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Question Number 131296 by EDWIN88 last updated on 03/Feb/21

$$Ifa{\sin }^{-1}\left(x\right)-b{\cos }^{-1}\left(x\right)=c$$$$thenthevalueofa{\sin }^{-1}\left(x\right)+b{\cos }^{-1}\left(x\right)$$$$\left(wheneverexists\right)isequalto?$$

Answered by liberty last updated on 03/Feb/21

$$\mathrm{We}\mathrm{have}{\sin }^{-1}\left(\mathrm{x}\right)+{\cos }^{-1}\left(\mathrm{x}\right)=\frac{\pi }{2}$$$$\mathrm{then}{\mathrm{bsin}}^{-1}\left(\mathrm{x}\right)+{\mathrm{bcos}}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{b}\pi }{2}...\left(\mathrm{i}\right)$$$${\mathrm{asin}}^{-1}\left(\mathrm{x}\right)-{\mathrm{bcos}}^{-1}\left(\mathrm{x}\right)=\mathrm{c}...\left(\mathrm{ii}\right)$$$$\therefore \mathrm{on}\mathrm{adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get}$$$${\sin }^{-1}\left(\mathrm{x}\right)=\frac{\frac{\mathrm{b}\pi }{2}+\mathrm{c}}{\mathrm{a}+\mathrm{b}}\mathrm{and}{\cos }^{-1}\left(\mathrm{x}\right)=\frac{\frac{\mathrm{a}\pi }{2}-\mathrm{c}}{\mathrm{a}+\mathrm{b}}$$$$\mathrm{Therefore}{\mathrm{asin}}^{-1}\left(\mathrm{x}\right)+{\mathrm{bcos}}^{-1}\left(\mathrm{x}\right)=\frac{\pi \mathrm{ab}+\mathrm{c}(\mathrm{a}-\mathrm{b})}{\mathrm{a}+\mathrm{b}}$$

Commented by EDWIN88 last updated on 03/Feb/21

$$$$$$\mathrm{sehr}\mathrm{interessante}\mathrm{Erklrung}\mathrm{Sir}$$

Answered by mr W last updated on 03/Feb/21

$$t={\sin }^{-1}x$$$$\frac{\pi }{2}-t={\cos }^{-1}x$$$$a{\sin }^{-1}\left(x\right)-b{\cos }^{-1}\left(x\right)=c$$$$\Rightarrow at-b(\frac{\pi }{2}-t)=c$$$$\Rightarrow (a+b)t=c+\frac{b\pi }{2}$$$$\Rightarrow t=\frac{1}{a+b}(c+\frac{b\pi }{2})$$$$a{\sin }^{-1}\left(x\right)+b{\cos }^{-1}\left(x\right)$$$$=at+b(\frac{\pi }{2}-t)$$$$=(a-b)t+\frac{b\pi }{2}$$$$=\frac{a-b}{a+b}(c+\frac{b\pi }{2})+\frac{b\pi }{2}$$$$=\frac{(a-b)c+ab\pi }{a+b}$$

Commented by EDWIN88 last updated on 03/Feb/21

$$nice$$

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