What traumatic event does Zitkala-Sa describe at the school?

The forcible cutting of her long hair

Why was the cutting of her hair so humiliating to Zitkala-Sa?

Among her people, shingled hair was worn by cowards and mourners

State the principal value range of \(\sin^{-1}x\).

The principal range is \([-\tfrac{\pi}{2},\tfrac{\pi}{2}]\), with domain \([-1,1]\).

State the principal value range of \(\cos^{-1}x\).

The principal range is \([0,\pi]\), with domain \([-1,1]\).

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

Since \(\sin\tfrac{\pi}{6}=\tfrac{1}{2}\) and \(\tfrac{\pi}{6}\) lies in the principal range, \(\sin^{-1}\!\left(\tfrac{1}{2}\right)=\tfrac{\pi}{6}\).

Prove \(\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}\).

Let \(\sin^{-1}x=\theta\); then \(x=\sin\theta=\cos(\tfrac{\pi}{2}-\theta)\), so \(\cos^{-1}x=\tfrac{\pi}{2}-\theta\), giving the sum \(\tfrac{\pi}{2}\).

Evaluate \(\tan^{-1}(1)\).

\(\tan\tfrac{\pi}{4}=1\) and \(\tfrac{\pi}{4}\) is in the principal range, so \(\tan^{-1}(1)=\tfrac{\pi}{4}\).

Simplify \(\cos^{-1}(\cos\tfrac{7\pi}{6})\).

Since \(\tfrac{7\pi}{6}\notin[0,\pi]\), use \(\cos\tfrac{7\pi}{6}=\cos(2\pi-\tfrac{7\pi}{6})=\cos\tfrac{5\pi}{6}\); thus the value is \(\tfrac{5\pi}{6}\).

State the value of \(\tan^{-1}x+\cot^{-1}x\).

For all real \(x\), \(\tan^{-1}x+\cot^{-1}x=\tfrac{\pi}{2}\).

Write the addition formula for \(\tan^{-1}\).

\(\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}\) provided \(xy<1\).

Evaluate \(\sin^{-1}(\sin\tfrac{2\pi}{3})\).

As \(\tfrac{2\pi}{3}\notin[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\), use \(\sin\tfrac{2\pi}{3}=\sin(\pi-\tfrac{2\pi}{3})=\sin\tfrac{\pi}{3}\); hence the value is \(\tfrac{\pi}{3}\).

Why are trig functions restricted before inverting?

They are periodic and not one-one over their full domains; restricting to a principal branch makes them bijective so a single-valued inverse exists.

Inverse Trigonometric Functions

Q: Evaluate \(\sin^{-1}(\sin\tfrac{2\pi}{3})\).

As \(\tfrac{2\pi}{3}\notin[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\), use \(\sin\tfrac{2\pi}{3}=\sin(\pi-\tfrac{2\pi}{3})=\sin\tfrac{\pi}{3}\); hence the value is \(\tfrac{\pi}{3}\).

Q: Why are trig functions restricted before inverting?

They are periodic and not one-one over their full domains; restricting to a principal branch makes them bijective so a single-valued inverse exists.

Class 12 Mathematics · Mathematics Part I · Chapter 2

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CBSE Class 12 · Mathematics

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Summary

Trigonometric functions are periodic and therefore not one-one over their full domains. To define inverses, each is restricted to a principal-value branch on which it is bijective. The chapter defines \(\sin^{-1}x,\cos^{-1}x,\tan^{-1}x\) and their reciprocals, listing the principal ranges: \(\sin^{-1}x\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\), \(\cos^{-1}x\in[0,\pi]\) and \(\tan^{-1}x\in(-\tfrac{\pi}{2},\tfrac{\pi}{2})\). Several identities follow, such as \(\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}\) and \(\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}\) (for \(xy<1\)). These functions and identities are essential tools for simplifying expressions, solving equations and integrating, where inverse trig functions appear as antiderivatives.

Principal value branchesInverse sine, cosine and tangentProperties and identitiesSimplification of inverse trig expressions

Key terms

Principal value branch

The restricted interval on which a trig function is bijective so its inverse is defined.

\(\sin^{-1}x\)

Inverse sine, with range \([-\tfrac{\pi}{2},\tfrac{\pi}{2}]\) and domain \([-1,1]\).

\(\cos^{-1}x\)

Inverse cosine, with range \([0,\pi]\) and domain \([-1,1]\).

\(\tan^{-1}x\)

Inverse tangent, with range \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\) and domain \(\mathbb{R}\).

Complementary identity

\(\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}\).

Addition formula

\(\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}\) when \(xy<1\).

Extra questions & answers

Try one from this chapter

The principal value range of \(\sin^{-1}x\) is:

Flashcards & interactive practice