Finding value of $\sin (15\, ^{\circ})$ with half angle identity

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The answer I got when trying to solve it was $\sqrt{\frac{1 - \sqrt3}{2} }$ but the book says it's $\sqrt{ \frac{2 - \sqrt3}{2}}$ and I don't know how the two on the top half gets there.

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2 Answers

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$$\sin15^\circ=\sqrt{\frac{1-\cos30^\circ}2}=\sqrt{\frac{1-\sqrt3/2}2}=\sqrt{\frac{2-\sqrt3}4}=\frac{\sqrt{2-\sqrt3}}2$$

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By the half angle formulas $$\sin\dfrac{\pi}{6}=2\sin\dfrac{\pi}{12}\cos\dfrac{\pi}{12}.$$ Therefore if you substitute $x=\sin\dfrac{\pi}{12},$ you will get the equation $$\dfrac{1}{2}=2x\sqrt{1-x^2}.$$ One way to solve this equation is squaring both sides. But, if you square an equation number of roots will be double. Therefore you should check the roots by backward substitution. $$1=16x^2(1-x^2)$$ $$3=(4x^2-2)^2$$ Since $$0\lt\sin\dfrac{\pi}{12}=x\lt\frac12$$ we have $4x^2-1\lt0.$ Therefore choose $$4x^2-2=-\sqrt3$$ $$x^2=\dfrac{2-\sqrt3}{4}=\dfrac{(\sqrt3-1)^2}{8}$$ Therefore choose $$x=\dfrac{\sqrt3-1}{2\sqrt2}.$$ Hence $$\sin\dfrac{\pi}{12}=\dfrac{\sqrt6-\sqrt2}{4}$$

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