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Question Number 147811 by mathdanisur last updated on 23/Jul/21

Answered by Olaf_Thorendsen last updated on 23/Jul/21

$$\mathrm{X}=\sin \left(\frac{\arccos x}{4}\right)$$$$\Rightarrow {\mathrm{X}}^2={\sin }^2\left(\frac{\arccos x}{4}\right)$$$${\mathrm{X}}^2=\frac{1}{2}(1-\cos \left(2\times \frac{\arccos x}{4}\right))$$$${\mathrm{X}}^2=\frac{1}{2}(1-\cos \left(\frac{\arccos x}{2}\right))$$$$0⩽\arccos x⩽\pi \Rightarrow 0⩽\frac{\arccos x}{2}⩽\frac{\pi }{2}$$$${\mathrm{X}}^2=\frac{1}{2}(1-\sqrt{{\cos }^2\left(\frac{\arccos x}{2}\right)})$$$${\mathrm{X}}^2=\frac{1}{2}(1-\sqrt{\frac{1}{2}(1+\cos \left(2\times \frac{\arccos x}{2}\right)})$$$${\mathrm{X}}^2=\frac{1}{2}(1-\sqrt{\frac{1}{2}(1+x)})$$$$\mathrm{X}=+\sqrt{\frac{1}{2}(1-\sqrt{\frac{x+1}{2}})}$$

Commented by mathdanisur last updated on 23/Jul/21

$$thankyouSir$$

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