How do you find sin(tan^-1(a/3))?

1 Answer
Sep 27, 2016

sin(tan^(-1)(a/3)) = a/(sqrt(a^2+9)

Explanation:

Note that the range of tan^(-1)(y) is (-pi/2, pi/2).

If theta in (-pi/2, pi/2) then cos theta > 0.

If a/3 > 0 then consider a right angled triangle with sides a/3, 1 and sqrt(a^2/9+1)

We find:

sin(tan^(-1)(a/3)) = (a/3)/sqrt(a^2/9+1) = a/(sqrt(a^2+9)

This identity continues to hold for a/3 < 0 since sin theta and tan theta are odd functions.

It also holds for a=9 since sin(tan^(-1)(0)) = sin(0) = 0