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I'm having some difficulty with evaluating the following expression:

$ \sin (\frac 1 2 \arccos(\frac 3 7))$

I've tried substituting $\arccos (\frac 37)$ with $x$. This way I get that $\cos(x)=\frac 37$ thus $1- 2\sin^2(\frac x2)=\frac 3 7$ and $\sin(\frac x2)=\sqrt\frac2 7$ or $\sin(\frac x2)=-\sqrt\frac2 7$

Which of these answers do I omit and why? Thanks in advance!

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You have to consider the range of $\arccos (x)$. We have that

$$\frac{1}{2}\arccos(x)\in \left[0,\frac{\pi}{2}\right]$$

On that domain, $\sin(x)$ is positive.

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  • $\begingroup$ Thank you so much! $\endgroup$ Commented Oct 5, 2019 at 8:42
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With some more trigonometry, it is much simpler:

Set $\theta=\arccos\frac 37$. We have a linearisation formula $$\sin^2\frac\theta 2=\frac{1-\cos\theta}2.$$ On the other hand $\;0\le\frac\theta2\le \frac\pi2$.

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