Question Statement

Find the principal values of each of the following without using a calculator:

(i) ${\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$ (ii) ${\sin }^{-1}(1)$ (iii) ${\tan }^{-1}(\sqrt{3})$ (iv) ${\cot }^{-1}\left(\frac{\sqrt{3}}{3}\right)$ (v) ${\sec }^{-1}\left(\frac{2\sqrt{3}}{3}\right)$ (vi) ${\mathrm{cosec}}^{-1}(-\sqrt{2})$ (vii) ${\cos }^{-1}\left(\frac{-1}{2}\right)$ (viii) ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ (ix) ${\mathrm{cosec}}^{-1}(-2)$

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Background and Explanation

To find the principal value of an inverse trigonometric function, we look for the angle $\theta$ within the function's restricted range (the principal value branch) that satisfies the trigonometric equation. For example, the range for ${\sin }^{-1}x$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$, while for ${\cos }^{-1}x$ it is $[0,\pi ]$.

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Solution

Part (i)

Let $\theta ={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$. By definition of inverse functions: $\begin{array}{rl}\cos \theta & =\frac{\sqrt{3}}{2}\\ \cos \theta & =\cos \left(\frac{\pi }{6}\right)\\ \Rightarrow \theta & =\frac{\pi }{6}\end{array}$ Since $\frac{\pi }{6}$ is in the range $[0,\pi ]$, the principal value is: Hence, ${\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{6}$

Part (ii)

Let $\theta ={\sin }^{-1}(1)$. $\begin{array}{rl}\theta & ={\sin }^{-1}(1)\\ \sin \theta & =1\end{array}$ We know that $\sin \left(\frac{\pi }{2}\right)=1$. Hence, ${\sin }^{-1}(1)=\frac{\pi }{2}$

Part (iii)

Let $\theta ={\tan }^{-1}\sqrt{3}$. $\begin{array}{rl}\tan \theta & =\sqrt{3}\\ \Rightarrow \tan \theta & =\tan \left(\frac{\pi }{3}\right)\end{array}$ So, $\theta =\frac{\pi }{3}$ Hence, ${\tan }^{-1}(\sqrt{3})=\frac{\pi }{3}$

Part (iv)

Let $\theta ={\cot }^{-1}\left(\frac{\sqrt{3}}{3}\right)$. Note that $\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$. $\begin{array}{rl}\theta & ={\cot }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ \cot \theta & =\frac{1}{\sqrt{3}}\\ \frac{1}{\tan \theta }& =\frac{1}{\sqrt{3}}\\ \tan \theta & =\sqrt{3}\\ \Rightarrow \tan \theta & =\tan \left(\frac{\pi }{3}\right)\end{array}$ So, $\theta =\frac{\pi }{3}$ Hence, ${\cot }^{-1}\left(\frac{\sqrt{3}}{3}\right)=\frac{\pi }{3}$

Part (v)

Let $\theta ={\sec }^{-1}\left(\frac{2\sqrt{3}}{3}\right)$. Note that $\frac{2\sqrt{3}}{3}=\frac{2}{\sqrt{3}}$. $\begin{array}{rl}\theta & ={\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)\\ \sec \theta & =\frac{2}{\sqrt{3}}\\ \frac{1}{\cos \theta }& =\frac{2}{\sqrt{3}}\\ \cos \theta & =\frac{\sqrt{3}}{2}\\ \Rightarrow \cos \theta & =\cos \frac{\pi }{6}\\ \Rightarrow \theta & =\frac{\pi }{6}\end{array}$ Hence, ${\sec }^{-1}\left(\frac{2\sqrt{3}}{3}\right)=\frac{\pi }{6}$

Part (vi)

Let $\theta ={\mathrm{cosec}}^{-1}(-\sqrt{2})$. $\begin{array}{rl}\mathrm{cosec}\theta & =-\sqrt{2}\\ \frac{1}{\sin \theta }& =\frac{-\sqrt{2}}{1}\\ \sin \theta & =\frac{-1}{\sqrt{2}}\end{array}$ Using the principal value range for sine $[-\frac{\pi }{2},\frac{\pi }{2}]$: $\sin \theta =\sin \left(\frac{-\pi }{4}\right)$ So, $\theta =\frac{-\pi }{4}$ Hence, ${\mathrm{cosec}}^{-1}(-\sqrt{2})=-\frac{\pi }{4}$

Part (vii)

Let $\theta ={\cos }^{-1}\left(\frac{-1}{2}\right)$. $\cos \theta =\frac{-1}{2}$ Since the range of ${\cos }^{-1}$ is $[0,\pi ]$, we find the angle in the second quadrant: $\begin{array}{rl}\theta & =\pi -\frac{\pi }{3}\\ & =\frac{3\pi -\pi }{3}\\ \theta & =\frac{2\pi }{3}\end{array}$ Hence, ${\cos }^{-1}\left(\frac{-1}{2}\right)=\frac{2\pi }{3}$

Part (viii)

Let $\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$. $\begin{array}{rl}\tan \theta & =\frac{1}{\sqrt{3}}\\ \tan \theta & =\tan \frac{\pi }{6}\end{array}$ $\Rightarrow \theta =\frac{\pi }{6}$ Hence, ${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$

Part (ix)

Let $\theta ={\mathrm{cosec}}^{-1}(-2)$. $\begin{array}{rl}\mathrm{cosec}\theta & =-2\\ \frac{1}{\sin \theta }& =\frac{-2}{1}\\ \sin \theta & =\frac{-1}{2}\end{array}$ Using the principal value range: $\begin{array}{rl}\sin \theta & =\sin \left(\frac{-\pi }{6}\right)\\ \theta & =\frac{-\pi }{6}\end{array}$ Hence, ${\mathrm{cosec}}^{-1}(-2)=-\frac{\pi }{6}$

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Key Formulas or Methods Used

• Principal Value Ranges:
  • ${\sin }^{-1}x\in [-\frac{\pi }{2},\frac{\pi }{2}]$
  • ${\cos }^{-1}x\in [0,\pi ]$
  • ${\tan }^{-1}x\in (-\frac{\pi }{2},\frac{\pi }{2})$
• Reciprocal Identities:
  • $\sec \theta =\frac{1}{\cos \theta }$
  • $\mathrm{cosec}\theta =\frac{1}{\sin \theta }$
  • $\cot \theta =\frac{1}{\tan \theta }$
• Rationalization: $\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$ and $\frac{2\sqrt{3}}{3}=\frac{2}{\sqrt{3}}$.

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Summary of Steps

1. Equate the inverse trigonometric expression to a variable $\theta$.
2. Rewrite the expression in terms of the standard trigonometric function (e.g., $\cos \theta =x$).
3. If necessary, use reciprocal identities to convert $\sec ,\mathrm{cosec},$ or $\cot$ into $\cos ,\sin ,$ or $\tan$.
4. Identify the reference angle from known exact values (e.g., $\frac{\pi }{6},\frac{\pi }{4},\frac{\pi }{3}$).
5. Adjust the angle to fit within the specific principal value branch of the inverse function.