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Precalculus : Trigonometric Functions

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Example Questions

Example Question #1 : Determine The Domain Of A Trigonometric Function

Which of the following is the correct domain of $cot(\theta)+3$ , where represents an integer?

Possible Answers:

$\textup{Everywhere except: }\pi k$

$\textup{Only Every: }\pi k$

$\textup{Everywhere except: }\frac{\pi}{2} k$

$\textup{Only Every: }2\pi k$

$\textup{Everywhere except: }2\pi k$

Correct answer:

$\textup{Everywhere except: }\pi k$

Explanation:

The cotangent graph only has a period of $\pi$ intervals and is most similar to the tangent graph. The domain of cotangent exists everywhere except every $\pi k$ value since an asymptote exists at those values in the domain.

The y-intercept of 3 shifts the cotangent graph up by three units, so this does not affect the domain.

Therefore, the graph exists everywhere except $\pi k$ , where is an integer.

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Example Question #2 : Trigonometric Functions

Please choose the best answer from the following choices.

What is the domain of the following function?

y=sin(x)

Possible Answers:

$\left ( 1,1\right )$

$\left [ 1,1\right ]$

$\left ( -\infty ,\infty \right )$

$\left [ -\infty ,\infty \right ]$

Correct answer:

$\left ( -\infty ,\infty \right )$

Explanation:

All x values make the function work. Thus, making the domain $\left ( -\infty ,\infty \right )$ . They're parentheses instead of brackets because parentheses are used when you can't actually use the specific value next to it. It is impossible to use infinity which makes parentheses appropriate. Brackets are used when you CAN use the specific value next to it.

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Example Question #2 : Determine The Domain Of A Trigonometric Function

Please choose the best answer from the following choices.

What is the domain of the following function:

$y=\cos (x)$

Possible Answers:

$\left [ -\infty ,\infty \right ]$

$\left ( -\infty ,\infty \right )$

$\left ( 0,1\right )$

$\left [ 0,1\right ]$

Correct answer:

$\left ( -\infty ,\infty \right )$

Explanation:

All x values work for the function. Thus, making the domain all real numbers. Parentheses are required because you can never actually use the number infinity.

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Example Question #1 : Trigonometric Functions

Please choose the best answer from the following choices.

What is the domain of $y=\tan (x)$ .

Possible Answers:

$\left [ -\pi /2,\pi /2\right ]$

$\left ( -\pi /2,\pi /2\right )$

$\left \{ x\mid x\neq \pi /2+k\pi \right \}$

$\left ( -\infty ,\infty \right )$

Correct answer:

$\left \{ x\mid x\neq \pi /2+k\pi \right \}$

Explanation:

If you look at a graph of the function, you can see that every curve has a vertical asymptote that repeats every $\pi$ radians in the positive and negative x-direction, starting at $-\pi/2$ radians. Also, the curve has a length that stretches $\pi$ radians which makes the domain $\left \{ x\mid x\neq \pi /2+k\pi \right \}$ .

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Example Question #4 : Determine The Domain Of A Trigonometric Function

What is the restriction of the domain of the function given by:

$y=3sec(3(x-\frac{\pi}{6}))$

For all the answer choices below, is any integer.

Possible Answers:

$x\neq \frac{\pi}{6}n$

$x\neq \frac{2\pi}{3}+\pi n$

$x\neq \frac{\pi}{3}n$

$x\neq \frac{\pi}{3}+\pi n$

Correct answer:

$x\neq \frac{\pi}{3}n$

Explanation:

y=sec(x) has restrictions on its domain such that

$x\neq \frac{\pi}{2}+\pi n$ , where is any integer.

To determine the domain for $y=3sec(3(x-\frac{\pi}{6}))$ ,

we equate the terms within the secant function and set them equal to the original domain restriction.

$3(x-\frac{\pi}{6})\neq \frac{\pi}{2}+\pi n$

Solving for ,

$x-\frac{\pi}{6}\neq\frac{\pi}{6}+\frac{\pi}{3}n$

$x\neq\frac{\pi}{3}+\frac{\pi}{3}n=\frac{\pi}{3}n$

The new domain restriction is:

$x\neq \frac{\pi}{3}n$ where is an integer

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Example Question #1 : Find The Value Of Any Of The Six Trigonometric Functions

Solve the following:

$cot\left(\frac{\pi}{2}\right)+tan(\pi)-sec\left(\frac{\pi}{3}\right)$

Possible Answers:

$-\frac{1}{2}$

$\frac{1}{2}$

Correct answer:

Explanation:

Rewrite $cot\left(\frac{\pi}{2}\right)+tan(\pi)-sec\left(\frac{\pi}{3}\right)$ in terms of sine and cosine functions.

