\ \\ & = (3^4)^{12} \\ \\ & = 3^{4 × 12} \\ \\ & = 3^{48} \\ \\ \end{align}$

$3^7 + 3^7 + 3^7$ Solution

⁰

¹

²

³

⁴

⁵

⁶

⁷

⁸

⁹

⁻

⁺

⁽

⁾

₀

₁

₂

₃

₄

₅

₆

₇

₈

₉

₋

₊

₍

₎

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & 3^7 + 3^7 + 3^7 \\ \\ & = 3(3^7) \\ \\ & = 3^1(3^7) \\ \\ & = 3^{1 + 7} \\ \\ & = 3^8 \\ \\ \end{align}$

Root Equation Review

Solve for the variable:

$\sqrt{3x + 4} = 4$ Solution

x =

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} \sqrt{3x + 4} & = 4 \\ \\ (\sqrt{3x + 4})^2 & = 4^2 \\ \\ 3x + 4 & = 16 \\ \\ 3x & = 12 \\ \\ x & = 4 \\ \\ \end{align}$

$\sqrt{x + 9} = x + 7$ Solution

Hint Clear Info

x₁ = x₂ =

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} \sqrt{x + 9} & = x + 7 \\ \\ (\sqrt{x + 9})^2 & = (x + 7)^2 \\ \\ (\sqrt{x + 9})^2 & = (x + 7)(x + 7) \\ \\ x + 9 & = x^2 + 7x + 7x + 49 \\ \\ x + 9 & = x^2 + 14x + 49 \\ \\ 0 & = x^2 + 13x + 40 \\ \\ 0 & = (x + 8)(x + 5) \\ \\ x & = -8,\ -5 \\ \\ \end{align}$

$x + \sqrt{x + 1} = 5$ Solution

Hint Clear Info

x₁ = x₂ =

Incorrect Attempts:

CHECK

Hint Unavailable

Make sure to move first $\begin{align} x + \sqrt{x + 1} & = 5 \\ \\ \sqrt{x + 1} & = 5 - x \\ \\ (\sqrt{x + 1})^2 & = (5 - x)^2 \\ \\ x + 1 & = (5 - x)(5 - x) \\ \\ x + 1 & = 25 - 5x - 5x + x^2 \\ \\ 0 & = 24 - 11x + x^2 \\ \\ 0 & = (x - 3)(x - 8) \\ \\ x & = 3,\ 8 \\ \\ \end{align}$

$\sqrt[3]{8x + 5} = -3$ Solution Video

x =

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} \sqrt[3]{8x + 5} & = -3 \\ \\ (\sqrt[3]{8x + 5})^3 & = (-3)^3 \\ \\ 8x + 5 & = -27 \\ \\ 8x & = -32 \\ \\ x & = -4 \\ \\ \end{align}$

$2\sqrt{3x + 3} - 3x = 0$ Solution Video

Hint Clear Info

x₁ = ━━x₂ =

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} \sqrt{3x + 3} & = 3x \\ \\ (2\sqrt{3x + 3})^2 & = (3x)^2 \\ \\ 4(3x + 3) & = 9x^2 \\ \\ 12x + 12 & = 9x^2 \\ \\ 9x^2 - 12x - 12 & = 0 \\ \\ 3(3x^2 - 4x - 4) & = 0 \\ \\ 3(3x + 2)(x - 2) & = 0 \\ \\ x & = \dfrac{-2}{3},\ 2 \\ \\ \end{align}$

Rational Equation Word Problems - Working Together

Solve the rational equation word problems, and remember the general form equation for problems on 'working together'.

Marcello takes 9 more minutes than Koji to make 10 pizzas. Working together, they can make 10 pizzas in 20 minutes. How long does it take Koji to make 10 pizzas when he works alone? Solution

t =

Hint Clear Info

Incorrect Attempts:

CHECK

minutes

Hint Unavailable

Let 'x' be the time for Koji to make 10 pizzas.
Let 'x + 9' be the time for Marcello to make 10 pizzas. $\begin{align} \dfrac{20}{x} + \dfrac{20}{x + 9} & = 1 \\ \\ \dfrac{20}{\cancel{x}}(\cancel{x}) + \dfrac{20}{x + 9}(x) & = 1(x) \\ \\ 20 + \dfrac{20x}{x + 9} & = x \\ \\ 20(x + 9) + \dfrac{20x}{\cancel{x + 3}}(\cancel{x + 9}) & = x(x + 9) \\ \\ 20x + 180 + 20x & = x^2 + 9x \\ \\ 0 & = x^2 - 31x - 180 \\ \\ \\ \\ x & = \xcancel{-5},\ 36 \\ \\ \end{align}$ Koji takes 36 minutes...

Sandi and John are computer programmers with different experience levels. It takes John 10 hours less than double the time it takes Sandi to write 2000 lines of code. Working together, it takes them 12 hours. How long does it take Sandi working alone? Solution

t =

Hint Clear Info

Incorrect Attempts:

CHECK

hours

Hint Unavailable

Let 'x' be the time for Sandi to write 2000 lines of code.
Let '2x - 10' be the time for John to write 2000 lines of code. $\begin{align} \dfrac{12}{\cancel{x}}(\cancel{x}) + \dfrac{12}{2x - 10}(x) & = 1(x) \\ \\ 12(2x - 10) + \dfrac{12x}{\cancel{2x - 10}}(\cancel{2x - 10}) & = x(2x - 10) \\ \\ 12(2x - 10) + 12x & = x(2x - 10) \\ \\ 24x - 120 + 12x & = 2x^2 - 10x \\ \\ 0 & = 2x^2 - 46x + 120 \\ \\ x & = 3, 20 \end{align}$ Takes Sandi 20 hours.. [Can't take Sandi 3 hours because then for John, 2(3) - 10 would be negative time]

Rational Functions: Factoring, X-intercept(s), and Y-intercept

Determine the 'x' and 'y' intercepts in coordinate form (x, y). Round your answers to one decimal place and list in increasing order, where applicable.

$f(x) = \dfrac{2x^3 - 5x^2 + 2x}{x^2 - 5x + 4}$ Solution

Hint Clear Info

x-intercepts:
y-intercept:

Incorrect Attempts:

CHECK

Hint Unavailable

Factor fully, $\begin{align} & = \dfrac{2x^3 - 5x^2 + 2x}{x^2 - 5x + 4} \\ \\ & = \dfrac{x(2x^2 - 5x + 2)}{x^2 - 5x + 4} \\ \\ & = \dfrac{x(x - 2)(2x - 1)}{(x - 1)(x - 4)} \\ \\ \end{align}$ x-intercepts: (0, 0), (0.5, 0), (2, 0).

Solve for y-intercept with ƒ(0), if any... $\begin{align} f(0) & = \dfrac{2(0)^3 - 5(0)^2 + 2(0)}{(0)^2 - 5(0) + 4} \\ \\ & = \dfrac{0}{4} \\ \\ & = 0 \\ \\ \end{align}$ y-intercept: (0, 0)

$f(x) = \dfrac{x + 1}{x - 3} - \dfrac{2x + 3}{x - 2}$ Solution

Hint Clear Info

x-intercepts:
y-intercept:

Incorrect Attempts:

CHECK

Hint Unavailable

Simplify & factor fully, $\begin{align} & = \dfrac{(x + 1)(x - 2)}{(x - 3)(x - 2)} - \dfrac{(2x + 3)(x - 3)}{(x - 2)(x - 3)} \\ \\ & = \dfrac{x^2 - x - 2}{(x - 2)(x - 3)} - \dfrac{2x^2 - 3x - 9}{(x - 2)(x - 3)} \\ \\ & = \dfrac{x^2 - x - 2 - [2x^2 - 3x - 9]}{(x - 2)(x - 3)} \\ \\ & = \dfrac{-x^2 + 2x + 7}{(x - 2)(x - 3)} \\ \\ \end{align}$ Using the quadratic equation to determine the x-intercepts... $\begin{align} x & = \dfrac{-(b) ± \sqrt{(b)^2 \ - \ 4(a)(c)}}{2(a)} \\ \\ & = \dfrac{-(2) ± \sqrt{(2)^2 \ - \ 4(-1)(7)}}{2(-1)} \\ \\ x_1 & = 1 - 2\sqrt{2} \\ \\ & = -1.8 \\ \\ x_2 & = 1 + 2\sqrt{2} \\ \\ & = 3.8 \\ \\ \end{align}$ x-intercepts: (-1.8, 0), (3.8, 0)

