evaluate sin(2tan^(inverse)9/40)

Question

I need help evaluating.

Answer 1

Let θ = Tan^-1(9/40)

 

Then tanθ = 9/40 and θ is acute

 

        Draw a right triangle with acute angle θ, side opposite θ of length 9 and side adjacent to θ of length 40.

 

        By the Pythagorean Theorem, the hypotenuse has length √1681.

 

So, sin[2Tan^-1(9/40] = sin(2θ) = 2sinθcosθ

 

                                             = 2(9/√1681)(40/√1681)

 

                                             = 720/1681

Answer 2

sin(2tan^-1(9/40))=2sin(tan^-1(9/40))cos(tan^-1(9/40))= (*)

 

[ 1/cos²(x)=1+tan²(x),  cos(x)=1/√(1+tan²(x)),

sin(x)=tan(x)/√(1+tan²(x)) and tan(tan^-1(x))=x]

 

=(*)=2(9/40)/√(1+(9/40)²)*1/√(1+(9/40)²)=

=(9/20)/(40²+9²)/40²=(9*1600)/(20*1681)=720/1681