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Question Number 157644 by Ar Brandon last updated on 26/Oct/21

$$\mathrm{Prove}\frac{1}{2}\sqrt{2-\sqrt{3}}=\frac{\sqrt{6}-\sqrt{2}}{4}$$

Answered by ajfour last updated on 26/Oct/21

$$v=\frac{1}{2}\sqrt{2-\sqrt{3}}$$$$v^2=\frac{1}{2}-\frac{\sqrt{3}}{4}=\frac{2-\sqrt{3}}{4}$$$$=\frac{8-\sqrt{48}}{16}={\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)}^2$$$$v=\frac{\sqrt{6}-\sqrt{2}}{4}$$$$$$

Commented by Ar Brandon last updated on 26/Oct/21

$$\mathrm{thanks}$$

Answered by mindispower last updated on 26/Oct/21

$$\frac{1}{2}\sqrt{2-\sqrt{3}}=\sqrt{\frac{1-\left(\frac{\sqrt{3}}{2}\right)}{2}}=\sqrt{\frac{1+cos\left(\frac{5\pi }{6}\right)}{2}}$$$$cos\left(2x\right)=2{cos}^2\left(x\right)-1used$$$$=\mid cos\left(\frac{5\pi }{12}\right)\mid =cos(\frac{\pi }{6}+\frac{\pi }{4})=\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{1}{2}.\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$$

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