Solve for y-intercept with ƒ(0), if any... $\begin{align} f(0) & = \dfrac{(0) + 1}{(0) - 3} - \dfrac{2(0) + 3}{(0) - 2} \\ \\ & = \dfrac{7}{6} \\ \\ & = 1.2 \\ \\ \end{align}$ y-intercept: (0, 1.2)

Rational Expressions: Restrictions

Simplify and state the restrictions. (Enter your answers in increasing order, and where applicable leave as fully reduced fraction, instead of repeating decimals)

$\dfrac{6x\ -\ 2}{18x^2\ -\ 12x\ +\ 2}$ Solution

Hint Clear Info

x ≠ ━━

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = \dfrac{2(3x\ -\ 1)}{2(9x^2 - 6x + 1)} \\ \\ & = \dfrac{2(3x\ -\ 1)}{2(3x\ -\ 1)(3x\ -\ 1)} \\ \\ & = \dfrac{2(3x\ -\ 1)}{2(3x\ -\ 1)^2} \\ \\ \end{align}$ $x \ne \dfrac{1}{3}$

$\dfrac{6a^2\ -\ 11ab\ -\ 7b^2}{3a^2\ -\ 16ab\ +\ 21b^2}$ Solution

Hint Clear Info

a ≠ ━━  ,
b ≠ ━━  ,   ━━

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = \dfrac{(2a + b)(3a - 7b)}{(3a - 7b)(a - 3b)} \\ \\ \end{align}$ $a \ne \dfrac{7b}{3},\ 3b \\ \\ b \ne \dfrac{1a}{3}, \ \dfrac{3a}{7}$

$\dfrac{11\ +\ 5x}{12x^4\ -\ 75x^2} \div \dfrac{3\ +\ 3x}{(7x\ -\ 1)}$ Solution Video

Hint Clear Info

x ≠ , , , ━━ ,

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = \dfrac{11\ +\ 5x}{3x^2(4x^2\ -\ 25)} \times \dfrac{(7x\ -\ 1)}{3\ +\ 3x} \\ \\ & = \dfrac{11\ +\ 5x}{3x^2(2x\ -\ 5)(2x\ +\ 5)} \times \dfrac{(7x\ -\ 1)}{3\ +\ 3x} \\ \\ & = \dfrac{(11\ +\ 5x)(7x\ -\ 1)}{3x^2(2x\ -\ 5)(2x\ +\ 5)(3\ +\ 3x)} \\ \\ \end{align}$

$x \ne -\dfrac{5}{2},\ \dfrac{5}{2}, \ 0,\ -1,\ \dfrac{1}{7}$

Enter in increasing order:

$x \ne -2.5,\ -1, \ 0,\ \dfrac{1}{7},\ 2.5$

(Don't forget to state a restriction on anything that was ever a denominator, so even though the 7x-1 gets flipped to the top, it still has to be included in the restriction list!)

$\dfrac{x^2\ -\ 5x\ -\ 6}{36\ -\ x^2}$ Solution

Hint Clear Info

x ≠ ,

Incorrect Attempts:

CHECK

Hint Unavailable

Horizontal and Vertical Asymptotes: Degree of Numerator < Degree of Denominator

Determine the horizontal and vertical asymptotes of the following functions. (Order from low to high, where applicable)

$f(x) = \ \dfrac{1}{(x\ +\ 3)(x\ -\ 2)}$ Solution

Hint Clear Info

x₁ = x₂ =
y =

Incorrect Attempts:

CHECK

Hint Unavailable

Vertical asymptote(s) @ x = -3, 2

Horizontal asymptote(s) @ y = 0

$f(x) = \ \dfrac{8}{x^2-4}$ Solution

Hint Clear Info

x₁ = x₂ =
y =

Incorrect Attempts:

CHECK

Hint Unavailable

Vertical asymptote(s) @ x = -2, 2

Horizontal asymptote(s) @ y = 0

$f(x) = \ \dfrac{3x}{2x^2\ -\ 11x\ -\ 6}$ Solution

Hint Clear Info

x₁ = x₂ =
y =

Incorrect Attempts:

CHECK

Hint Unavailable

Vertical asymptote(s) @ x = -0.5, 6

Horizontal asymptote(s) @ y = 0

Horizontal and Vertical Asymptotes: Degree of Numerator = Degree of Denominator

Determine the horizontal and vertical asymptotes of the following functions.

$f(x) = \ \dfrac{12x\ +\ 4}{3x\ +\ 2}$ Solution

Hint Clear Info

Vertical asymptote(s) @ x = ━━
Horizontal asymptote(s) @ y =

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = \dfrac{12x\ +\ 4}{3x\ +\ 2} \\ \\ & = \dfrac{4(3x\ +\ 1)}{3x\ +\ 2} \\ \\ \end{align}$ Vertical asymptote(s) @ x = -2/3

Horizontal asymptote(s) @ y = 12/3 = 4

$f(x) = \ \dfrac{25\ -\ 5x^2}{2x^2\ +\ 2x\ -\ 4}$ Solution

Hint Clear Info

Vertical asymptote(s): x = ,
Horizontal asymptote(s): y =

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = \dfrac{25\ -\ 5x^2}{2x^2\ +\ 2x\ -\ 4} \\ \\ & = \dfrac{5(5\ -\ x^2)}{(x\ +\ 2)(2x\ -\ 2)} \\ \\ \end{align}$ Vertical asymptote(s) @ x = -2, 1

Horizontal asymptote(s) @ y = -5/2 = -2.5

$f(x) = \ \dfrac{5x\ + \ 3x^2\ -\ 7x^3}{(4x^2\ -\ 9)(x\ +\ 1)}$ Solution

Hint Clear Info

Vertical asymptote(s): x = , ,
Horizontal asymptote(s): y = ━━

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = \dfrac{5x\ + \ 3x^2\ -\ 7x^3}{(4x^2\ -\ 9)(x\ +\ 1)} \\ \\ & = \dfrac{x(5\ +\ 3x\ -\ 7x^2)}{(2x\ +\ 3)(2x\ -\ 3)(x\ +\ 1)} \\ \\ \end{align}$ Vertical asymptote(s) @ x = -3/2, 3/2, -1
(order as: x = -1.5, -1, 1.5)

Horizontal asymptote(s) @ y = -7/4

$f(x) = \ \dfrac{2x^2\ -\ 2x}{x^2\ -\ 1}$ Solution

Hint Clear Info

Vertical asymptote(s): x =
Horizontal asymptote(s): y =

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = \dfrac{2x^2\ -\ 2x}{x^2\ -\ 1} \\ \\ & = \dfrac{2x(x\ -\ 1)}{(x\ +\ 1)(x\ -\ 1)} \\ \\ & = \dfrac{2x\cancel{(x\ -\ 1)}}{(x\ +\ 1)\cancel{(x\ -\ 1)}} \\ \\ \end{align}$ Vertical asymptote(s) @ x = -1 only.

! Careful. There is no vertical asymptote at x = +1 because it cancels out on the top and the bottom. This is a hole!

Horizontal asymptote(s) @ y = 2

$f(x) = \ \dfrac{3(2x\ -\ 2)^2}{(7\ -\ 2x)(3x\ +\ 1)}$ Solution

Hint Clear Info

Vertical asymptote(s) @ x = ━━ ,   ━━
Horizontal asymptote(s) @ y =

Incorrect Attempts:

CHECK

Hint Unavailable

Horizontal and Vertical Asymptotes: Degree of Numerator > Degree of Denominator... (numerator = denominator + 1)

Determine the oblique (slant) and vertical asymptotes of the following functions.

$f(x) = \ \dfrac{x^2 \ + \ 4x \ -\ 6}{x \ + \ 2}$ Solution

Hint Clear Info

Vertical asymptote(s): x =
Oblique asymptote(s): y =

Incorrect Attempts:

CHECK

Hint Unavailable

$f(x) = \ \dfrac{-4x^2\ +\ 2x\ +\ 2}{2x\ +\ 3}$ Solution

Hint Clear Info

Vertical asymptote(s) @ x = ━━
Oblique asymptote(s) @ y =

Incorrect Attempts:

CHECK

Hint Unavailable

$f(x) = \ \dfrac{-3x^2\ +\ 4x\ +\ 2x^3}{2x\ +\ 4x^2}$ Solution

Hint Clear Info

Vertical asymptote(s) @ x = ━━ ,
Oblique asymptote(s) @ y = ━━ x -

Incorrect Attempts:

CHECK

Hint Unavailable

$f(x) = \ \dfrac{x^3\ +\ 2x^2}{x^2\ -\ 9}$ Solution

Hint Clear Info

Vertical asymptote(s) @ x = ,
Oblique asymptote(s) @ y =

Incorrect Attempts:

CHECK

Hint Unavailable

Vertical asymptote: x = -3, 3

Horizontal asymptote: none

Oblique (slant) asymptote: y = x + 2

Positive and Negative Intervals

Given the graph of the function

Determine the positive and negative intervals for the diagram of the function above. Solution

1. Positive: -2 ≤ x ≤ 0 and Negative: 0 ≤ x ≤ 2
2. Positive: -1 ≤ x ≤ 1 and Negative: 1 ≤ x ≤ -1
3. Positive: 1 ≤ x ≤ -1 and Negative: -1 ≤ x ≤ 1
4. Positive: x ≤ -2, 0 ≤ x ≤ 2 and Negative: x ≥ 2, -2 ≤ x ≤ 0

Positive and negative intervals occur are based on where the function exists above or below the x-axis for positive or negative values of y.

