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Trigonometry : Unit Circle and Radians

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

Example Questions

Example Question #5 : Angles In The Unit Circle

Which of the following is NOT a special angle on the unit circle?

Possible Answers:

$75^\circ$

$150^\circ$

$315^\circ$

$90^\circ$

$330^\circ$

Correct answer:

$75^\circ$

Explanation:

For an angle to be considered a special angle, the angle must be able to produce a 30-60-90 or a 45-45-90 triangle.

The only angle that is not capable of the special angles formation is $75 \textup{ degrees}$ .

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Example Question #71 : Unit Circle And Radians

If $0\leq x\leq\pi$ and $cos(5x)=\frac{-1}{2}$ , then =

Possible Answers:

$\frac{\pi}{5}$

$\frac{6\pi}{15}$

$\frac{2\pi}{15}$

The solution does not lie in the given interval.

Correct answer:

$\frac{2\pi}{15}$

Explanation:

We first make the substitution k=5x .

In the interval $0\leq k\leq \pi$ , the equation $\cos k=\frac{-1}{2}$ has the solution $k=\frac{2\pi}{3}$ .

Solving for ,

$\begin{align*} 5x&=\frac{2\pi}{3}\\ x&=\frac{2\pi}{15} \end{align*}$ .

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Example Question #1 : Angles In The Unit Circle

What is the value of $\sin(\pi/6)$ from the unit circle?

Possible Answers:

-1/2

$\sqrt{3}/2$

$-\sqrt{3}/2$

1/2

Correct answer:

1/2

Explanation:

From the unit circle, the value of

$\sin(\pi/6)=1/2$ .

This can be found using the coordinate pair associated with the angle $\pi/6$ which is $(\sqrt3/2,1/2)$ .

Recall that the (x,y) pair are $(\cos,\sin)$ .

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Example Question #3 : Angles In The Unit Circle

What is

$\arcsin\left(-\frac{1}{2}\right)$ ,

using the unit circle?

Possible Answers:

$5\pi/6$

$\pi/6$

$11\pi/6$

$\pi/3$

Correct answer:

$11\pi/6$

Explanation:

Recall that the unit circle can be broken down into four quadrants. Each quadrant has similar coordinate pairs basic on the angle. The only difference between the actual coordinate pairs is the sign on them. In quadrant I all signs are positive. In quadrant II only sine and cosecant are positive. In quadrant III tangent and cotangent are positive and quadrant IV only cosine and secant are positive.

$\frac{11\pi}{6}$ has the reference angle of $\frac{\pi}{6}$ and lies in quadrant IV therefore (cos, sin)=(+,-) .

From the unit circle, the coordinate point of

$(\sqrt3/2,-1/2)$ corresponds with the angle $\frac{11\pi}{6}$ .

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Example Question #5 : Angles In The Unit Circle

Give the exact value.

Use the unit circle to find:

$\sin\frac{5\pi}{6}$

Possible Answers:

$\frac{\sqrt{2}}{2}$

$-\frac{1}{2}$

$\frac{1}{2}$

$\frac{\sqrt{3}}{2}$

None of the above

Correct answer:

$\frac{1}{2}$

Explanation:

Locate $\frac{5\pi}{6}$ on the unit circle.

Sine is related to the why y value of the coordinate point because it is opposite/Hyp.

In other words the (x,y) pair of the point located on the unit circle that extends from the origin is $(\cos(\theta),\sin(\theta))$ .

The coordinate pair for $\frac{5\pi}{6}$ is $\left(\frac{-\sqrt{3}}{2},\frac{1}{2}\right)$ thus,

$\sin\frac{5\pi}{6}=\frac{1}{2}$

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Example Question #71 : Unit Circle And Radians

Give the exact value.

Use the unit circle to find:

$\cos \frac{4\pi}{3}$

Possible Answers:

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

$\frac{1}{2}$

$-\frac{1}{2}$

None of the above

$-\frac{\sqrt{2}}{2}$

Correct answer:

$-\frac{1}{2}$

Explanation:

Locate $\frac{4\pi}{3}$ on the unit circle.

Cosine is related to the why x value of the points, because it is Adj/Hyp.

In other words the (x,y) pair of the point located on the unit circle that extends from the origin is $(\cos(\theta),\sin(\theta))$ .

The coordinate point of $\frac{4\pi}{3}$ is $\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right )$ thus,

$\cos \frac{4\pi}{3}=-\frac{1}{2}$ .

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Example Question #71 : Unit Circle And Radians

Give the exact value.

Consider the unit circle. What is the value of the given trigonometric function?

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

Possible Answers:

$-\frac{\sqrt{3}}{3}$

$\sqrt{3}$

$-\sqrt{3}$

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

Correct answer:

$-\sqrt{3}$

Explanation:

Use the unit circle to locate the following:

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

Recall that the (x,y) pair of the point located on the unit circle that extends from the origin is the following:

$(\cos(\theta),\sin(\theta))$

Since the tangent is equal to the sine divided by the cosine, we need to find the following:

$\sin \frac{2\pi}{3}=\frac{\sqrt{3}}{2}$

$\cos \frac{2\pi}{3}=-\frac{1}{2}$

Now, we will use this information to solve the problem.

$\tan \frac{2\pi}{3}=\frac{\sin\frac{2\pi}{3}}{\cos \frac{2\pi}{3}}$

Substitute the calculated values:

$\frac{\sin\frac{2\pi}{3}}{\cos \frac{2\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}$

Solve. Remember that dividing by a fraction is the same as multiplying by its reciprocal.

$\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\frac{\sqrt{3}}{2}\times-\frac{2}{1}$

$\frac{\sqrt{3}}{\not{2}}\times-\frac{\not{2}}{1}=-\sqrt{3}$

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Trigonometry : Unit Circle and Radians

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #5 : Angles In The Unit Circle

Example Question #71 : Unit Circle And Radians

Example Question #1 : Angles In The Unit Circle

Example Question #3 : Angles In The Unit Circle

Example Question #5 : Angles In The Unit Circle

Example Question #71 : Unit Circle And Radians

Example Question #71 : Unit Circle And Radians

All Trigonometry Resources

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