# Question 859bc

Jun 20, 2017

Using the half-angle formula for cosine:

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

Therefore...with
$\theta = \arcsin \left(\frac{3}{5}\right)$

cos((1/2)arcsin(3/5)) = +-sqrt(((1+cos(arcsin(3/5)))/2)#

Now, set up a right triangle having $\theta$ as one of the angles that is not the right angle. Since $\theta = \arcsin \left(\frac{3}{5}\right)$, we realize that $\theta$ is an angle whose sine is $\frac{3}{5}$. Since "sine equals opposite over hypotenuse," mark the hypotenuse of the triangle with a 5, and the side opposite angle $\theta$ with a 3.

This is a 3-4-5 triangle. The length of the remaining side is 4.
The cosine of the angle is adjacent over hypotenuse:

$\cos \left(\arcsin \left(\frac{3}{5}\right)\right) = \frac{4}{5}$.

Now $1 + \frac{4}{5} = \frac{9}{5}$. Dividing $\frac{9}{5}$ by 2 gives $\frac{9}{10}$.

This is the radicand. Back to the half-angle formula,

$\cos \left(\left(\frac{1}{2}\right) \arcsin \left(\frac{3}{5}\right)\right) = \pm \sqrt{\left(\frac{1 + \cos \left(\arcsin \left(\frac{3}{5}\right)\right)}{2}\right)} = \pm \sqrt{\frac{9}{10}}$.

If we wish to rationalize the radical, $\sqrt{\frac{9}{10}}$ = $\frac{3}{\sqrt{10}}$ = $\frac{3 \sqrt{10}}{10}$.

So...

$\cos \left(\left(\frac{1}{2}\right) \arcsin \left(\frac{3}{5}\right)\right) = \frac{3 \sqrt{10}}{10}$.

Since $\arcsin x$ is in Quadrant 1 when $x > 0$, we need not worry about the $= -$ symbol. We know that the angle is in the first quadrant where cosine is positive. Observe that I removed the $= -$ symbol for that reason.