Determine the intervals of increase and decrease for the diagram of the function above. Solution

1. Increase: -2 ≤ x ≤ 0 and Decrease: 0 ≤ x ≤ 2
2. Increase: -1 < x < 1 and Decrease: -1 > x > 1
3. Increase: 1 ≤ x ≤ -1 and Decrease: -1 ≤ x ≤ 1
4. Increase: x < -2, 0 < x < 2 and Decrease: x > 2, -2 < x < 0

Think of the slope of the tangent at any one point. The intervals of increase have a tangent with a positive slope, and the intervals of decrease have a tangent with a negative slope.

Graphing the Reciprocals of Linear Functions

For the reciprocal of the function $f(x) = 4x - 5$, determine:

The vertical and horizontal asymptotes, and the domain and range. Solution Video

Hint Clear Info

Domain:{    XER | x ≠ ━━}
Range: { YER | y ≠ }

Incorrect Attempts:

CHECK

Hint Unavailable

The positive/negative intervals. Solution

Hint Clear Info

Positive:{    XER |         ━━}
Negative:{    YER |         ━━}

Incorrect Attempts:

CHECK

Hint Unavailable

To find positive/negative intervals substitute x-values on either sides of the vertical asymptotes into ƒ(x), and check if function is positive or negative...

Since the vertical asymptote is x = 5/4, check the intervals: x<5/4, and x>5/4...

For x < 5/4... $\begin{align} f(x) & = \dfrac{1}{4x - 5} \\ \\ f(1) & = \dfrac{1}{4(1) - 5} \\ \\ & = -1 \\ \\ \end{align}$ The function is negative: {XER | x < 5/4}

For x > 5/4... $\begin{align} f(x) & = \dfrac{1}{4x - 5} \\ \\ f(2) & = \dfrac{1}{4(2) - 5} \\ \\ & = +\dfrac{1}{3} \\ \\ \end{align}$ The function is positive: {XER | x > 5/4}

The increasing/decreasing intervals. Solution

Hint Clear Info

Decreasing:{    XER | x ≠ ━━}

Incorrect Attempts:

CHECK

Hint Unavailable

Make a sketch to visualize the increasing/decreasing intervals, using...

• HA: y = 0
• VA: x = $\frac{5}{4}$
• As x → -∞, y → 0
• As x → +∞, y → 0

From this you can see...

1. Increasing: none
2. Decreasing: {XER / x ≠ $\frac{5}{4}$}, or (−∞, $\frac{5}{4}$) ∪ ($\frac{5}{4}$, ∞)

Graphing the Reciprocals of Quadratic Functions

For the reciprocal of the function $f(x) = 2x^2 + 11x + 12$, determine:

The vertical and horizontal asymptotes, and the domain and range. Solution

Hint Clear Info

Domain:{    XER | x ≠ , ━━}
Range: { YER | y ≠ }

Incorrect Attempts:

CHECK

Hint Unavailable

Determine where a point of inflection (turning point) exists. Solution

1. $\dfrac{-12}{5}$
2. $\dfrac{-13}{8}$
3. $\dfrac{-8}{3}$
4. $\dfrac{-11}{4}$

The turning point exists between the vertical asymptotes, -4 and -3/2... $\begin{align} x & = \dfrac{x_1 + x_2}{2} \\ \\ x & = \dfrac{-\frac{3}{2} + (-4)}{2} \\ \\ x & = \dfrac{-11}{4} \\ \\ \end{align}$

The positive/negative intervals. Solution

Hint Clear Info

Positive:{    XER | ━━}
Negative:{    XER | ━━}

Incorrect Attempts:

CHECK

Hint Unavailable

Make an interval chart using the asymptotes: -4 and -3/2

	(-∞, -4)	(-4, -3/2)	(-3/2, ∞)
(2x + 3)	—	—	+
(x + 4)	—	+	+
Result:	+	—	+

Positive: { -4 > x > $\frac{-3}{2}$ },     or (-∞, -4) ∪ ($\frac{-3}{2}$, ∞)
Negative: { -4 < x < $\frac{-3}{2}$ },     or (-4, $\frac{-3}{2}$)

The increasing/decreasing intervals, using interval notation: ( ) [ ]. Solution

Hint Clear Info

Increasing: (    -∞ , ) ∪ ( , ━━ )
Decreasing: (     ━━ , ━━ ) ∪ ( ━━ , ∞ )

Incorrect Attempts:

CHECK

Hint Unavailable

Make a sketch to visualize the increasing/decreasing intervals, using...

• HA: y = 0
• VA: x = -4, $-\frac{3}{2}$
• As x → -∞, y → 0
• As x → +∞, y → 0

Determine the axis of symmetry of the local maximum, either from the vertex of the original function or from the midpoint of the vertical asymptotes. $\begin{align} & = \dfrac{-4 + \tfrac{-3}{2}}{2} \\ \\ & = \dfrac{\tfrac{-8}{2} - \tfrac{3}{2}}{2} \\ \\ & = \dfrac{-11}{4} \\ \\ \end{align}$ After a sketch, you can see...

1. Increasing: {XER / x < -4, -4 < x < $\tfrac{-11}{4}$},     or (-∞, -4) ∪ (-4, $\tfrac{-11}{4}$)
2. Decreasing: {XER / $\tfrac{-11}{4}$ < x < $\tfrac{-3}{2}$, x > $\tfrac{-3}{2}$},     or ($\tfrac{-11}{4}$, $\tfrac{-3}{2}$) ∪ ($\tfrac{-3}{2}$, ∞)

Graphing the Reciprocals of Quadratic Functions

For the reciprocal of the function $f(x) = 3x^2 + 10x + 3$, determine:

The vertical and horizontal asymptotes, and the domain and range. Solution

The x-intercepts, positive/negative intervals, and increasing/decreasing intervals. Solution

Determine the end behaviour of the function. Solution

1. x → -∞, y → 0, x → +∞, y → 0
2. x → -∞, y → +∞, x → +∞, y → -∞
3. x → -∞, y → 0, x → +∞, y → +∞
4. x → -∞, y → -∞, x → +∞, y → -∞
5. x → -∞, y → -∞, x → +∞, y → 0

From the previous sketch, see that: x → -∞, y → 0, x → +∞, y → 0

Reciprocals of Quadratic Functions: The Relationship of Roots and Vertical Asymptotes

For the reciprocal of the following functions, determine if any vertical asymptotes exist.

$f(x) = x^2 - 8x - 9$ Solution

$f(x) = (x-8)^2 + 4$ Solution

Reciprocal Function Equations

Determine an equation of a reciprocal function given the following information.

Real root at (-1, 0) and a hole at x = 1 Solution

$f(x) = \ \dfrac{x^2\ -\ 1}{x\ -\ 1}$

Hole at x = 3, horizontal asymptote at y = 0, and vertical asymptote at x = -5. Solution

$f(x) = \ \dfrac{x\ -\ 3}{(x\ +\ 5)(x\ -\ 3)}$

A vertical asymptote exists at x = 10, and a x-intercept of x = 5, given the following function: Solution $f(x) = \ \dfrac{(2x\ +\ a)^2}{(3x\ -\ b)^2}$

Linear & Quadratic Inequalities Practice

Solve

$5x \le 8x - 6$ Solution

$\begin{align} 5x & \le 8x - 6 \\ \\ -3x & \le -6 \\ \\ 3x & \ge 6 \\ \\ x & \ge 2 \\ \\ \end{align}$

$-4x + 5 < 2x < 5$ Solution

$\frac{5}{6} < x < \frac{5}{2}$

$-3 \le -3x \le 6$ Solution

$-2 \le x \le 1$

or

$1 \ge x \ge -2$

$3x \le x - 2 \le 2x + 1$ Solution

$-1 \ge x \ge -3$

$-9x < 6 + x < 10$ Solution

$-\frac{3}{5} < x< 4$

$0 > 4x^2 + 36x$ Solution

$\begin{align} 0 & > 4x^2 + 36x \\ \\ 0 & > 4x(x + 9) \\ \\ \end{align}$ $-9 > x > 0$

Rational Inequalities

Solve algebraically.

