Question #3504b

Apr 10, 2017

$\cos \left(\frac{1}{2} \cdot {\sin}^{-} 1 \left(\frac{\sqrt{3}}{2}\right)\right) = \frac{\sqrt{3}}{2}$

Explanation:

I'm assuming the function is this;

$\cos \left(\frac{1}{2} \cdot {\sin}^{-} 1 \left(\frac{\sqrt{3}}{2}\right)\right)$

Then we can simplify this as,

$\cos \left(\frac{1}{2} \cdot \frac{\pi}{3}\right)$ since, $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

$\implies \cos \left(\frac{\pi}{6}\right)$

$\implies \frac{\sqrt{3}}{2}$

Apr 10, 2017

$\frac{\sqrt{3}}{2}$

Explanation:

$\cos \left(\frac{1}{2} \left(\textcolor{b l u e}{\arcsin \left(\frac{\sqrt{3}}{2}\right)}\right)\right)$

$\arcsin \left(\textcolor{red}{\frac{\sqrt{3}}{2}}\right) = \textcolor{b l u e}{60}$ degrees, because $\sin \textcolor{b l u e}{60} = \textcolor{red}{\frac{\sqrt{3}}{2}}$

$= \cos \left(\frac{1}{2} \left(\textcolor{b l u e}{60}\right)\right)$

$= \cos \left(30\right)$

$= \frac{\sqrt{3}}{2}$

Apr 17, 2017

More generally:

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

So:

$\cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \left(\theta\right)}{2}}$

Then:

$\cos \left(\frac{1}{2} \arcsin \left(\frac{\sqrt{3}}{2}\right)\right) = \sqrt{\frac{1 + \cos \left(\arcsin \left(\frac{\sqrt{3}}{2}\right)\right)}{2}}$

Note that $\cos \left(\arcsin \left(\frac{\sqrt{3}}{2}\right)\right) = \frac{1}{2}$, since this refers to a $1 - \sqrt{3} - 2$ right triangle.

$\cos \left(\frac{1}{2} \arcsin \left(\frac{\sqrt{3}}{2}\right)\right) = \sqrt{\frac{1 + \frac{1}{2}}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$