How would I rewirte sin(tan^-1 x) as an equivalent expression that only involves x ?

Answer 1

Just set up a right triangle wit one leg = x, the other leg = 1 , so hypotenuse = sqrt(x^2+1)
So sin must be: x/(sqrt(x^2+1)

Answer 2

let tan^-1(x) = y

tan y = x

sin(y)/ cos(y) = x

sin^2(y)/cos^2(y) = x^2

sin^2(y) = x^2(cos^2(y)

sin^2(y) = x^2(1- sin^2(y)

sin^2(y) = x^2 - x^2 sin^2(y)

sin^2(y) + x^2(sin^2(y) = x^2

sin^2(y)[1 + x^2] = x^2

sin^2(y) = x^2/(1+x^2)

sin(y) = x/sqrt(1+x^2)

sin(tan^-1(x) = sin y = x/sqrt(1+x^2)