$\dfrac{10x^2\ +\ 2}{2x^2\ +\ 1} \ < \ 3$ Solution

Move the terms to the same side, simplify, and factor, $\begin{align} \dfrac{10x^2\ +\ 2}{2x^2\ +\ 1} & < 3 \\ \\ \dfrac{10x^2\ +\ 2}{2x^2\ +\ 1} - 3 & < 0 \\ \\ \dfrac{10x^2\ +\ 2}{2x^2\ +\ 1} - 3\dfrac{(2x^2\ +\ 1)}{(2x^2\ +\ 1)} & < 0 \\ \\ \dfrac{10x^2\ +\ 2 - 6x^2 - 3}{2x^2\ +\ 1} & < 0 \\ \\ \dfrac{4x^2\ -\ 1}{2x^2\ +\ 1} & < 0 \\ \\ \dfrac{(2x + 1)(2x - 1)}{2x^2\ +\ 1} & {\color{Red}{<}} 0 \\ \\ \end{align}$ To find where the function ƒ(x) is negative (< 0)... Don't worry about the binomial that results in the imaginary number (2x² + 1). Set up an interval chart with ranges between any of the roots and vertical asymptotes (if any),

	(-∞, $-\tfrac{1}{2}$)	($-\tfrac{1}{2}$, $\tfrac{1}{2}$)	($\tfrac{1}{2}$, ∞)
(2x + 1)	—	+	+
(2x - 1)	—	—	+
result:	(—)(—) = +	(+)(—) = —	(+)(+) = +

The rational function is negative in the regions, $-\dfrac{1}{2} < x < \dfrac{1}{2}$

$\dfrac{5}{9 - x^2} \ < \ -\dfrac{5}{x - 3}$ Solution

Move the terms to the same side, simplify, and factor, $\begin{align} \dfrac{5}{9 - x^2} & < -\dfrac{5}{x - 3} \\ \\ \dfrac{5}{(3 + x)(3 - x)} & < -\dfrac{5}{x - 3} \\ \\ \dfrac{5}{(3 + x)(3 - x)} + \dfrac{5}{x - 3} & < 0 \\ \\ \dfrac{5}{(3 + x)(3 - x)} + \dfrac{5}{x - 3} \times \dfrac{(-1)}{(-1)} & < 0 \\ \\ \dfrac{5}{(3 + x)(3 - x)} + \dfrac{-5}{(-x + 3)} & < 0 \\ \\ \dfrac{5}{(3 + x)(3 - x)} - \dfrac{5}{(3 - x)} & < 0 \\ \\ \dfrac{5}{(3 + x)(3 - x)} - \dfrac{\color{Green}{5}}{(3 - x)} \times \dfrac{\color{Green}{(3 + x)}}{(3 + x)} & < 0 \\ \\ \dfrac{5 - \color{Green}{5(3 + x)}}{(3 + x)(3 - x)} & < 0 \\ \\ \dfrac{5 - 15 - 5x}{(3 + x)(3 - x)} & < 0 \\ \\ \dfrac{-5(2 + x)}{(3 + x)(3 - x)} & {\color{Red}{<}} \ 0 \\ \\ \end{align}$ To find where the function ƒ(x) is negative (< 0)... Set up an interval chart with the ranges between any of the roots (x = -2) and vertical asymptotes (x = -3, +3)

	(-∞, -3)	(-3, -2)	(-2, 3)	(3, ∞)
(2 + x)	—	—	+	+
(3 + x)	—	+	+	+
(3 - x)	+	+	+	—
result:	+	—	+	—

The rational function is negative in the regions, $-3 < x < -2, \quad x > 3$

Converting Degrees and Radians

Convert 4.30 radians into degrees. Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

degrees

Hint Unavailable

Radian Definition

Determine the arc length of a circle created by a 4.5 radian angle if the radius of the circle is 10cm. Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

cm

Hint Unavailable

Exact Values of Trig Ratios: Something to Watch

The exact value of cos$\left( \dfrac{3\pi}{4} \right)$ is $\dfrac{-1}{\sqrt{2}}$ Solution Video

1. True
2. False

Although this seems correct, the answer cannot have a square root in the denominator!
To derationalize the denominator, multiply the top and bottom by the square root:
$\begin{align} cos \dfrac{3\pi} {4} & = \dfrac{-1} {\sqrt{2}} \\ \\ & = \dfrac{-1} {\sqrt{2}} × \dfrac{\sqrt{2}} {\sqrt{2}} \\ \\ & = \dfrac{-1\sqrt{2}} {(\sqrt{2})(\sqrt{2})} \\ \\ & = \dfrac{-\sqrt{2}} {2} \\ \\ \end{align}$

Special Acute Reflex Angles in the CAST System

Given that $\sin \theta = \dfrac{-1}{2}$, determine $\ \cos \theta$, $\tan \theta$, $\ \sec \theta$, $\ \csc \theta$, and $\ \cot \theta$ for $\theta \le 270˚$ Solution Video

Hint Clear Info

$\sqrt{\quad}$ $\sqrt{\quad}$ $\sqrt{\quad}$ $\cos\theta = \dfrac{}{\quad\quad\quad\quad}, \tan\theta = \dfrac{}{\quad\quad\quad\quad}, \sec\theta = \dfrac{}{\quad\quad\quad\quad\quad}, \csc\theta = \quad\quad, \cot\theta =$

Incorrect Attempts:

CHECK

Hint Unavailable

Characteristics of Graphs of Trig Parent Functions

The domain range, and period of the three main trig functions are given below. Fill in the blanks. Solution

	sin(θ)	cos(θ)	tan(θ)
Domain	x∈ℝ	x∈ℝ	x ≠$\frac{\pi}{2}$ + n·π, n∈I, n = {...-1, 0, 1, ...}
Range	y∈ℝ, -1 ≤ y ≤ 1	__________	y∈ℝ
Period	2π	2π	π

Hint Clear Info

y∈ℝ,       ≤ y ≤

Incorrect Attempts:

CHECK

Hint Unavailable

Make a quick sketch of cos(θ)... see that the maximum is +1 and the minimum is -1, so the range is: y∈ℝ, -1 ≤ y ≤ 1

State the x-intercepts in periodic notation in terms of pi, for the function f(x) = sin(θ). Solution

Remember, or make a quick sketch, and see that some of the x-intercepts are: 0, π, 2π, so the x-intercepts are.. π·n, n∈I (n∈I means 'n' is all real integers)

Transformations of Trig Functions

Which of the following set of transformations would transform the graph of ƒ(x) = cos(x) into the function below. Solution g(x) = 2cos(3x - $\pi$) + 1

1. Vertical stretch by a factor of 2, horizontally compressed by 3, translated $\pi$ right, and translated 1 unit up.
2. Vertical stretch by a factor of 2, horizontally compressed $\dfrac{1}{3}$, translated $\dfrac{\pi}{3}$ right, and translated 1 unit up.
3. Vertical stretch by a factor of 2, horizontally compressed $\dfrac{1}{3}$, translated $\pi$ right, and translated 1 unit up.
4. Vertical stretch by a factor of 2, horizontally compressed by 3, translated $\dfrac{\pi}{3}$ right, and translated 1 unit up.

Vertical stretch by a factor of 2, horizontally compressed $\dfrac{1}{3}$, translated $\dfrac{\pi}{3}$ right, and translated 1 unit up.

Period of Trig Functions

Given the period of a trig function equals 20, determine the exact value of k, in radians. Solution y = a · sin [k(x - c)] + d

ƒ(x) =

Hint Clear Info

$\dfrac{\pi}{\quad\quad}$

Incorrect Attempts:

CHECK

radians

Hint Unavailable

For the period, (T = 20) $\begin{align} \text{Period} & = \dfrac{2\pi}{k} \\ \\ k & = \dfrac{2\pi}{\text{T}} \\ \\ k & = \dfrac{2\pi}{20} \\ \\ k & = \dfrac{\pi}{10} \\ \\ \end{align}$

Solving for the Independent Variable in the argument of a Trig Function

Given the function below...

