<meta http-equiv="refresh" content="1; url=/nojavascript/"> Exact Values for Inverse Sine, Cosine, and Tangent ( Read ) | Trigonometry | CK-12 Foundation

# Exact Values for Inverse Sine, Cosine, and Tangent

%
Best Score
Practice Exact Values for Inverse Sine, Cosine, and Tangent
Best Score
%

# Inverse Trigonometric Functions

Trig Riddle: I am an angle that measures between $0^\circ$ and $360^\circ$ . My tangent is $-\sqrt{3}$ . What angle am I?

### Guidance

Earlier in the unit we learned how to find the measure of an acute angle in a right triangle using the inverse trigonometric ratios on the calculator. Now we will extend this inverse concept to finding the possible angle measures given a trigonometric ratio on the unit circle. We say possible, because there are an infinite number of possible angles with the same ratios. Think of the unit circle. For which angles does $\sin \theta=\frac{1}{2}$ ? From the special right triangles, we know that the reference angle must be $30^\circ$ or $\frac{\pi}{6}$ . But because sine is positive in the first and second quadrants, the angle could also be $150^\circ$ or $\frac{5 \pi}{6}$ . In fact, we could take either of these angles and add or subtract $360^\circ$ or $2 \pi$ to it any number of times and still have a coterminal angle for which the sine ratio would remain $\frac{1}{2}$ . For problems in this concept we will specify a finite interval for the possible angles measures. In general, this interval will be $0 \le \theta < 360^\circ$ for degree measures and $0 \le \theta < 2 \pi$ for radian measures.

Inverse Trigonometric Ratios on the Calculator

When you use the calculator to find an angle given a ratio, the calculator can only give one angle measure. The answers for the respective functions will always be in the following quadrants based on the sign of the ratio.

Trigonometric Ratio Positive Ratios Negative Ratios
Sine $0 \le \theta \le 90$ or $0 \le \theta \le \frac{\pi}{2}$ $-90 \le \theta \le 0$ or $-\frac{\pi}{2} \le \theta \le 0$
Cosine $0 \le \theta \le 90$ or $0 \le \theta \le \frac{\pi}{2}$ $90 < \theta \le 180^\circ$ or $\frac{\pi}{2} < \theta \le \pi$
Tangent $0 \le \theta \le 90$ or $0 \le \theta \le \frac{\pi}{2}$ $-90 \le \theta < 0$ or $-\frac{\pi}{2} \le \theta < 0$

#### Example A

Use your calculator to find all solutions on the interval $0 \le \theta < 360^\circ$ . Round your answers to the nearest tenth.

a. $\cos^{-1} (0.5437)$

b. $\tan^{-1}(-3.1243)$

c. $\csc^{-1}(3.0156)$

Solution: For all of these, we must first make sure the calculator is in degree mode.

a. Type in $2^{nd} \text{COS}$ , to get $\cos^{-1}($ on your calculator screen. Next, type in the ratio to get $\cos^{-1} (0.5437)$ on the calculator and press ENTER. The result is $57.1^\circ$ . This is an angle in the first quadrant and a reference angle. We want to have all the possible angles on the interval $0 \le \theta < 360^\circ$ . To find the second angle, we need to think about where else cosine is positive. This is in the fourth quadrant. Since the reference angle is $57.1^\circ$ , we can find the angle by subtracting $57.1^\circ$ from $360^\circ$ to get $302.9^\circ$ as our second angle. So $\cos^{-1}(0.5437)=57.1^\circ, 302.9^\circ$ .

b. Evaluate $\tan^{-1}(-3.1243)$ on the calculator using the same process to get $-72.3^\circ$ . This is a $72.3^\circ$ reference angle in the fourth quadrant. Since we want all possible answers on the interval $0 \le \theta < 360^\circ$ , we need angles with reference angles of $72.3^\circ$ in the second and fourth quadrants where tangent is negative.

$2^{nd}$ quadrant: $180^\circ - 72.3^\circ = 107.7^\circ$ and $4^{th}$ quadrant: $360^\circ - 72.3^\circ = 287.7^\circ$

So, $\tan^{-1}(-3.1243)=107.7^\circ, 287.7^\circ$

c. This time we have a reciprocal trigonometric function. Recall that $\sin \theta=\frac{1}{\csc \theta}$ . In this case, $\csc \theta=3.0156$ so $\sin \theta=\frac{1}{3.0156}$ and therefore $\csc^{-1}(3.0156)=\sin^{-1} \left(\frac{1}{3.0156}\right)=19.4^\circ$ from the calculator. Now, we need to find our second possible angle measure. Since sine (and subsequently, cosecant) is positive in the second quadrant, that is where our second answer lies. The reference angle is $19.4^\circ$ so the angle is $180^\circ -19.4^\circ=160.6^\circ$ . So, $\csc^{-1}(3.0156)=19.4^\circ, 160.6^\circ$ .

