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Question Number 32538 by GAUTHAM last updated on 27/Mar/18

Commented by prof Abdo imad last updated on 28/Mar/18

$$wehave{sin}^2\left(\frac{\pi }{12}\right)=\frac{1-cos\left(\frac{\pi }{6}\right)}{2}=\frac{1-\frac{\sqrt{3}}{2}}{2}$$$$=\frac{2-\sqrt{3}}{4}\Rightarrow sin\left(\frac{\pi }{12}\right)=\frac{\sqrt{2-\sqrt{3}}}{2}\Rightarrow$$$$arcsin\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right)=\frac{\pi }{12}.$$

Answered by rahul 19 last updated on 27/Mar/18

$${\sin }^{-1}\sqrt{\frac{2-\sqrt{3}}{4}}$$$$\Rightarrow {\sin }^{-1}\frac{\sqrt{3}-1}{2\sqrt{2}}$$$$\Rightarrow 15°$$

Commented by MJS last updated on 27/Mar/18

$$\mathrm{to}\mathrm{me}\mathrm{it}\mathrm{looks}\mathrm{like}{\sin }^{-1}\left(\frac{\sqrt{2-\sqrt{3}}}{4}\right)$$$$\mathrm{and}\mathrm{then}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{hard}...$$

Commented by rahul 19 last updated on 27/Mar/18

$$well,yesthenit{can}^{\prime }tbesolveeasily.$$$$soithoughtthereiswholeroot...$$

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