Principal Values of Inverse Trigonometric Functions — Worked Solutions

Ques­tion 1: Six prin­ci­pal val­ues#

The prob­lem#

Find the prin­ci­pal val­ues of: (a) $\displaystyle \sin^{-1}\frac{1}{\sqrt{2}}$; (b) $\displaystyle \cos^{-1}\frac{\sqrt{3}}{2}$; (c) $\displaystyle \tan^{-1}\frac{1}{\sqrt{3}}$; (d) $\csc^{-1}(-2)$; (e) $\displaystyle \sec^{-1}\frac{2}{\sqrt{3}}$; (f) $\cot^{-1}(-\sqrt{3})$.

Under­stand­ing the prob­lem#

The prin­ci­pal value of $\sin^{-1}a$ is the one angle in the prin­ci­pal range of $\sin^{-1}$ whose sine is $a$; sim­i­larly for the oth­ers. The ranges are:

• $\sin^{-1}$: $\displaystyle \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$; $\cos^{-1}$: $[0, \pi]$; $\tan^{-1}$: $\displaystyle \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$;
• $\csc^{-1}$: $\displaystyle \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] - \{0\}$; $\sec^{-1}$: $\displaystyle [0, \pi] - \left\{\tfrac{\pi}{2}\right\}$; $\cot^{-1}$: $(0, \pi)$.

The idea#

For each part: (1) find a stan­dard angle with the right value; (2) if the value is neg­a­tive, move to the cor­rect side using the range; (3) con­firm the angle lies in the prin­ci­pal range.

Step-by-step solu­tion#

Part (a)

Step 1. $\displaystyle \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$, and $\displaystyle \frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

$$\displaystyle \sin^{-1}\frac{1}{\sqrt{2}} = \frac{\pi}{4}$$

Part (b)

Step 1. $\displaystyle \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, and $\displaystyle \frac{\pi}{6} \in [0, \pi]$.

$$\displaystyle \cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6}$$

Part (c)

Step 1. $\displaystyle \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$, and $\displaystyle \frac{\pi}{6} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

$$\displaystyle \tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}$$

Part (d)

Step 1. $\csc\theta = -2$ means $\displaystyle \sin\theta = -\frac{1}{2}$.

Step 2. In the range $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$, a neg­a­tive sine means a neg­a­tive angle: $\displaystyle \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$.

$$\displaystyle \csc^{-1}(-2) = -\frac{\pi}{6}$$

Part (e)

Step 1. $\displaystyle \sec\theta = \frac{2}{\sqrt{3}}$ means $\displaystyle \cos\theta = \frac{\sqrt{3}}{2}$, so $\displaystyle \theta = \frac{\pi}{6}$, which is in $\displaystyle [0, \pi] - \left\{\frac{\pi}{2}\right\}$.

$$\displaystyle \sec^{-1}\frac{2}{\sqrt{3}} = \frac{\pi}{6}$$

Part (f)

Step 1. $\displaystyle \cot\frac{\pi}{6} = \sqrt{3}$. For $-\sqrt{3}$ we need an angle in $(0, \pi)$ with neg­a­tive cotan­gent, that is, in the sec­ond quad­rant.

Step 2. $\displaystyle \cot\left(\pi - \frac{\pi}{6}\right) = -\cot\frac{\pi}{6} = -\sqrt{3}$.

$$\displaystyle \cot^{-1}(-\sqrt{3}) = \frac{5\pi}{6}$$

Check­ing the answer#

Every answer lies in its prin­ci­pal range, and sub­sti­tut­ing back gives the right value, for exam­ple $\displaystyle \cot\frac{5\pi}{6} = \frac{\cos(5\pi/6)}{\sin(5\pi/6)} = \frac{-\sqrt{3}/2}{1/2} = -\sqrt{3}$. ✓

Answer#

(a) $\displaystyle \frac{\pi}{4}$; (b) $\displaystyle \frac{\pi}{6}$; (c) $\displaystyle \frac{\pi}{6}$; (d) $\displaystyle -\frac{\pi}{6}$; (e) $\displaystyle \frac{\pi}{6}$; (f) $\displaystyle \frac{5\pi}{6}$.