#### Example B

Use your calculator to find $\theta$ , to two decimal places, where $0 \le \theta < 2 \pi$ .

a. $\sec \theta = 2.1647$

b. $\sin \theta =-1.0034$

c. $\cot \theta =-1.5632$

Solution: For each these, we will need to be in radian mode on the calculator.

a. Since $\cos \theta=\frac{1}{\sec \theta}, \sec^{-1}(2.1647)=\cos^{-1} \left(\frac{1}{2.1647}\right)=1.09$ radians. This is a first quadrant value and thus the reference angle as well. Since cosine (and subsequently, secant) is also positive in the fourth quadrant, we can find the second answer by subtracting from $2 \pi$ : $2 \pi -1.09=5.19$ .

Hence, $\sec^{-1}(2.1647)=1.09, 5.19$

b. From the calculator, $\sin^{-1}(-0.3487)=-0.36$ radians, a fourth quadrant reference angle of 0.36 radians. Now we can use this reference angle to find angles in the third and fourth quadrants within the interval given for $\theta$ .

$3^{rd}$ quadrant: $\pi+0.36=3.50$ and $4^{th}$ quadrant: $2 \pi -0.36=5.92$

So, $\sin^{-1}(-0.3487)=3.50, 5.92$

c. Here, $\tan \theta=\frac{1}{\cot \theta}$ , so $\cot^{-1}(-1.5632)=\tan^{-1}\left(-\frac{1}{1.5632}\right)=-0.57$ , a fourth quadrant reference angle of 0.57 radians. Since the ratio is negative and tangent and cotangent are both negative in the $2^{nd}$ and $4^{th}$ quadrants, those are the angles we must find.

$2^{nd}$ quadrant: $\pi - 0.57=2.57$ and $4^{th}$ quadrant: $2 \pi-0.57=5.71$

So, $\cot^{-1}(-1.5632)=2.57, 5.71$

#### Example C

Without using a calculator, find $\theta$ , where $0 \le \theta < 2 \pi$ .

a. $\sin \theta=-\frac{\sqrt{3}}{2}$

b. $\cos \theta=\frac{\sqrt{2}}{2}$

c. $\tan \theta=-\frac{\sqrt{3}}{3}$

d. $\csc \theta=-2$

Solution:

a. From the special right triangles, sine has the ratio $\frac{\sqrt{3}}{2}$ for the reference angle $\frac{\pi}{3}$ . Now we can use this reference angle to find angles in the $3^{rd}$ and $4^{th}$ quadrant where sine is negative.

$3^{rd}$ quadrant: $\pi + \frac{\pi}{3}=\frac{4 \pi}{3}$ and $4^{th}$ quadrant: $2 \pi -\frac{\pi}{3}=\frac{5 \pi}{3}$

So, $\theta=\frac{4 \pi}{3}, \frac{5 \pi}{3}.$

b. From the special right triangles, cosine has the ratio $\frac{\sqrt{2}}{2}$ for the reference angle $\frac{\pi}{4}$ . Since cosine is positive in the first and fourth quadrants, one answer is $\frac{\pi}{4}$ and the second answer ( $4^{th}$ quadrant) will be $2 \pi -\frac{\pi}{4}=\frac{7 \pi}{4}$ . So, $\theta=\frac{\pi}{4}, \frac{7 \pi}{4}$ .

c. From the special right triangles, tangent has the ratio $\frac{\sqrt{3}}{3}$ for the reference angle $\frac{\pi}{6}$ . Since tangent is negative in the second and fourth quadrants, we will subtract $\frac{\pi}{6}$ from $\pi$ and $2 \pi$ to find the angles.

$\pi -\frac{\pi}{6}=\frac{5 \pi}{6}$ and $2 \pi -\frac{\pi}{6}=\frac{11 \pi}{6}$ . So, $\theta = \frac{5 \pi}{6}, \frac{11 \pi}{6}$ .

d. First, consider that if $\csc \theta=-2$ , then $\sin \theta=-\frac{1}{2}$ . Next, from special right triangles, we know that sine is $\frac{1}{2}$ for a $\frac{\pi}{6}$ reference angle. Finally, find the angles with a reference angle of $\frac{\pi}{6}$ in the third and fourth quadrants where sine is negative. $\pi + \frac{\pi}{6}=\frac{7 \pi}{6}$ and $2 \pi -\frac{\pi}{6}=\frac{11 \pi}{6}$ . So, $\theta=\frac{7 \pi}{6}, \frac{11 \pi}{6}$ .

