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

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Answer 1 · 2016-09-27T23:24:10

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

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#