Com­mon mis­take to avoid#

Giv­ing $\displaystyle \cot^{-1}(-\sqrt{3}) = -\frac{\pi}{6}$. That angle has the right cotan­gent but is out­side $(0, \pi)$.

Ques­tion 2: A sum of two prin­ci­pal val­ues#

The prob­lem#

Find $\displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) + \sin^{-1}(-1)$.

Under­stand­ing the prob­lem#

Find each prin­ci­pal value sep­a­rately, then add them.

The idea#

For $\cos^{-1}$ of a neg­a­tive num­ber, the answer lies in the sec­ond quad­rant: $\cos^{-1}(-a) = \pi - \cos^{-1}a$. For $\sin^{-1}$ of a neg­a­tive num­ber, the answer is a neg­a­tive angle.

Step-by-step solu­tion#

Step 1. $\displaystyle \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, so

$$\displaystyle \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$

Step 2. $\displaystyle \sin\left(-\frac{\pi}{2}\right) = -1$, and $\displaystyle -\frac{\pi}{2}$ is in $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

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

Step 3. Add.

$$\displaystyle \frac{5\pi}{6} - \frac{\pi}{2} = \frac{5\pi}{6} - \frac{3\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$$

Check­ing the answer#

$\displaystyle \cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ ✓ and $\displaystyle \frac{5\pi}{6} \in [0, \pi]$ ✓.

Answer#

$\displaystyle \frac{\pi}{3}$.

Ques­tion 3: A dif­fer­ence of prin­ci­pal val­ues#

The prob­lem#

Find $\tan^{-1}\sqrt{3} - \sec^{-1}(-\sqrt{2})$.

Under­stand­ing the prob­lem#

Again find each prin­ci­pal value and com­bine. The sec­ond one has a neg­a­tive argu­ment, so be care­ful with the range of $\sec^{-1}$.

The idea#

$\sec^{-1}(-\sqrt{2})$ is the angle in $\displaystyle [0, \pi] - \left\{\tfrac{\pi}{2}\right\}$ whose cosine is $\displaystyle -\frac{1}{\sqrt{2}}$; that angle is in the sec­ond quad­rant.

Step-by-step solu­tion#

Step 1. $\displaystyle \tan\frac{\pi}{3} = \sqrt{3}$, so $\displaystyle \tan^{-1}\sqrt{3} = \frac{\pi}{3}$.

Step 2. $\sec\theta = -\sqrt{2}$ means $\displaystyle \cos\theta = -\frac{1}{\sqrt{2}}$. In $[0, \pi]$ that is

$$\displaystyle \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$

Step 3. Sub­tract, using the com­mon denom­i­na­tor $12$.

$$\displaystyle \frac{\pi}{3} - \frac{3\pi}{4} = \frac{4\pi}{12} - \frac{9\pi}{12} = -\frac{5\pi}{12}$$

Check­ing the answer#

$\displaystyle \sec\frac{3\pi}{4} = \frac{1}{\cos(3\pi/4)} = \frac{1}{-1/\sqrt{2}} = -\sqrt{2}$ ✓. A neg­a­tive final answer is fine: the ques­tion asks for a dif­fer­ence, not a prin­ci­pal value.

Answer#

$\displaystyle -\frac{5\pi}{12}$.

Ques­tion 4: Inverse of a func­tion of an angle#

The prob­lem#

Find $\displaystyle \sin^{-1}\left(\sin\frac{3\pi}{4}\right)$, $\displaystyle \cos^{-1}\left(\cos\frac{5\pi}{3}\right)$, $\displaystyle \tan^{-1}\left(\tan\frac{7\pi}{6}\right)$.