Solve for x algebraically, in radians, rounded to the nearest hundredth decimal place. Solution

x =

Hint Clear Info

Incorrect Attempts:

CHECK

rad

Hint Unavailable

$\begin{align} 5 & = 13\sin \left( \dfrac{2\pi}{5} x + 1 \right) - 1.5 \\ \\ 13\sin \left( \dfrac{2\pi}{5} x + 1 \right) & = 5 + 1.5 \\ \\ \sin \left( \dfrac{2\pi}{5} x + 1 \right) & = \dfrac{6.5}{13} \\ \\ \left( \dfrac{2\pi}{5} x + 1 \right) & = \sin^{-1} \left( \dfrac{6.5}{13} \right) \\ \\ \dfrac{2\pi}{5} x & = \sin^{-1} \left( \dfrac{1}{2} \right) - 1 \\ \\ x & = \left( \sin^{-1} \left( \dfrac{1}{2} \right) - 1 \right) \dfrac{5}{2\pi} \\ \\ x & = \left( \dfrac{\pi}{6} - 1 \right) \dfrac{5}{2\pi} \\ \\ x & = -0.38 \\ \\ \end{align}$

Determine the next value of 'x', rounded to the nearest hundredth decimal place, where the function equals 5, over the interval: 0 ≤ x ≤ 0.5π Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Using the cast system, sin is positive (1/2), at π/6 and 5π/6... So $\begin{align} 5 & = 13\sin \left( \dfrac{2\pi}{5} x + 1 \right) - 1.5 \\ \\ 13\sin \left( \dfrac{2\pi}{5} x + 1 \right) & = 5 + 1.5 \\ \\ \sin \left( \dfrac{2\pi}{5} x + 1 \right) & = \dfrac{6.5}{13} \\ \\ \left( \dfrac{2\pi}{5} x + 1 \right) & = \sin^{-1} \left( \dfrac{6.5}{13} \right) \\ \\ \dfrac{2\pi}{5} x & = \sin^{-1} \left( \dfrac{1}{2} \right) - 1 \\ \\ x & = \left( \sin^{-1} \left( \dfrac{1}{2} \right) - 1 \right) \dfrac{5}{2\pi} \\ \\ x & = \left( \dfrac{\pi}{6} - 1 \right) \dfrac{5}{2\pi} \\ \\ x & = -0.38 \\ \\ \end{align}$ And, $\begin{align} 5 & = 13\sin \left( \dfrac{2\pi}{5} x + 1 \right) - 1.5 \\ \\ 13\sin \left( \dfrac{2\pi}{5} x + 1 \right) & = 5 + 1.5 \\ \\ \sin \left( \dfrac{2\pi}{5} x + 1 \right) & = \dfrac{6.5}{13} \\ \\ \left( \dfrac{2\pi}{5} x + 1 \right) & = \sin^{-1} \left( \dfrac{6.5}{13} \right) \\ \\ \dfrac{2\pi}{5} x & = \sin^{-1} \left( \dfrac{1}{2} \right) - 1 \\ \\ x & = \left( \sin^{-1} \left( \dfrac{1}{2} \right) - 1 \right) \dfrac{5}{2\pi} \\ \\ x & = \left( \dfrac{5\pi}{6} - 1 \right) \dfrac{5}{2\pi} \\ \\ x & = 1.29 \\ \\ \end{align}$

Features of Trig Functions

Determine the features of the function below. Reduce your answers fully.

The amplitude Solution

a =

Hint Clear Info

Incorrect Attempts:

CHECK

ƒ(x) = a·sin[k(x - d)] + c

ƒ(x) = a·sin[k(x - d)] + c

a = 75

The vertical translation Solution

Hint Clear Info

Incorrect Attempts:

CHECK

units up

Hint Unavailable

ƒ(x) = a·sin[k(x - d)] + c

10 units up

The equation of the axis of symmetry. Solution

y =

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

ƒ(x) = a·sin[k(x - d)] + c

y = 10

The period Solution

period =

Hint Clear Info

–––––

Incorrect Attempts:

CHECK

Hint Unavailable

ƒ(x) = a·sin[k(x - d)] + c

k = $\dfrac{5\pi}{12}$ $\begin{align} Period & = \dfrac{2\pi}{k} \\ \\ & = \dfrac{2\pi}{\left(\dfrac{5\pi}{12}\right)} \\ \\ & = 2\pi × \dfrac{12}{5\pi} \\ \\ & = 2\cancel{\pi} × \dfrac{12}{5\cancel{\pi}} \\ \\ & = \dfrac{2 × 12}{5} \\ \\ & = \dfrac{24}{5} \\ \\ \end{align}$

The horizontal translation Solution

Hint Clear Info

$\dfrac{\quad\quad \pi}{\quad\quad\quad\quad}$

Incorrect Attempts:

CHECK

units right

Hint Unavailable

$\begin{align} P(t) & = 75sin\left(\frac{5\pi}{12}t - 6\right) + 10 \\ \\ & = 75sin\left(\frac{5\pi}{12}\left(t - n\right)\right) + 10 \\ \\ \\ \\ \\ \\ \frac{5\pi}{12}n & = 6 \\ \\ n & = \frac{6(12)}{5\pi} \\ \\ & = \frac{72}{5\pi} \\ \\ & = 75sin\left(\frac{5\pi}{12}\left(t - \frac{72}{5\pi}\right)\right) + 10 \\ \\ \end{align}$ $\cfrac{72\pi}{5}$ units right.

The y-intercept. Solution

Hint Clear Info

(   ,   )

Incorrect Attempts:

CHECK

Hint Unavailable

The y-intercept when t = 0 $\begin{align} P(t) & = 75sin\left(\frac{5\pi}{12}t - 6\right) + 10 \\ \\ y & = 75sin\left(\frac{5\pi}{12}(0) - 6\right) + 10 \\ \\ y & = 75sin\left(\cancel{\frac{5\pi}{12}(0)} - 6\right) + 10 \\ \\ y & = 75sin\left(- 6\right) + 10 \\ \\ y & = 31 \\ \\ \end{align}$ The point is (0, 31)

The maximum y-value. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

The maximum value occurs at a + c

ƒ(x) = a•sin[k(x - d)] + c

$\begin{align} P(t) & = 75sin\left(\frac{5\pi}{12}t - 6\right) + 10 \\ \\ a + c & = 75 + 10 \\ \\ & = 85 \\ \\ \end{align}$ y = 85

Word Problems with Trig Functions: Ferris Wheel

A large ferris wheel at a circus has a diameter of 50m with a platform to mount it at the lowest point 2m above ground level. One full ride around the wheel takes 4 minutes. Find the following

The amplitude Solution

a =

Hint Clear Info

Incorrect Attempts:

CHECK

meters

Hint Unavailable

A point on the wheel oscillates between a maximum height of 52m, and a minimum height of 2m. $\begin{align} Amplitude & = \frac{maximum\ -\ minimum}{2} \\ \\ & = \frac{52 - 2}{2} \\ \\ & = \frac{50}{2} \\ \\ & = 25\ m \\ \\ \end{align}$

The vertical translation Solution

Hint Clear Info

Incorrect Attempts:

CHECK

meters

Hint Unavailable

= axis of symmetry. $\begin{align} & = minimum + amplitude \\ \\ & = 2\ m + 25\ m \\ \\ & = 27\ m \\ \\ \end{align}$

The period Solution

period =

Hint Clear Info

Incorrect Attempts:

CHECK

minutes

Hint Unavailable

"One full ride around the wheel takes 4 minutes." The period is the time to complete one cycle. Therefore the period is 4 minutes.

The 'k' value Solution

k =

Hint Clear Info

$\dfrac{\pi}{\quad\quad\quad}$

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} T & = \frac{2\pi}{k} \\ \\ k & = \frac{2\pi}{T} \\ \\ & = \frac{2\pi}{4} \\ \\ & = \frac{\pi}{2} \\ \\ \end{align}$

The equation of the trig function, H(t) with height starting at the bottom of the ferris wheel at t = 0. Solution

H(t) =

Hint Clear Info

$\cos\dfrac{\pi}{\quad\quad} (x) +$

Incorrect Attempts:

CHECK

Hint Unavailable

If the ferris wheel starts at the bottom, then height = 2m when t = 0. $\begin{align} H(t) & = a \cos[k(x - d)] + c \\ \\ H(t) & = -25 \cos \frac{\pi}{2} (x) + 27 \\ \\ \end{align}$

Word Problems with Trig Functions: Clock

The hour hand on a clock rotates at 1 revolution per 12 hours. The clock has a diameter of 4cm and the point of interest to model is located at the tip of the hour hand, at the perimeter of the clock. Model the following using a sine function with start time at 12:00 noon.

