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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

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6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Graphs Of Inverse Trigonometric Functions

Which of the following is the graph of $cos^{-1}(x) + 2$ with $D: -1 \leq x \leq 1$ ?

Possible Answers:

Screen shot 2020 08 27 at 2.14.48 pm

Screen shot 2020 08 27 at 2.14.55 pm

Screen shot 2020 08 27 at 2.13.29 pm

Screen shot 2020 08 27 at 2.14.41 pm

Correct answer:

Screen shot 2020 08 27 at 2.13.29 pm

Explanation:

We first need to think about the graph of the function $cos^{-1}(x)$ .

Screen shot 2020 08 27 at 2.13.25 pm

Using the formula y = D + Acos(B(x - C)) where is the vertical shift, we have to perform a transformation of moving the function $cos^{-1}(x)$ up two units on the graph.

Screen shot 2020 08 27 at 2.13.29 pm

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Example Question #1 : Graphs Of Inverse Trigonometric Functions

Which of the following is the correct graph and range of the inverse function of 2sin(x) with $D:\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$ ?

Possible Answers:

$R: \frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$

Screen shot 2020 08 27 at 3.51.45 pm

$R: \frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$

Screen shot 2020 08 27 at 3.51.52 pm

$R:-\pi \leq y \leq \pi$

Screen shot 2020 08 27 at 3.51.45 pm

$R:-\pi \leq y \leq \pi$

Screen shot 2020 08 27 at 3.51.52 pm

Correct answer:

$R: \frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$

Screen shot 2020 08 27 at 3.51.52 pm

Explanation:

First, we must solve for the inverse of 2sin(x)

y = 2sin(x)

x = 2sin(y)

$\frac{x}{2} = sin(y)$

$sin^-1(\frac{x}{2}) = y$

So now we are trying to find the range of and plot the function $y = sin^{-1}(\frac{x}{2})$ . Let’s start with the graph of $sin^{-1}(x)$ . We know the domain is [-1,1] .

Screen shot 2020 08 27 at 3.51.45 pm

Now using the formula $y = D + Acos^{-1}(B(x-C))$ where $\frac{2\pi}{B}$ = Period, the period of $y = sin^{-1}(\frac{x}{2})$ is $4\pi$ . And so we perform a transformation to the graph of $sin^{-1}(x)$ to change the period from $2\pi$ to $4\pi$ .

Screen shot 2020 08 27 at 3.51.52 pm

We can see that the graph has a range of $[\frac{-\pi}{2},\frac{\pi}{2}]$

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Example Question #7 : Graphs Of Inverse Trigonometric Functions

True or False: The domain for arctan(x) will always be all real numbers no matter the value of or any transformations applied to the tangent function.

Possible Answers:

False

True

Correct answer:

True

Explanation:

This is true because just as the range of tan(x) is all real numbers due to the vertical asymptotes of the function, the function arctan(x) extends to all values of but is limited in its values of . No matter the transformations applied, all values of will still be reached.

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Example Question #3 : Graphs Of Inverse Trigonometric Functions

Which of the following is the graph of $\frac{1}{2}cos^{-1}(x) + 1$ ?

Possible Answers:

Screen shot 2020 08 27 at 1.57.17 pm

Screen shot 2020 08 27 at 1.54.54 pm

Screen shot 2020 08 27 at 1.55.06 pm

Screen shot 2020 08 27 at 1.55.00 pm

Correct answer:

Screen shot 2020 08 27 at 1.55.06 pm

Explanation:

First, we must consider the graph of $cos^{-1}(x)$ .

Screen shot 2020 08 27 at 1.54.54 pm

Using the formula $D + Acos^{-1}(B(x-C))$ we can apply the transformations step-by-step. First we will transform the amplitude, so $A = \frac{1}{2}$ so we must shorten the amplitude to $\frac{1}{2}$ .

Screen shot 2020 08 27 at 1.55.00 pm

Now we must apply a vertical shift of one unit since D=1 . This leaves us with our answer.

