Question

(f) (2 points) \\( \tan ^ { - 1 } ( 1 ) \\) (g) (3 points) \\( \sin ^ { - 1 } \left( \sin \left( \frac { 5 \pi } { 4 } \
ight) \
ight) \\)

Explanation:

Step1: Recall the range of $\tan^{-1}x$

The range of \(y = \tan^{-1}x\) is \((-\frac{\pi}{2},\frac{\pi}{2})\). We know that \(\tan\theta=1\) when \(\theta=\frac{\pi}{4}+k\pi,k\in\mathbb{Z}\). In the range \((-\frac{\pi}{2},\frac{\pi}{2})\), when \(k = 0\), \(\theta=\frac{\pi}{4}\). So, \(\tan^{-1}(1)=\frac{\pi}{4}\).

Step2: Recall the range of \(\sin^{-1}x\)

The range of \(y=\sin^{-1}x\) is \([-\frac{\pi}{2},\frac{\pi}{2}]\). We know that \(\sin(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}\). Let \(y = \sin^{-1}(\sin(\frac{5\pi}{4}))\). We can rewrite \(\sin(\frac{5\pi}{4})=\sin(-\frac{\pi}{4}+\pi)=-\sin(\frac{\pi}{4})\) (using the formula \(\sin(A + B)=\sin A\cos B+\cos A\sin B\) with \(A=\pi\) and \(B =-\frac{\pi}{4}\), \(\sin(\pi-\alpha)=\sin\alpha\) is not applicable here. Another way: \(\sin^{-1}(\sin\theta)=\theta\) when \(\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]\). \(\sin(\frac{5\pi}{4})=\sin(-\frac{\pi}{4}+ 2\pi-\pi)=\sin(-\frac{\pi}{4})\) (using \(\sin(x + 2k\pi)=\sin x\) and \(\sin(\pi + x)=-\sin x\)). Since \(-\frac{\pi}{4}\in[-\frac{\pi}{2},\frac{\pi}{2}]\), then \(\sin^{-1}(\sin(\frac{5\pi}{4}))=-\frac{\pi}{4}\)

Answer:

\(\tan^{-1}(1)=\frac{\pi}{4}\), \(\sin^{-1}(\sin(\frac{5\pi}{4}))=-\frac{\pi}{4}\)