Calculate $\sin\frac{\pi}{16}$ from given trigonometric identities
There is the choice between four trigonometric identities$$\sin(4\phi)=8\cos^{3}(\phi)\sin(\phi)-4\cos(\phi)\sin(\phi)$$$$\cos(4\phi)=8\cos^{4}(\phi) -8\cos^{2}(\phi)+1$$$$\sin^{4}(\phi)=1/8(\cos(4\phi) -4\cos(2\phi)+3)$$$$\cos^{4}(\phi)=1/8(\cos(4\phi)+4\cos(2\phi)+3)$$to calculate $\sin\frac{\pi}{16}$. When using the first I got stuck trying to eliminate cos expressions and bring $\sin(4\phi)$ and $\sin(\phi)$ on different sides of the equation, as in$$\sin(4\phi)=\sin (2\phi)(2-4\sin^{2}(\phi))$$What way to go ?
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$\begingroup$Hint: Note that $\cos(\pi/4) = \dfrac{1}{\sqrt{2}}$, Use the second one of your list and put $y = \cos^2(\pi/16)$ to get a quadratic equation in $y$, and then use the identity $\sin(\pi/16) = \sqrt{1-\cos^2(\pi/16)}= \sqrt{1-y}$ to finish.
$\endgroup$ $\begingroup$Use the second identity:
$$ 8\cos^4\left(\frac{\pi}{16}\right) - 8\cos^2\left(\frac{\pi}{16}\right) + 1 = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt2}{2} $$
Let $x = \cos^2\left(\frac{\pi}{16}\right)$ then
$$ 8x^2-8x = \frac{\sqrt{2}-2}{2} $$
Completing the square:$$ 4x^2 - 4x + 1 = \frac{\sqrt{2}-2}{4}+1 = \frac{2+\sqrt{2}}{4} $$
$$ (2x-1)^2 = \frac{2+\sqrt{2}}{4} $$
Note that $2x-1 = 2\cos^2\left(\frac{\pi}{16}\right)-1 = \cos \left(\frac{\pi}{8}\right) > 0$ so we take the positive root
$$ 2x-1 = \frac{\sqrt{2+\sqrt{2}}}{2} $$
$$ x = \frac{2+\sqrt{2+\sqrt{2}}}{4} $$
Then we have
$$ \sin\left(\frac{\pi}{16}\right) = \sqrt{1-x} = \frac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2} $$
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