Screen shot 2020 08 27 at 1.55.06 pm

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Example Question #1 : Solve A Trigonometric Function By Squaring Both Sides

True or False: All solutions found from squaring both sides of a trigonometric function are valid should be given as a final answer.

Possible Answers:

True

False

Correct answer:

False

Explanation:

This is not true. This is because when squaring both sides and then plugging back into the original equations, some of our solutions may be extraneous solutions. Therefore, when solving a trigonometric equation by squaring both sides, all solutions found must be plugged back into the original equation and validated.

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Example Question #651 : Trigonometry

True or False: You should always solve for a trigonometric equation by squaring both sides. This will always be the most efficient method.

Possible Answers:

False

True

Correct answer:

False

Explanation:

We should only square by both sides when all other identities are not able to be used in an equation. Quite often, you will find that a trigonometric identity can be used to simplify an equation. Squaring both sides is ultimately trying to produce a trigonometric identity in order to solve for the equation.

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Example Question #1 : Solve A Trigonometric Function By Squaring Both Sides

Solve the following equation by squaring both sides: $1 + cot(\theta) = csc(\theta)$

Possible Answers:

$\theta = \frac{\pi}{2}$

$\theta = 0$

$\theta = 0, \pi$

$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$

Correct answer:

$\theta = \frac{\pi}{2}$

Explanation:

We begin with our original equation:

$1 + cot(\theta) = csc(\theta)$

$(1+cot(\theta))^2 = (csc(\theta))^2$

$1 + 2cot(\theta) + cot^2(\theta) = csc^2(\theta)$

$1 + 2cot(\theta) + cot^2(\theta) = 1 + cot^2(\theta)$ (Pythagorean Identity)

$2cot(\theta) = 0$

$cot(\theta) = 0$

Looking at the unit circle we see that $cot(\theta) = 0$ at $\frac {\pi}{2}$ and $\frac{3\pi}{2}$ . We must plug these back into our original equation to validate them.

Checking $\theta = \frac{\pi}{2}$

$1 + cot(\frac{\pi}{2}) = csc(\frac{\pi}{2})$

1 + 0 = 1

1 = 1

Checking $\theta = \frac{3\pi}{2}$

$1 + cot(\frac{3\pi}{2}) = \frac{3\pi}{2}$

1 + 0 = -1

$1 \neq -1$

And so our only solution is $\theta = \frac{\pi}{2}$

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Example Question #3 : Solve A Trigonometric Function By Squaring Both Sides

Solve the following equation by squaring both sides: $-\sqrt{3} + tan(\theta) = sec(\theta)$

Possible Answers:

$\theta = \frac {\sqrt\pi}{6}, \frac{7\pi}{6}$

$\theta = \frac{\sqrt11\pi}{6}$

$\theta = \frac{\sqrt7\pi} {6}$

$\theta = \frac {\sqrt \pi} {6}$

Correct answer:

$\theta = \frac {\sqrt \pi} {6}$

Explanation:

We begin with our original equation:

$-\sqrt{3} + tan(\theta) = sec(\theta)$

$(-\sqrt{3} + tan(\theta))^2$ $= (sec(\theta))^2 (sec(\theta))^2sec(\theta))^2c(\theta))^2(\theta))^2(\theta))^2(\tan))^2))^2^2$

$3 - 2\sqrt{3}tan(\theta) + tan^2(\theta) = sec^2(\theta)$

$3 - 2\sqrt{3}tan(\theta) + tan^2(\theta) = 1 + tan^2(\theta)$ (Pythagorean Identity)

$-2\sqrt{3}tan(\theta) = -2$

$tan(\theta) = \frac{1}{\sqrt{3}}$

From the unit circle, we see that $tan(\theta) = \frac{1}{\sqrt{3}}$$ when $$\theta = \frac{\pi}{6}, \frac{7\pi}{6}$ . We must check both of these solutions in the original equation.