Concept Problem Revisit

From the special right triangles, tangent has the ratio $\sqrt{3}$ for the reference angle $60^\circ$ . Since tangent is negative in the second and fourth quadrants, we will subtract $60^\circ$ from $180^\circ$ and $360^\circ$ to find the angles.

$180^\circ - 60^\circ=120^\circ$ and $360^\circ - 60^\circ = 300^\circ$ . So, I am the angle that measures either $120^\circ$ or $300^\circ$ .

### Guided Practice

1. Use your calculator to find all solutions on the interval $0 \le \theta < 360^\circ$ . Round your answers to the nearest tenth.

a. $\sin^{-1}(0.7821)$

b. $\cot^{-1}(-0.6813)$

c. $\sec^{-1}(4.0159)$

2. Use your calculator to find $\theta$ , to two decimal places, where $0 \le \theta < 2 \pi$ .

a. $\cos \theta=-0.9137$

b. $\tan \theta=5.0291$

c. $\csc \theta=2.1088$

3. Without using a calculator, find $\theta$ , where $0 \le \theta < 2 \pi$ .

a. $\cos \theta=-\frac{\sqrt{3}}{2}$

b. $\cot \theta=\frac{\sqrt{3}}{3}$

c. $\sin \theta=-1$

1. a. $51.5^\circ$ and $180^\circ - 51.5^\circ=128^\circ$

b. $\cot^{-1}(-0.6813)=\tan^{-1} \left(-\frac{1}{0.6813} \right)=-55.7^\circ, 180^\circ-55.7^\circ=124.3^\circ$ and $360^\circ -55.7^\circ=304.3^\circ$

c. $\sec^{-1}(4.0159)=\cos^{-1} \left(\frac{1}{4.0159}\right)=75.6^\circ$ and $360^\circ -75.6^\circ=284.4^\circ$

2. a. $\cos^{-1}(-0.9137)=2.72$ and $\pi+2.72=30.34$

b. $\tan^{-1}(5.0291)=1.37$ and $\pi+1.37=4.51$

c. $\csc^{-1}(2.1088)=\sin^{-1} \left(\frac{1}{2.1088}\right)=0.49$ and $\pi-0.49=2.65$

3. a. $\cos^{-1} \left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}$ , since the ratio is negative, $\theta=\pi -\frac{\pi}{6}=\frac{5 \pi}{6}$ and $\pi + \frac{\pi}{6}=\frac{7 \pi}{6}$

b. $\cot^{-1}\left(\frac{\sqrt{3}}{3}\right)=\tan^{-1} \sqrt{3}=\frac{\pi}{3}, \theta =\frac{\pi}{3}$ , and $\pi + \frac{\pi}{3}=\frac{4 \pi}{3}$

c. $\sin^{-1}(-1)=\frac{3 \pi}{2}, \theta=\frac{3 \pi}{2}$

### Practice

For problems 1-6, use your calculator to find all solutions on the interval $0 \le \theta < 360^\circ$ . Round your answers to the nearest tenth.

1. $\cos^{-1}(-0.2182)$
2. $\sec^{-1}(10.8152)$
3. $\tan^{-1}(-20.2183)$
4. $\sin^{-1}(0.8785)$
5. $\csc^{-1}(-6.9187)$
6. $\cot^{-1}(0.8316)$

For problems 7-12, use your calculator to find $\theta$ , to two decimal places, where $0 \le \theta < 2 \pi$ .

1. $\sin \theta=-0.6153$
2. $\cos \theta=0.1382$
3. $\cot \theta=-2.8135$
4. $\sec \theta=-8.8775$
5. $\tan \theta=0.9990$
6. $\csc \theta=12.1385$

For problems 13-18, find $\theta$ , without using a calculator, where $0 \le \theta < 2 \pi$ .

1. $\sin \theta=0$
2. $\cos \theta=-\frac{\sqrt{2}}{2}$
3. $\tan \theta=-1$
4. $\sec \theta=\frac{2 \sqrt{3}}{3}$
5. $\sin \theta=\frac{1}{2}$
6. $\cot \theta=\text{undefined}$
7. $\cos \theta=-\frac{1}{2}$
8. $\csc \theta=\sqrt{2}$
9. $\tan \theta=\frac{\sqrt{3}}{3}$