Question

I need help with I just can't seem to get anywhere. this is as far as I have got:

Solve for b
arcsin(b)+ 2arctan(b)=pi

arcsin(b)=pi-2arctan(b)
b=sin(pi-2arctan(b))

Sub in Sin difference identity
let 2U=(2arctan(b))

sin(a-b)=sinacosb-cosasinb
=(sin(pi))(cos(2U))-(cos(pi))(sin(2U))
=(0)(cos(2U))-(-1)(sin(2U))
=(sin(2u))

b=sin(2arctan(b))

Now what should I do?





sin(2arctan(b))=
2 sin(arctan(b))cos(arctanb)

If you draw a right triangle with the lenghts of the two sides at right angles b and 1, so that one of the angeles becomes arctan(b), you see that

sin(arctan(b)) = b/sqrt[1+b^2]

cos(arctan(b)) = 1/sqrt[1+b^2]