Under­stand­ing the prob­lem#

It is tempt­ing to say the inverse "undoes" the func­tion and write $\displaystyle \frac{3\pi}{4}$, $\displaystyle \frac{5\pi}{3}$ and $\displaystyle \frac{7\pi}{6}$. That is only true when the angle is already in the prin­ci­pal range. Here none of them is.

The idea#

First eval­u­ate the inner value (or rewrite the angle), then find the angle in the prin­ci­pal range with the same value, as in the "Watch the range" exam­ples.

Step-by-step solu­tion#

Part (a)

Step 1. $\displaystyle \frac{3\pi}{4}$ is not in $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Eval­u­ate the inner sine.

$$\displaystyle \sin\frac{3\pi}{4} = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$

Step 2. The prin­ci­pal value of $\displaystyle \sin^{-1}\frac{1}{\sqrt{2}}$ is $\displaystyle \frac{\pi}{4}$.

Part (b)

Step 1. $\displaystyle \frac{5\pi}{3}$ is not in $[0, \pi]$. Eval­u­ate the inner cosine.

$$\displaystyle \cos\frac{5\pi}{3} = \cos\left(2\pi - \frac{\pi}{3}\right) = \cos\frac{\pi}{3} = \frac{1}{2}$$

Step 2. $\displaystyle \cos^{-1}\frac{1}{2} = \frac{\pi}{3}$.

Part (c)

Step 1. $\displaystyle \frac{7\pi}{6}$ is not in $\displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Tan­gent has period $\pi$, so sub­tract $\pi$.

$$\displaystyle \tan\frac{7\pi}{6} = \tan\left(\frac{7\pi}{6} - \pi\right) = \tan\frac{\pi}{6}$$

Step 2. $\displaystyle \frac{\pi}{6}$ is in the range, so $\displaystyle \tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \frac{\pi}{6}$.

Check­ing the answer#

Each answer lies in its prin­ci­pal range and has the same func­tion value as the orig­i­nal angle: $\displaystyle \sin\frac{\pi}{4} = \sin\frac{3\pi}{4}$, $\displaystyle \cos\frac{\pi}{3} = \cos\frac{5\pi}{3}$, $\displaystyle \tan\frac{\pi}{6} = \tan\frac{7\pi}{6}$. ✓

Answer#

$\displaystyle \frac{\pi}{4}$; $\displaystyle \frac{\pi}{3}$; $\displaystyle \frac{\pi}{6}$.

Ques­tion 5: Domain, range and graph of $\cos^{-1}x$#

The prob­lem#

State the domain and range of $\cos^{-1}x$ and sketch its graph.

Under­stand­ing the prob­lem#

The domain is the set of inputs $x$ for which $\cos^{-1}x$ is defined; the range is the set of out­puts. The sketch should show the shape and the end points.

The idea#

$\cos^{-1}$ is the inverse of $\cos$ restricted to $[0, \pi]$. Its graph is that piece of the cosine graph reflected in the line $y = x$.

Step-by-step solu­tion#

Step 1. Domain. Cosine only takes val­ues from $-1$ to $1$, so $\cos^{-1}x$ is defined for $x \in [-1, 1]$.

Step 2. Range. By the choice of prin­ci­pal branch, the out­puts are angles in $[0, \pi]$.

Step 3. Key points to plot.

$$\displaystyle \cos^{-1}(-1) = \pi, \qquad \cos^{-1}0 = \frac{\pi}{2}, \qquad \cos^{-1}1 = 0$$

Step 4. Draw axes with $x$ from $-1$ to $1$ and $y$ from $0$ to $\pi$. Plot $(-1, \pi)$, $\displaystyle \left(0, \frac{\pi}{2}\right)$ and $(1, 0)$, and join them with a smooth curve that falls steadily from left to right. The curve is steep­est at the two ends (nearly ver­ti­cal) and passes through $\displaystyle \left(0, \frac{\pi}{2}\right)$.

