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October 20, 2014

October 20, 2014

Posted by **Jen** on Friday, July 27, 2007 at 12:16am.

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]

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