# How do you solve sin(arcsin(3/5) + arccos(3/5))?

May 15, 2015

Picture a right angled triangle with sides 3, 4 and 5 (since ${3}^{2} + {4}^{2} = {5}^{2}$). Call the smallest angle $\alpha$ and the next smallest $\beta$.

$\sin \alpha = \frac{3}{5}$, being the length of the opposite side divided by the length of the hypotenuse.

$\cos \beta = \frac{3}{5}$, being the length of the adjacent side divided by the length of the hypotenuse.

So $\arcsin \left(\frac{3}{5}\right) + \arccos \left(\frac{3}{5}\right) = \alpha + \beta = \frac{\pi}{2}$ (or ${90}^{o}$) since together with the right angle, the internal angles of the triangle must add up to $\pi$ (${180}^{o}$).

$\sin \left(\frac{\pi}{2}\right) = 1$.

So $\sin \left(\arcsin \left(\frac{3}{5}\right) + \arccos \left(\frac{3}{5}\right)\right) = 1$.