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

1 Answer
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 = 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#.