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Precalculus : Find the value of any of the six trigonometric functions

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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 #2 : 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 #3 : Find The Value Of Any Of The Six Trigonometric Functions

Q2 new

Find the value of .

Possible Answers:

120

106

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 #1 : 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:

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

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

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

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

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

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 #5 : 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:

$\frac{1}{2}$

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

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

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

What is the value of $tan(\pi)+sin(\pi)$ ?

Possible Answers:

$\infty$

$\frac{\sqrt2}{2}$

Correct answer:

Explanation:

Convert $tan(\pi)+sin(\pi)$ in terms of sine and cosine.

$tan(\pi)+sin(\pi)= \frac{sin(\pi)}{cos(\pi)}+sin(\pi)$

Since theta is $\pi$ radians, the value of $sin(\pi)$ is the y-value of the point on the unit circle at $\pi$ radians, and the value of $cos(\pi)$ corresponds to the x-value at that angle.

The point on the unit circle at $\pi$ radians is (-1,0) .

Therefore, $sin(\pi)=0$ and $cos(\pi)=-1$ . Substitute these values and solve.

$\frac{sin(\pi)}{cos(\pi)}+sin(\pi)=\frac{0}{-1}+0=0$

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Example Question #7 : Find The Value Of Any Of The Six Trigonometric Functions

Solve: $\frac{1}{sin(\frac{\pi}{4})}$

Possible Answers:

$\frac{\sqrt2}{2}$

$\sqrt2$

$2\sqrt2$

$\frac{1}{2}$

Correct answer:

$\sqrt2$

Explanation:

First, solve the value of $sin\left(\frac{\pi}{4}\right)$ .

On the unit circle, the coordinate at $\frac{\pi}{4}$ radians is $\left(\frac{\sqrt2}{2}, \frac{\sqrt2}{2}\right)$ . The sine value is the y-value, which is $\frac{\sqrt2}{2}$ . Substitute this value back into the original problem.

$\frac{1}{sin(\frac{\pi}{4})}= \frac{1}{\frac{\sqrt2}{2}}= \frac{2}{\sqrt2}$

Rationalize the denominator.

$\frac{2}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}= \frac{2\sqrt2}{2}= \sqrt2$

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Example Question #8 : Find The Value Of Any Of The Six Trigonometric Functions

Find the exact answer for: cos(60)-sin(30)+cos(30)

Possible Answers:

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

$\frac{1}{2}$

$\frac{\sqrt3}{2}$

$\sqrt3$

Correct answer:

$\frac{\sqrt3}{2}$

Explanation:

To evaluate cos(60)-sin(30)+cos(30) , solve each term individually.

cos(60) refers to the x-value of the coordinate at 60 degrees from the origin. The x-value of this special angle is $\frac{1}{2}$ .

sin(30) refers to the y-value of the coordinate at 30 degrees. The y-value of this special angle is $\frac{1}{2}$ .

cos(30) refers to the x-value of the coordinate at 30 degrees. The x-value is $\frac{\sqrt3}{2}$ .

Combine the terms to solve cos(60)-sin(30)+cos(30) .

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

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Example Question #9 : Find The Value Of Any Of The Six Trigonometric Functions

Find the value of

$\sin(300)$ .

Possible Answers:

$-\sqrt2$

$-\frac{1}{2}$

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

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

$-\sqrt3}$

Correct answer:

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

Explanation:

The value of sin(300) refers to the y-value of the coordinate that is located in the fourth quadrant.

This angle 300 is also from the origin.

Therefore, we are evaluating sin(-60) .

$sin(-60)=-sin(60)=-\frac{\sqrt3}{2}$

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Example Question #10 : Find The Value Of Any Of The Six Trigonometric Functions

Simplify the following expression:

$sin(-270^{\circ})$

Possible Answers:

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

Correct answer:

Explanation:

Simplify the following expression:

$sin(-270^{\circ})$

Begin by locating the angle on the unit circle. -270 should lie on the same location as 90. We get there by starting at 0 and rotating clockwise $270^{\circ}$

So, we know that

$sin(-270^{\circ})=sin(90^{\circ})$

And since we know that sin refers to y-values, we know that

$sin(90^{\circ})=1$

So therefore, our answer must be 1

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Precalculus : Find the value of any of the six trigonometric functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

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

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

Example Question #3 : 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 #5 : 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 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #8 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #9 : Find The Value Of Any Of The Six Trigonometric Functions

Example Question #10 : Find The Value Of Any Of The Six Trigonometric Functions

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