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How to evaluate $\sin^2\left(\frac{1}{2}\arccos a\right)$ in terms of $a$

I first have attempted to work form the inside but then the $\frac{1}{2}$ had messed that process up

Then i have attempted to do the power reduction formula but that didnt work

I was wondering how would i do this?

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    $\begingroup$ $$\sin^2(x) = \frac{1-\cos (2x)}{2}$$ $\endgroup$ – Jack D'Aurizio Jun 11 '17 at 16:40
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HINT:

$$\sin^2\left(x\right)=\frac{1-\cos\left(2x\right)}{2}$$

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Use $$\sin^2(\frac{\theta}{2})=\frac{1-\cos(\theta)}{2}$$

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Let $u = \arccos (a)$. Then, using user35508's for the half angle formula, $\sin^2(u/2) = \frac {1 - \cos(u)}{2}$. Substituting $u$ to the above we get $\frac {1 - \cos(\arccos a)}{2}$. Since $\cos (\arccos (a))$ = $a$, we get $\frac {1-a}{2}$.

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