The amplitude Solution

a =

Hint Clear Info

Incorrect Attempts:

CHECK

cm

Hint Unavailable

Amplitude = diameter / 2

= 2 cm

The period Solution

T =

Hint Clear Info

Incorrect Attempts:

CHECK

hours

Hint Unavailable

The period (T) is the total time for one revolution, which is given as 12 hours.

T = 12 hours

The vertical translation Solution

c =

Hint Clear Info

Incorrect Attempts:

CHECK

ƒ(x) = a·sin[k(x - d)] + c

If the axis of symmetry is located in the center of the clock, then the vertical translation could be assumed to be 0.

The horizontal translation on a sin graph (assuming the hour hand starts at 12:00) Solution

Hint Clear Info

Incorrect Attempts:

CHECK

hours

Hint Unavailable

Sketch the function and see that the that it is transformed from a parent function of sin by 3 hours to the left.

The horizontal translation would be -3, hours.

The equation of the (sin) trig function. Solution

y =

Hint Clear Info

$\sin\dfrac{\pi}{\quad\quad} (x) -$

Incorrect Attempts:

CHECK

ƒ(x) = a·sin[k(x - d)] + c

You need to determine the k-value for the equation: $\begin{align} k & = \dfrac{2\pi}{T} \\ \\ & = \dfrac{2\pi}{12\ hours} \\ \\ & = \dfrac{\pi}{6} \\ \\ \end{align}$ $y = 2\sin \left(\dfrac{\pi}{6} \right)(t) - 3$

Word Problems with Trig Functions: Tides

Narragansett Bay has a maximum tide height 5m higher than the lowest point of 1m. A low tide point occurs at 2:00PM. If the time between adjacent high and low tides is 6 hours, 15 minutes, determine the following...

The amplitude Solution

a =

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

amplitude

= (max - min)/2

= (6 - 1)/2

= (5)/2

= 2.5

The k-value Solution

k =

Hint Clear Info

$\dfrac{\quad\quad \pi}{\quad\quad\quad\quad}$

Incorrect Attempts:

CHECK

Hint Unavailable

"If the time between adjacent high and low tides is 6 hours, 15 minutes"... this is half a cycle. Therefore the period is double this... 12 hours and 30 minutes... 12.5 hours...

Period = 2π/k
12.5 = 2π/k
k = 2π/12.5
k = 4π/25

The vertical translation Solution

c =

Hint Clear Info

Incorrect Attempts:

CHECK

ƒ(x) = a·cos[k(x - d)] + c

c = (max + min)/2

= (6 + 1)/2

= (7)/2

= 3.5

The horizontal translation, d (using a positive cos function, starting at 12:00 noon) Solution

d =

Hint Clear Info

Incorrect Attempts:

CHECK

ƒ(x) = a·cos[k(x - d)] + c

If we are to use a cos function instead of a sine function, then we will use the maximum point of the tide as our starting point. We will set time as zero at 12:00 noon. So, with a low at 2:00PM there is a high tide 6 hours, 15 minutes later at 8:15PM. This is 8 hours, 15 minutes or 8.25 hours after noon. So, the horizontal translation will be 8.25 right

d = -8.25

The equation of the trig function Solution

H(t) =

Hint Clear Info

$\cos \Bigg[ \dfrac{\quad\quad\quad\pi}{\quad\quad\quad} \Bigg(t - \ \quad\quad \Bigg) \Bigg] +$

Incorrect Attempts:

CHECK

H(t) = a·cos[k(t - d)] + c

H(t) = a·cos[k(t - d)] + c

= $2.5 \cos\left[\dfrac{4\pi}{25}(t - 8.25)\right] + 3.5$

Word Problems with Trig Functions: Seasonal Temperatures

A function modelling the average temperature T, in degrees Celsius, in a certain ountry over the course of one year is given below where t is the time in days. Determine the following

The maximum temperature, and the month this occurs in. [3] Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Maximum Temperature
= 16 + 13
= 29˚ Celsius

Solve to find this occurs in April. $\begin{align} T(t) & = 16\sin\left(\frac{2\pi}{365}(t - 99)\right) + 13 \\ \\ 29 & = 16\sin\left(\frac{2\pi}{365}(t - 99)\right) + 13 \\ \\ \cancel{16} & = \cancel{16}\sin\left(\frac{2\pi}{365}(t - 99)\right) \\ \\ 1 & = \sin\left(\frac{2\pi}{365}(t - 99)\right) \\ \\ \frac{2\pi}{365}(t - 99) & = \sin ^{-1} (1) \\ \\ \frac{2\cancel{\pi}}{365}(t - 99) & = \frac{\cancel{\pi}}{2} \\ \\ t & = \frac{365}{2(2)} + 99 \\ \\ t & = 190.25 \\ \\ \end{align}$ The 190th day is ~ 6.3 months ~ June...

What is the average temperature for the entire year? Solution

Hint Clear Info

Incorrect Attempts:

CHECK

˚C

Think of the axis of symmetry

The average is the axis of symmetry, which is 13 (degrees Celsius).

Predict the temperature on May 21^st, the 140^th day of the year. Solution

T =

Hint Clear Info

Incorrect Attempts:

CHECK

˚C

Hint Unavailable

Substitute t = 140 into the equation... $\begin{align} T(t) & = 16\sin\left(\frac{2\pi}{365}(140 - 99)\right) + 13 \\ \\ & = 23.4 \\ \\ \end{align}$ This evaluates to 23.4 degrees Celsius.

Equivalent Trigonometric Expressions

Given the cofunction identity,

sin23˚ is equivalent to: Solution

1. cos(87˚)
2. cos(77˚)
3. cos(67˚)
4. cos(57˚)

sin(x˚)
= cos(90˚ - x˚)
= cos(90˚ - 23˚)
= cos(67˚)

$\sin \left(\dfrac{\pi}{3}\right)$ is equivalent to cos( __ ) ? Solution

Hint Clear Info

$\cos\bigg(\dfrac{\pi}{\quad\quad}\bigg)$

Incorrect Attempts:

CHECK

Hint Unavailable

Use the cofunction identity and make a common denominator: $\begin{align} \sin\theta & = \cos\left(\dfrac{\pi}{2} - \theta \right) \\ \\ \sin \left(\dfrac{\pi}{3}\right) & = \cos\left(\dfrac{\pi}{2} - \dfrac{\pi}{3} \right) \\ \\ & = \cos\left(\dfrac{3\pi}{6} - \dfrac{2\pi}{6} \right) \\ \\ & = \cos\left(\dfrac{3\pi - 2\pi}{6}\right) \\ \\ & = \cos\left(\dfrac{\pi}{6}\right) \\ \\ \end{align}$

The following is correct. Solution $\cos\left(\dfrac{\pi}{2} - \theta \right) = \cos\left(-\left(\theta - \dfrac{\pi}{2}\right)\right)$

1. True
2. False

$\begin{align} & \cos\left(\dfrac{\pi}{2} - \theta \right) \\ \\ & = \cos\left( - \theta + \dfrac{\pi}{2} \right) \\ \\ & = \cos\left( - \left(\theta - \dfrac{\pi}{2} \right)\right) \\ \\ \end{align}$

The following is correct. If... Solution $\sin\left(\dfrac{\pi}{2} - \theta \right) = \cos\left( \theta \right)$ Then, $\sin\left(\dfrac{\pi}{2} + \pi - \theta \right) = \cos\left( \pi + \theta \right)$

1. True
2. False

It is an equivalent horizontal shift, by a factor of pi, $\begin{align} \sin\left(\dfrac{\pi}{2} - \theta \right) & = \cos\left( \theta \right) \\ \\ \sin\left(\dfrac{\pi}{2} + \pi - \theta \right) & = \cos\left( \pi + \theta \right) \\ \\ \sin\left(\dfrac{3\pi}{2} - \theta \right) & = \cos\left( \pi + \theta \right) \end{align}$

Exact Value with Compound Angle Formula (In Degrees and Radians)

Evaluate each of the following with an exact value, using the compound angle formulas.

cos(450˚) Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Compound angle formula:

cos(x˚ + y˚) = (cos x˚)(cos y˚) - (sin x˚)(sin y˚) $\begin{align} & \cos(450˚) \\ \\ & = \cos(360˚ + 90˚) \\ \\ & = (\cos\,360˚)(\cos\,90˚) - (\sin\,360˚)(\sin\,90˚) \\ \\ & = (1)(0) - (0)(1) \\ \\ & = \cancel{(1)(0)} - \cancel{(0)(1)} \\ \\ & = 0 \\ \\ \end{align}$

sin(225˚) Solution

Hint Clear Info

$- \dfrac{\sqrt{ \quad}}{\quad}$

Incorrect Attempts:

