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Question Number 14366 by RasheedSindhi last updated on 31/May/17

$$\mathrm{Determine}$$$$\mathrm{a})\cos \left({\sin }^{-1}\mathrm{x}\right)$$$$\mathrm{b})\sin \left({\cos }^{-1}\mathrm{x}\right)$$$$\mathrm{c}){\cos }^{-1}\left(\sin \mathrm{x}\right)$$$$\mathrm{d}){\sin }^{-1}\left(\cos \mathrm{x}\right)$$$$\mathrm{e}){\sin }^{-1}\left(\sin \mathrm{x}\right)$$$$\mathrm{f})\sin \left({\sin }^{-1}\mathrm{x}\right)$$$$\mathrm{g}){\sin }^{-1}\left({\sin }^{-1}\mathrm{x}\right)$$$$\mathrm{h})\sin \left(\sin \mathrm{x}\right)$$

Commented by Tinkutara last updated on 31/May/17

$$\sin \left({\sin }^{-1}x\right)=x(\mathrm{Principal}\mathrm{value}\mathrm{in}$$$$\mathrm{the}\mathrm{interval}[\frac{-\pi }{2},\frac{\pi }{2}])$$

Commented by RasheedSindhi last updated on 01/Jun/17

$$\mathrm{Thank}\mathrm{you}.$$

Answered by mrW1 last updated on 31/May/17

$$a)$$$$t={\sin }^{-1}x$$$$\sin t=x$$$$\cos \left({\sin }^{-1}x\right)=\cos t=\pm \sqrt{1-x^2}$$$$b)$$$$\sin \left({\cos }^{-1}x\right)=\pm \sqrt{1-\cos {\left({\cos }^{-1}x\right)}^2}$$$$=\pm \sqrt{1-x^2}$$$$c)$$$$t=\sin x$$$${\sin }^{-1}t=x$$$${\cos }^{-1}\left(\sin x\right)={\cos }^{-1}t=\frac{\pi }{2}-{\sin }^{-1}t=\frac{\pi }{2}-x$$$$d)$$$${\sin }^{-1}\left(\cos x\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\cos x\right)=\frac{\pi }{2}-x$$$$e)$$$${\sin }^{-1}\left(\sin x\right)=x$$$$f)$$$$\sin \left({\sin }^{-1}x\right)=x$$$$g)$$$${\sin }^{-1}\left({\sin }^{-1}x\right)isdefinedwhen$$$$-1⩽{\sin }^{-1}x⩽1$$$$\Rightarrow -\sin \left(1\right)⩽x⩽\sin \left(1\right)\approx 0.84$$$${\sin }^{-1}\left({\sin }^{-1}x\right)\approx {\sin }^{-1}\left(\frac{x}{\sin \left(1\right)}\right)\approx {\sin }^{-1}\left(\frac{x}{0.84}\right)$$$$h)$$$$sin\left(\sin x\right)isdefinedforx\in R$$$$\sin \left(\sin x\right)\approx \sin \left(1\right)\sin x\approx 0.84\sin x$$

Commented by RasheedSindhi last updated on 01/Jun/17

$$\mathcal{T}\mathrm{h}a\mathrm{n}k\mathrm{s}\mathrm{Sir}!$$

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