# How do you evaluate sin(arcsin(1/2) + arccos (0)) ?

Jun 22, 2018

Under the multivalued interpretation,

$\sin \left(\arcsin \left(\frac{1}{2}\right) + \arccos \left(0\right)\right) = \pm \setminus \frac{\sqrt{3}}{2}$

#### Explanation:

Let's use the multivalued interpretation of $\arcsin$ and $\arccos .$

The real question is what is

$\sin \arccos \left(\frac{a}{b}\right)$ and $\cos \arcsin \left(\frac{a}{b}\right)$

When we have the cosine of an inverse sine or the sine of an inverse cosine the denominator $b$ is the hypotenuse and the other side is $\sqrt{{b}^{2} - {a}^{2}} ,$ with indeterminate sign. So,

$\sin \arccos \left(\frac{a}{b}\right) = \pm \setminus \frac{\sqrt{{b}^{2} - {a}^{2}}}{b}$

$\cos \arcsin \left(\frac{a}{b}\right) = \pm \frac{\sqrt{{b}^{2} - {a}^{2}}}{b}$

$\sin \left(\arcsin \left(\frac{1}{2}\right) + \arccos \left(0\right)\right)$

$= \sin \arcsin \left(\frac{1}{2}\right) \cos \arccos 0 + \cos \arcsin \left(\frac{1}{2}\right) \sin \arccos 0$

$= \frac{1}{2} \left(0\right) + \left(\pm \setminus \frac{\sqrt{3}}{2}\right) \left(\pm 1\right)$

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