Trigonometry - Angles in the Unit Circle

Give the exact value.

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

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

$-\sqrt{3}$

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

$\sqrt{3}$

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

Explanation

Use the unit circle to locate the following:

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

Recall that the 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.

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

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

What are the ways to write 360o and 720o in radians?

$\small 2\pi,4\pi$

$\small \frac{\pi}{2},\pi$

$\small \pi,2\pi$

$\small \frac{\pi}{4},\frac{\pi}{2}$

$\small \small 0,\pi$

Explanation

$\small 360^o=2\pi$ on the unit circle.

$\small \small \small 720^o=4\pi$ on the unit circle.

Which of the following is not an angle in the unit circle?

$\frac{\pi}{5}$

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

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

$\pi$

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

Explanation

The unit circle is in two increments: $\frac{\pi}{6}$ : $(\frac{\pi}{6},\frac{2\pi}{6},\frac{3\pi}{6})$ , etc. and $\frac{\pi}{4}$ : $(\frac{\pi}{4},\frac{2\pi}{4},\frac{3\pi}{4})$ , etc. The only answer choice that is not a multiple of either $\frac{\pi}{6}$ or $\frac{\pi}{4}$ is $\frac{\pi}{5}$ .

What is the equivalent of $72^{\circ}$ , in radians?

$\frac{2\pi }{5}$

$\frac{\pi }{5}$

$\frac{\pi }{10}$

$\frac{2\pi}{10}$

$\pi$

Explanation

To convert from degrees to radians, use the equality

$2\pi = 360^{\circ}$ or $\pi = 180^{\circ}$ .

From here we use $\frac{\pi}{180^{\circ}}$ as a unit multiplier to convert our degrees into radians.

$\left (72^{\circ} \right )\left ( \frac{\pi }{180^{\circ}} \right ) = \frac{2\pi}{5}$

What is the value of ?

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

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

$\frac{1}{2}$

Explanation

To help with this one, draw a 45-45-90 triangle. With legs equal to 1 and a hypotenuse of $\sqrt2$ .

Then, use the definition of tangent as opposite over adjacent to find the value.

Since the legs are congruent, we get that the ratio is 1.

What is ?

$\frac{1}{2}$

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

Explanation

Recall that on the unit circle, sine represents the y coordinate of the unit circle.

Then, since we are at 90˚, we are at the positive y axis, the point (0,1).

At this point on the unit circle, the y value is 1.

Thus .

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

$75^\circ$

$90^\circ$

$150^\circ$

$315^\circ$

$330^\circ$

Explanation

For an angle to be considered a special angle, the angle must be able to produce a or a triangle.

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

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

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

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

$\frac{\pi}{5}$

The solution does not lie in the given interval.

Explanation

We first make the substitution .

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 ,

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

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

$\sqrt{3}/2$

$-\sqrt{3}/2$

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 pair are $(\cos,\sin)$ .

What is

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

using the unit circle?

$11\pi/6$

$\pi/6$

$5\pi/6$

$\pi/3$

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 .