CHECK

Hint Unavailable

Compound angle formula:

sin(x˚ + y˚) = (sin x˚)(cos y˚) + (cos x˚)(sin y˚) $\begin{align} & = \sin(225˚) \\ \\ & = \sin(180˚ + 45˚) \ \ or \ \ \sin(270˚ - 45˚) \\ \\ & = \sin\,180˚ \cos\,45˚ + \cos\,180˚ \sin\,45˚ \\ \\ & = (0)\left(\dfrac{1}{\sqrt{2}}\right) + (-1)\left(\dfrac{1}{\sqrt{2}}\right)\\ \\ & = \cancel{(0)\left(\dfrac{1}{\sqrt{2}}\right)} + (-1)\left(\dfrac{1}{\sqrt{2}}\right)\\ \\ & = -\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} \\ \\ & = \dfrac{-\sqrt{2}}{2} \\ \\ \end{align}$ Note this could also work or be done with sin(270˚ - 45˚)... you would get the same answer.

sin$\left(\dfrac{11\pi}{6}\right)$ Solution

1. $\dfrac{\sqrt{3}}{3}$
2. $\dfrac{\sqrt{3}}{2}$
3. $-\dfrac{1}{2}$
4. $-\dfrac{2\sqrt{3}}{3}$

Compound angle formula:

sin(x˚ - y˚) = (sin x˚)(cos y˚) - (cos x˚)(sin y˚) $\begin{align} & \sin \left(\dfrac{11\pi}{6}\right) \\ \\ & = \sin \left(\dfrac{12\pi}{6} - \dfrac{1\pi}{6} \right) \\ \\ & = \sin \left(2\pi - \dfrac{\pi}{6} \right) \\ \\ & = \sin \left(360˚ - 30˚ \right) \\ \\ & = (\sin 360˚)(\cos 30˚) - (\cos 360˚)(\sin 30˚) \\ \\ & = (0)\left( \dfrac{1}{2} \right) - (1)\left( \dfrac{1}{2} \right) \\ \\ & = \cancel{(0)\left( \dfrac{1}{2} \right)} - (1)\left( \dfrac{1}{2} \right) \\ \\ & = -\dfrac{1}{2} \\ \\ \end{align}$

Compound Angle Formula (aka Addition Subtraction Formula)

Simplify the trig ratio cos(3x)cos(4x) - sin(3x)sin(4x) Solution

1. cos(x)
2. cos(-x)
3. cos(7x)
4. cos(-7x)
5. cos(12x)

$\begin{align} & \cos(3x)\,\cos(4x) - \sin(3x)\,\sin(4x) \\ \\ & = \cos(3x + 4x) \\ \\ & = \cos(7x) \\ \\ \end{align}$

Compound Angle Formula (aka Addition Subtraction Formula)

Compound angles can be evaluated by first converting each measure to an exact value.

Evaluate. Solution $\cos\left(\dfrac{3\pi}{2}\right)\cos\left(\dfrac{\pi}{3}\right) \ - \ \sin\left(\dfrac{3\pi}{2}\right) \sin\left(\dfrac{\pi}{3}\right)$

Hint Clear Info

$\dfrac{\sqrt{ \quad}}{\quad}$

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} &\cos\left(\dfrac{3\pi}{2}\right)\cos\left(\dfrac{\pi}{3}\right) \ - \ \sin\left(\dfrac{3\pi}{2}\right) \sin\left(\dfrac{\pi}{3}\right) \\ \\ & = \cos(270)\cdot \cos(60) - \sin(270)\cdot \sin(60) \\ \\ & = \cancel{(0) \cdot \cos(60)} - \sin(270)\cdot \sin(60) \\ \\ & = 0 - \sin(270)\cdot \sin(60) \\ \\ & = - (-1)\cdot \dfrac{\sqrt{3}}{2} \\ \\ & = \dfrac{\sqrt{3}}{2} \\ \\ \end{align}$

$\cos \left(\dfrac{11\pi}{6}\right) \ = \ \cos\left(\dfrac{3\pi}{2}\right)\cos\left(\dfrac{\pi}{3}\right) \ - \ \sin\left(\dfrac{3\pi}{2}\right) \sin\left(\dfrac{\pi}{3}\right)$ Solution

1. True
2. False

Use compound angle formula. $\begin{align} & = \cos\left(\dfrac{3\pi}{2}\right)\cos\left(\dfrac{\pi}{3}\right) \ - \ \sin\left(\dfrac{3\pi}{2}\right) \sin\left(\dfrac{\pi}{3}\right) \\ \\ & = \cos \left(\dfrac{3\pi}{2} + \dfrac{\pi}{3}\right) \\ \\ & = \cos \left(\dfrac{9\pi}{6} + \dfrac{2\pi}{6}\right) \\ \\ & = \cos \left(\dfrac{11\pi}{6}\right) \\ \\ \end{align}$

Compound Angle Formula (aka Addition Subtraction Formula)

Determine an equivalent expression to the following. Solution $\sin\left(\dfrac{\pi}{4} \ - \ \pi\right)$

1. $\! \sin\left(\dfrac{\pi}{4}\right) \sin\left(\pi\right) - \cos\left(\dfrac{\pi}{4}\right)\cos\left(\pi\right)$
2. $\! \sin\left(\dfrac{\pi}{4}\right) \sin\left(\pi\right) + \cos\left(\dfrac{\pi}{4}\right)\cos\left(\pi\right)$
3. $\! \sin\left(\dfrac{\pi}{4}\right) \cos\left(\pi\right) - \cos\left(\dfrac{\pi}{4}\right)\sin\left(\pi\right)$
4. $\! \sin\left(\dfrac{\pi}{4}\right) \cos\left(\pi\right) + \cos\left(\dfrac{\pi}{4}\right)\sin\left(\pi\right)$

$\! \sin\left(\dfrac{\pi}{4}\right) \cos\left(\pi\right) - \cos\left(\dfrac{\pi}{4}\right)\sin\left(\pi\right)$

Compound Angle Formula (aka Addition Subtraction Formula)

Determine an equivalent expression to the following. Solution $cos\left(\dfrac{5\pi}{3}\right)$

1. $\! \cos\left(\dfrac{3\pi}{2} + \dfrac{\pi}{6}\right)$
2. $\! \cos\left(\dfrac{3\pi}{2} - \dfrac{\pi}{6}\right)$
3. $\! \sin\left(\dfrac{3\pi}{2} + \dfrac{\pi}{6}\right)$
4. $\! \sin\left(\dfrac{3\pi}{2} - \dfrac{\pi}{6}\right)$

$\begin{align} \cos\left(\dfrac{5\pi}{3}\right) & = \cos\left(\dfrac{10\pi}{6}\right) \\ \\ & = \cos\left(\dfrac{9\pi}{6} + \dfrac{\pi}{6}\right) \\ \\ & = \cos\left(\dfrac{3\pi}{2} + \dfrac{\pi}{6}\right) \quad\quad\quad\quad\quad \text{← Using radian measures always on The Unit Circle.} \\ \\ & = ... \\ \\ \end{align}$

Compound Angle Formula (aka Addition Subtraction Formula)

Determine the exact value of $\sin\dfrac{\pi}{12}$ Solution

Hint Clear Info

$\dfrac{\sqrt{\begin{matrix} \\ \quad \end{matrix}} - \sqrt{\begin{matrix} \\ \quad \end{matrix}}}{\quad}$

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & \sin \left(\dfrac{\pi}{12} \right) \\ \\ & = \sin \left(\dfrac{\pi}{4} - \dfrac{\pi}{6} \right) \\ \\ & = \sin\left(\dfrac{\pi}{4}\right)\cos\left(\dfrac{\pi}{6}\right) - \cos\left(\dfrac{\pi}{4}\right)\sin\left(\dfrac{\pi}{6}\right) \\ \\ & = \left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{\sqrt{3}}{2} \right) - \left( \dfrac{1}{\sqrt{2}} \right)\left( \dfrac{1}{2} \right) \\ \\ & = \left( \dfrac{\sqrt{3}}{2\sqrt{2}} \right) - \left( \dfrac{1}{2\sqrt{2}} \right) \\ \\ & = \left( \dfrac{\sqrt{3}\ -\ 1}{2\sqrt{2}} \right) \\ \\ & = \left( \dfrac{\sqrt{3}\ -\ 1}{2\sqrt{2}} \right) × \dfrac{\sqrt{2}}{\sqrt{2}} \\ \\ & = \dfrac{\sqrt{6}\ -\ \sqrt{2}}{2(2)} \\ \\ & = \dfrac{\sqrt{6}\ -\ \sqrt{2}}{4} \\ \\ \end{align}$

