Chapter 2: Inverse Trigonometric Functions

Find the inverse of \(f(x)=2x+3\).

1. Find the inverse of the bijection \(f(x)=2x+3\) from \(\mathbb{R}\) to \(\mathbb{R}\).
2. Write the output as \(y\) and solve for the input.
   $$y=2x+3\quad\Longrightarrow\quad x=\frac{y-3}{2}$$
3. Swap the variable names to express the inverse.
   $$f^{-1}(x)=\frac{x-3}{2}$$
4. Check by composition.
   $$f^{-1}(f(x))=\frac{2x+3-3}{2}=x$$

Evaluate \(\sin^{-1}\left(-\frac{1}{2}\right)\).

1. Evaluate \(\sin^{-1}\left(-\frac{1}{2}\right)\) using the principal range.
2. Find the angle in \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) whose sine is \(-\frac{1}{2}\).
   $$\sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2}$$
3. The angle lies in the required principal interval.
   $$-\frac{\pi}{6}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$

Evaluate \(\cos^{-1}\left(-\frac{1}{2}\right)\).

1. Evaluate \(\cos^{-1}\left(-\frac{1}{2}\right)\) in the principal range.
2. Find the angle in \([0,\pi]\) whose cosine is \(-\frac{1}{2}\).
   $$\cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2}$$
3. The angle is in the principal interval.
   $$\frac{2\pi}{3}\in[0,\pi]$$

Evaluate \(\cosec\left(\frac{\pi}{6}\right)\).

1. Find the cosecant value at \(x=\frac{\pi}{6}\), which lies in the restricted interval.
2. Use the reciprocal relationship.
   $$\cosec\left(\frac{\pi}{6}\right)=\frac{1}{\sin\left(\frac{\pi}{6}\right)}=\frac{1}{\frac{1}{2}}=2$$
3. This confirms that the restricted branch reaches a value outside \((-1,1)\).
   $$2\in\mathbb{R}\setminus(-1,1)$$

Find the principal value of \(\cosec^{-1}(2)\).

1. Find the principal value of \(\cosec^{-1}(2)\).
2. Let \(\cosec^{-1}(2)=y\); then use the reciprocal relation.
   $$\sin y=\frac{1}{2}$$
3. The angle in the principal range with this sine is \(\frac{\pi}{6}\).
   $$y=\frac{\pi}{6}$$

Find the principal value of \(\sec^{-1}(-2)\).

1. Find the principal value of \(\sec^{-1}(-2)\).
2. Let \(\sec^{-1}(-2)=y\); the corresponding cosine is:
   $$\cos y=-\frac{1}{2}$$
3. The angle in the principal range is \(\frac{2\pi}{3}\).
   $$y=\frac{2\pi}{3}$$

Find the principal value of \(\tan^{-1}(1)\).

1. Find the principal value of \(\tan^{-1}(1)\).
2. Solve the equivalent equation in the principal interval.
   $$\tan y=1,\quad y\in(-\frac{\pi}{2},\frac{\pi}{2})$$
3. The required angle is:
   $$y=\frac{\pi}{4}$$

Find the principal value of \(\cot^{-1}(-\frac{1}{\sqrt{3}})\).

1. Find the principal value of \(\cot^{-1}(-\frac{1}{\sqrt{3}})\).
2. Use the cotangent reference angle and select the angle in \((0,\pi)\).
   $$\cot(\pi-\frac{\pi}{3})=-\cot\frac{\pi}{3}=-\frac{1}{\sqrt{3}}$$
3. Therefore, the principal angle is:
   $$y=\frac{2\pi}{3}$$

Find the principal value of \(\sin^{-1}(\frac{1}{\sqrt{2}})\).

1. Find the principal value of \(\sin^{-1}(\frac{1}{\sqrt{2}})\).
2. The principal angle must lie in \([-\frac{\pi}{2},\frac{\pi}{2}]\) and have sine \(\frac{1}{\sqrt{2}}\).
   $$\sin y=\frac{1}{\sqrt{2}}$$
3. The required angle is:
   $$y=\frac{\pi}{4}$$

Show that \(\sin^{-1}(2x\sqrt{1-x^2})=2\sin^{-1}x\) for \(-1/\sqrt{2}\le x\le1/\sqrt{2}\), and equals \(2\cos^{-1}x\) for \(1/\sqrt{2}\le x\le1\).

1. For the first identity, set \(\theta=\sin^{-1}x\); the given range ensures \(2\theta\) is in the principal range of \(\sin^{-1}\).
   $$\begin{aligned}-\frac{1}{\sqrt{2}}\le x\le\frac{1}{\sqrt{2}}&\Longrightarrow -\frac{\pi}{4}\le\theta\le\frac{\pi}{4}\\\sin^{-1}\left(2x\sqrt{1-x^2}\right)&=\sin^{-1}(2\sin\theta\cos\theta)\\&=\sin^{-1}(\sin 2\theta)=2\theta=2\sin^{-1}x\end{aligned}$$
2. For the second identity, set \(\theta=\cos^{-1}x\); then \(2\theta\) is again in the principal range of \(\sin^{-1}\).
   $$\begin{aligned}\frac{1}{\sqrt{2}}\le x\le1&\Longrightarrow 0\le\theta\le\frac{\pi}{4}\\\sin^{-1}\left(2x\sqrt{1-x^2}\right)&=\sin^{-1}(2\cos\theta\sin\theta)\\&=\sin^{-1}(\sin 2\theta)=2\theta=2\cos^{-1}x\end{aligned}$$

Express \(\tan^{-1}\left(\dfrac{\cos x}{1-\sin x}\right)\) in simplest form for \(-\dfrac{3\pi}{2}<x<\dfrac{\pi}{2}\).

1. Rewrite the quotient using half-angle expressions, factor, and cancel.
   $$\frac{\cos x}{1-\sin x}=\frac{\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2}=\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}$$
2. The quotient is \(\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\); the angle lies in the principal range of \(\tan^{-1}\) for the given interval.
   $$\tan^{-1}\left(\frac{\cos x}{1-\sin x}\right)=\tan^{-1}\left(\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)=\frac{\pi}{4}+\frac{x}{2}$$

Express \(\cot^{-1}\left(\dfrac{1}{\sqrt{x^2-1}}\right)\) in simplest form for \(x>1\).

1. Let \(x=\sec\theta\) with \(0<\theta<\dfrac{\pi}{2}\); then the radical becomes a tangent.
   $$\sqrt{x^2-1}=\sqrt{\sec^2\theta-1}=\tan\theta$$
2. Use the reciprocal and the principal range of \(\cot^{-1}\).
   $$\cot^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=\cot^{-1}(\cot\theta)=\theta=\sec^{-1}x$$

Find the value of \(\sin^{-1}(\sin\frac{3\pi}{5})\).

1. \(\frac{3\pi}{5}\) is not in the principal-value range \([-\frac{\pi}{2},\frac{\pi}{2}]\).
   $$\frac{3\pi}{5}\notin\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$
2. Use sine symmetry to obtain an angle in the principal range with the same sine.
   $$\sin\frac{3\pi}{5}=\sin\left(\pi-\frac{3\pi}{5}\right)=\sin\frac{2\pi}{5}$$
3. Since \(\frac{2\pi}{5}\) is in the principal range, the inverse sine returns that angle.
   $$\frac{2\pi}{5}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$