$\frac{cos(\frac{\pi}{2})}{sin(\frac{\pi}{2})}+\frac{sin(\pi)}{cos(\pi)}-\frac{1}{cos(\frac{\pi}{3})}$

Since these angles are special angles from the unit circle, the values of each term can be determined from the x and y coordinate points at the specified angle.

Solve each term and simplify the expression.

$\frac{0}{1}+\frac{0}{-1}-\frac{1}{\frac{1}{2}}= 0+0-2=-2$

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Example Question #1 : Find The Value Of Any Of The Six Trigonometric Functions

Q1 new

Find the value of .

Possible Answers:

Correct answer:

Explanation:

Using trigonometric relationships, one can set up the equation

$cos(\theta)=\frac{adjacent}{hypotenuse}$

$cos(30^\circ)=\frac{25}{x}$ .

Solving for ,

$\\x\cdot cos(30^\circ)=25\\ \\x=\frac{25}{cos(30^\circ)}\\ \\x=28.8675... \\ \\x=29$

Thus, the answer is found to be 29.

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Example Question #7 : Trigonometric Functions

Q2 new

Find the value of .

Possible Answers:

106

120

Correct answer:

106

Explanation:

Using trigonometric relationships, one can set up the equation

$sin(\theta)=\frac{opposite}{hypotenuse}$ .

Plugging in the values given in the picture we get the equation,

$sin(25) = \frac{45}{x}$ .

Solving for ,

$\\x\cdot sin(25)=45\\ \\x=\frac{45}{sin(25)}\\ \\x=106.479...\\ \\x=106$ .

Thus, the answer is found to be 106.

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Example Question #2 : Find The Value Of Any Of The Six Trigonometric Functions

Find all of the angles that satistfy the following equation:

$sin(\theta )=\frac{\sqrt{3}}{2}$

Possible Answers:

$\theta= \frac{\pi k}{3}$

$\theta= \frac{\pi}{3}+2\pi k$ OR $\theta= \frac{2\pi}{3}+2\pi k$

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

$\theta= \frac{2\pi}{3}+2\pi k$

$\theta= \frac{\pi}{3}+\pi k$

Correct answer:

$\theta= \frac{\pi}{3}+2\pi k$ OR $\theta= \frac{2\pi}{3}+2\pi k$

Explanation:

The values of $\theta$ that fit this equation would be:

$\frac{\pi}{3}$ and $\frac{2\pi}{3}$

because these angles are in QI and QII where sin is positive and where

$sin(\theta)=\frac{\sqrt{3}}{2}$ .

This is why the answer

$\theta= \frac{\pi k}{3}$

is incorrect, because it includes inputs that provide negative values such as:

$sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}$

Thus the answer would be each $2\pi$ multiple of $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ , which would provide the following equations:

$\theta= \frac{\pi}{3}+2\pi k$ OR $\theta= \frac{2\pi}{3}+2\pi k$

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Example Question #2 : Find The Value Of Any Of The Six Trigonometric Functions

Evaluate: $sin\left(\frac{\pi}{2}\right)-sin\left(\frac{\pi}{3}\right)+sin\left(\frac{\pi}{4}\right)$

Possible Answers:

$\frac{1}{2}$

$\frac{2-\sqrt3+\sqrt2}{2}$

$\frac{2-\sqrt5}{2}$

Correct answer:

$\frac{2-\sqrt3+\sqrt2}{2}$

Explanation:

To evaluate $sin(\frac{\pi}{2})-sin(\frac{\pi}{3})+sin(\frac{\pi}{4})$ , break up each term into 3 parts and evaluate each term individually.

$sin(\frac{\pi}{2})=1$

$sin(\frac{\pi}{3})=\frac{\sqrt3}{2}$

$sin(\frac{\pi}{4})=\frac{\sqrt2}{2}$

Simplify by combining the three terms.

$1-\frac{\sqrt3}{2}+\frac{\sqrt2}{2}= \frac{2-\sqrt3+\sqrt2}{2}$

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Precalculus : Trigonometric Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Determine The Domain Of A Trigonometric Function

Example Question #2 : Trigonometric Functions

Example Question #2 : Determine The Domain Of A Trigonometric Function

Example Question #1 : Trigonometric Functions

Example Question #4 : Determine The Domain Of A Trigonometric Function

Example Question #1 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #1 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #7 : Trigonometric Functions

Example Question #2 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #2 : Find The Value Of Any Of The Six Trigonometric Functions

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