Exact Value of Compound Angle Formula Using Principle Angles of Special Related Acute Angles

Evaluate each of the following with an exact value, using the compound angle formulas. These questions use the principle angles of related acute angles that are special triangle angles.

cos(285˚) Solution

Hint Clear Info

$\dfrac{\sqrt{\begin{matrix} \\ \quad \end{matrix}} + \quad \ }{}$

Incorrect Attempts:

CHECK

Hint Unavailable

Use the compound angle formula: cos(x˚ + y˚) = (cos x˚)(cos y˚) - (sin x˚)(sin y˚)

You need to find combinations of angles that have 'Related Special Acute Angles' so you can lookup the exact values. $\begin{align} & \cos(285˚) \\ \\ & = \cos(150˚ + 135˚) \quad\quad\quad or \quad\quad\quad \cos(60˚ + 225˚)\\ \\ & = (\cos 150˚)(\cos 135˚) - (\sin 150˚)(\sin 135˚) \\ \\ & = \left(\dfrac{-\sqrt{3}}{2}\right)\left(\dfrac{-1}{2}\right) - \left(\dfrac{1}{2}\right)\left(\dfrac{-1}{2}\right) \\ \\ & = \dfrac{\sqrt{3}}{4} + \dfrac{1}{4} \\ \\ & = \dfrac{\sqrt{3} + 1}{4} \\ \\ \end{align}$ Note this could also work or be done with cos(60˚ + 225˚)... you would get the same answer.

Compound Angle Formula (aka Addition Subtraction Formula)

Determine the exact value of cos165˚ Solution

Hint Clear Info

$\dfrac{-\sqrt{\begin{matrix} \\ \quad \end{matrix}} - \sqrt{\begin{matrix} \\ \quad \end{matrix}}}{\quad}$

Incorrect Attempts:

CHECK

Hint Unavailable

Denationalized the denominator in cos45˚ and sin45˚ $\begin{align} &cos165˚ \\ \\ & = cos(120˚ + 45˚) \quad\quad\quad\quad\quad\quad 120˚ \ \text{has an acute angle of} \ 60˚ \ \text{in Quadrant II} \\ \\ & = cos120˚cos45˚ - sin120˚sin45˚ \\ \\ & = \left(\dfrac{-1}{2}\right)\left( \dfrac{\sqrt{2}}{2} \right) - \left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{\sqrt{2}}{2} \right) \\ \\ & = \dfrac{-\sqrt{2}}{4} - \dfrac{\sqrt{6}}{4} \\ \\ & = \dfrac{-\sqrt{2}\ -\ \sqrt{6}}{4} \\ \\ \end{align}$

Compound Angle Formula (aka Addition Subtraction Formula)

Determine the exact value of $\dfrac{2 \tan\,52.5˚}{1 - \tan^2\,52.5˚}$ Solution

Hint Clear Info

$-\sqrt{\begin{matrix} \\ \quad \end{matrix}} -$

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} \tan\,(2x) & = \dfrac{2 \tan\,x}{1 - \tan^2\, x} \\ \\ \\ \\ \\ \\ \tan\,\Big[2(52.5˚)\Big] & = \dfrac{2\tan\,52.5˚}{1 - \tan^2\, 52.5˚} \\ \\ \\ \\ \\ \\ \\ \\ ∴ x & = 52.5˚ \\ \\ \end{align}$ $\begin{align} & \dfrac{2 \tan\,52.5˚}{1 - \tan^2\, 52.5˚} \\ \\ & = \tan 2(52.5˚) \\ \\ & = \tan(105˚) \\ \\ & = \tan(60˚ + 45˚) \\ \\ & = \dfrac{\tan 60˚ + \tan 45˚}{1 - \tan 60˚ \, \tan 45˚} \\ \\ & = \dfrac{\sqrt{3} + 1}{1 - (\sqrt{3})(1)} \\ \\ & = \dfrac{\sqrt{3} + 1}{1 - \sqrt{3}} \\ \\ & = \dfrac{(\sqrt{3} + 1)}{(1 - \sqrt{3})} × \dfrac{(1 + \sqrt{3})}{(1 + \sqrt{3})} \\ \\ & = \dfrac{\sqrt{3} + 3 + 1 + \sqrt{3}}{1 + \sqrt{3} - \sqrt{3} - 3} \\ \\ & = \dfrac{2\sqrt{3} + 4}{-2} \\ \\ & = -\dfrac{\cancel{2}(\sqrt{3} + 2)}{\cancel{2}} \\ \\ & = -\dfrac{\sqrt{3} + 2}{1} \\ \\ & = -\sqrt{3} - 2 \\ \\ \end{align}$

Compound Angle Formula (aka Addition Subtraction Formula)

Determine the exact value that is equivalent to the following. Solution $\sin\left(\dfrac{11\pi}{12}\right)$

Hint Clear Info

$\dfrac{\sqrt{\begin{matrix} \\ \quad \end{matrix}} - \sqrt{\begin{matrix} \\ \quad \end{matrix}}}{\quad}$

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & = sin\left(\dfrac{11\pi}{12}\right) \\ \\ & = sin\left(\dfrac{21\pi}{12} - \dfrac{10\pi}{12}\right) \\ \\ & = sin\left(\dfrac{7\pi}{4} - \dfrac{5\pi}{6}\right) \quad\quad\quad\quad\quad \text{← Using radian measures always on The Unit Circle.} \\ \\ & = sin\left(\dfrac{7\pi}{4}\right)cos\left(\dfrac{5\pi}{6}\right) - cos\left(\dfrac{7\pi}{4}\right)sin\left(\dfrac{5\pi}{6}\right) \quad\quad\quad\quad \text{← Related acute angles are: 45˚ and 30˚} \\ \\ & = \left(\dfrac{-1}{\sqrt{2}}\right)\left(\dfrac{-\sqrt{3}}{2}\right) - \left(\dfrac{1}{\sqrt{2}}\right)\left(\dfrac{1}{2}\right) \\ \\ & = \dfrac{\sqrt{3}}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}} \\ \\ & = \dfrac{\sqrt{3} - 1}{2\sqrt{2}} \\ \\ & = \dfrac{(\sqrt{3} - 1)}{2\sqrt{2}} × \dfrac{\sqrt{2}}{\sqrt{2}}\\ \\ & = \dfrac{\sqrt{6} - \sqrt{2}}{2(2)} \\ \\ & = \dfrac{\sqrt{6} - \sqrt{2}}{4} \\ \\ \end{align}$

Double Angle Formula Intro

Determine which one of the following is not a double angle formula. Solution

1. $\sin(2x) = 2\sin x\,\cos x$
2. $\cos(2x) = \cos^2x - \sin^2x$
3. $\cos(2x) = 2\cos^2x - 1$
4. $\cos(2x) = 1 - 2\sin^2x$
5. $\tan(2x) = \dfrac{2\tan x}{1\ -\ \tan^2x}$
6. $\sin(2x) = 2\sin^2x - 1$

There are 5 main double angle formulas...

• $\sin(2x) = 2\sin x\,\cos x$
• $\cos(2x) = \cos^2x - \sin^2x$
• $\cos(2x) = 2\cos^2x - 1$
• $\cos(2x) = 1 - 2\sin^2x$
• $\tan(2x) = \dfrac{2\tan x}{1\ -\ \tan^2x}$

Which of the following is incorrect? Solution

1. sin²(x) = sin(x) · sin(x)
2. sin²(x) = [sin(x)]²
3. sin²(x) = sin(x²)
4. None of the above are incorrect

Incorrect: sin²(x) = sin(x²)

Determine 2·sin²(x) given, Solution $\sin(x) = \dfrac{3}{5}$

Hint Clear Info

━━

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} & 2\sin ^2(x) \\ \\ & = 2\left( \sin (x) \right)^2 \\ \\ & = 2\left( \dfrac{3}{5} \right)^2 \\ \\ & = \dfrac{2(9)}{25} \\ \\ & = \dfrac{18}{25} \end{align}$

Double Angle Formula

Given: $\tan(\theta) = \sqrt{3}$

Determine tan(2θ) Solution

Hint Clear Info

$\sqrt{\begin{matrix} \\ \quad \ \end{matrix}}$

Incorrect Attempts:

CHECK

Hint Unavailable

Just use the double angle formula, $\begin{align} \tan(2\theta) & = \dfrac{2\tan(\theta)}{1 - \tan^2(\theta)} \\ \\ & = \dfrac{2\sqrt{3}}{1 - (\sqrt{3})^2} \\ \\ & = \dfrac{2\sqrt{3}}{-2} \\ \\ & = -\sqrt{3} \end{align}$