Check­ing the answer#

Since $\cos$ is decreas­ing on $[0, \pi]$, its inverse must also be decreas­ing, which matches the falling curve. ✓

Answer#

Domain $[-1, 1]$, range $[0, \pi]$; the graph is a decreas­ing curve from $(-1, \pi)$ through $\displaystyle \left(0, \tfrac{\pi}{2}\right)$ to $(1, 0)$.

Ques­tion 6: Domain of $\sin^{-1}(2x - 3)$#

The prob­lem#

Find the domain of $\sin^{-1}(2x - 3)$.

Under­stand­ing the prob­lem#

$\sin^{-1}u$ is defined only when $-1 \le u \le 1$. Here $u = 2x - 3$, so you need all $x$ that keep $2x - 3$ in that inter­val.

The idea#

Solve the dou­ble inequal­ity $-1 \le 2x - 3 \le 1$.

Step-by-step solu­tion#

Step 1. Write the con­di­tion.

$$-1 \le 2x - 3 \le 1$$

Step 2. Add $3$ to all three parts.

$$2 \le 2x \le 4$$

Step 3. Divide by $2$.

$$1 \le x \le 2$$

Check­ing the answer#

$x = 1$: $2x - 3 = -1$ ✓; $x = 2$: $2x - 3 = 1$ ✓; $x = 3$: $2x - 3 = 3$, out­side, so unde­fined ✓.

Answer#

The domain is $[1, 2]$.

Ques­tion 7: When is $\sin^{-1}(\sin x) = x$?#

The prob­lem#

Is $\sin^{-1}(\sin x) = x$ for all real $x$? For which $x$?

Under­stand­ing the prob­lem#

$\sin x$ is defined for every real $x$, so $\sin^{-1}(\sin x)$ always makes sense. The ques­tion is whether it gives back $x$.

The idea#

$\sin^{-1}$ always returns a value in $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. So it can return $x$ only if $x$ itself is in that inter­val.

Step-by-step solu­tion#

Step 1. Counter-exam­ple: $\displaystyle x = \frac{5\pi}{6}$. Then $\displaystyle \sin x = \frac{1}{2}$ and $\displaystyle \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \ne \frac{5\pi}{6}$. So the answer to "for all $x$?" is no.

Step 2. If $\displaystyle x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then $x$ is the unique angle in the prin­ci­pal range with sine equal to $\sin x$, so $\sin^{-1}(\sin x) = x$.

Step 3. If $x$ is out­side that inter­val, the out­put (which is in the inter­val) can­not equal $x$.

Check­ing the answer#

This agrees with the lesson's exam­ple $\displaystyle \sin^{-1}\left(\sin\tfrac{5\pi}{6}\right) = \tfrac{\pi}{6}$.

Answer#

No. $\sin^{-1}(\sin x) = x$ only for $\displaystyle x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Ques­tion 8: Sine of an expres­sion with an inverse#

The prob­lem#

Eval­u­ate $\displaystyle \sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right)$.

Under­stand­ing the prob­lem#

Work from the inside: first find the prin­ci­pal value of $\displaystyle \sin^{-1}\left(-\frac{1}{2}\right)$, then sim­plify the bracket, then take the sine.

The idea#

$\sin^{-1}$ of a neg­a­tive num­ber is a neg­a­tive angle in $\displaystyle \left[-\frac{\pi}{2}, 0\right)$.

Step-by-step solu­tion#

Step 1. $\displaystyle \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$, so $\displaystyle \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$.

Step 2. Sim­plify the bracket. Sub­tract­ing a neg­a­tive angle adds it.

$$\displaystyle \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) = \frac{2\pi}{6} + \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$$

Step 3. Take the sine.

$$\displaystyle \sin\frac{\pi}{2} = 1$$

Check­ing the answer#

The largest pos­si­ble value of sine is $1$, reached at $\displaystyle \frac{\pi}{2}$, which matches. ✓

Answer#

$1$.