How do you find the exact value of sin[arcsin(1/2) + arccos (0) ] ?

Apr 29, 2018

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

Explanation:

Recall the following definitions of $a r c \sin \mathmr{and} a r c \cos$ functions :

$a r c \sin x = \theta , | x | \le 1 \iff \sin \theta = x , | \theta | \le \frac{\pi}{2}$.

$a r c \cos x = \theta , | x | \le 1 \iff \cos \theta = x , \theta \in \left[0 , \pi\right]$.

Now, $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2} , \mathmr{and} , | \frac{\pi}{6} | \le \frac{\pi}{2} \therefore \arcsin \left(\frac{1}{2}\right) = \frac{\pi}{6}$.

Similarly, $a r c \cos 0 = \frac{\pi}{2}$.

$\therefore \sin \left[\arcsin \left(\frac{1}{2}\right) + a r c \cos 0\right] = \sin \left(\frac{\pi}{6} + \frac{\pi}{2}\right)$,

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

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