Checking $\theta = \frac{\pi}{6}$

$-\sqrt{3} + tan(\frac{\pi}{6}) = sec(\frac{\pi}{6})$

$-\sqrt{3} + \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3}$

$\frac{\sqrt{3}}{3} \neq \frac{5\sqrt{3}}{3}$

Checking $\theta = \frac{7\pi}{6}$

$-\sqrt{3} + tan(\frac{7\pi}{6}) = sec(\frac{7\pi}{6})$

$-\sqrt{3} + \frac{\sqrt{3}}{3} = -\frac{2\sqrt{3}}{3}$

$\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{3}$

So we see our only solution is $\theta = \frac{\sqrt{\pi}}{6}$

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Example Question #4 : Solve A Trigonometric Function By Squaring Both Sides

Solve the following equation by squaring both sides: $tan(\theta) + 1 = sec(\theta)$

Possible Answers:

$\theta = \pi$

$\theta = 0, \pi$

$\theta = 1$

$\theta = 0$

Correct answer:

$\theta = 0$

Explanation:

We begin with our original equation

$tan(\theta) + 1 = sec(\theta)$

$(tan(\theta) + 1)^2 = (sec(\theta))^2$

$tan^2(\theta) + 2tan(\theta) + 1 = sec^2(\theta)$

$tan^2(\theta) + 2tan(\theta) + 1 = tan^2(\theta) + 1$ (Pythagorean Identity)

$2tan(\theta) = 0$

$tan(\theta) = 0$

$\frac{sin(\theta)}{cos(\theta)} = 0$ (substitution)

Using this form, we see we really only need to consider when $sin(\theta) = 0$ at and $\pi$ . Now we must plug these values into the original equation to check and see if they are both acceptable solutions to our problem.

Checking $\theta = 0$ :

tan(0) + 1 = sec(0)

0 + 1 = 1

1 = 1

Checking $\theta = \pi$

$tan(\pi) + 1 = sec(\pi)$

0 + 1 = -1

$1 \neq -1$

By checking our solutions we see the only solution to our equation is $\theta = 0$ .

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Example Question #5 : Solve A Trigonometric Function By Squaring Both Sides

Solve the following equation by squaring both sides: $cos(\theta) + sin(\theta) = 1$

Possible Answers:

$\theta = 0$

$\theta = 3\pi$

$\theta = \pi$

$\theta = 0, \pi$

Correct answer:

$\theta = 0$

Explanation:

We begin with our original equation.

$cos(\theta) + sin(\theta) = 1$

$(cos(\theta) + sin(\theta))^2 = 1^2$

$cos^2(\theta) + 2cos(\theta)sin(theta) + sin^2(\theta) = 1$

$cos^2(\theta) + sin^2(\theta) + 2cos(\theta)sin(\theta) = 1$

$1 + 2cos(\theta)sin(\theta) = 1$ (Pythagorean Identity)

$2cos(\theta)sin(\theta) = 0$

$sin(2\theta) = 0$ (Double-Angle Formula)

We know that $sin(2\theta)$ will be equal to for when $\theta$ is any multiple of and when $\theta = 0$ . We need to check both solutions (we will simply check $\pi$ for simplicity) to make sure they are valid solutions.

Checking $\theta = 0$ :

cos(0) + sin(0) = 1

1 + 0 = 1

1 = 1

Checking $\theta = \pi$

$cos(\pi) + sin(\pi) = 1$

-1 + 0 = 1

$-1 \neq 1$

By checking our solutions, we see the only solution to this equation is $\theta = 0$

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #1 : Graphs Of Inverse Trigonometric Functions

Example Question #1 : Graphs Of Inverse Trigonometric Functions

Example Question #7 : Graphs Of Inverse Trigonometric Functions

Example Question #3 : Graphs Of Inverse Trigonometric Functions

Example Question #1 : Solve A Trigonometric Function By Squaring Both Sides

Example Question #651 : Trigonometry

Example Question #1 : Solve A Trigonometric Function By Squaring Both Sides

Example Question #3 : Solve A Trigonometric Function By Squaring Both Sides

Example Question #4 : Solve A Trigonometric Function By Squaring Both Sides

Example Question #5 : Solve A Trigonometric Function By Squaring Both Sides

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