Question

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

Answer 1

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 = 3/5`, being the length of the opposite side divided by the length of the hypotenuse.

`cos beta = 3/5`, being the length of the adjacent side divided by the length of the hypotenuse.

So `arcsin(3/5)+arccos(3/5) = alpha + beta = 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 (pi/2) = 1`.

So `sin(arcsin(3/5)+arccos(3/5)) = 1`.