Solving trigonometric equation

1. Apr 15, 2010

thereddevils

This is part of the identity proving question .

from $$\sin C=\frac{2}{\sqrt{29}}$$ , how can i reach $$C=\frac{1}{2}\sin^{-1} (\frac{20}{29})$$ ?

2. Apr 15, 2010

tiny-tim

Hi thereddevils!

(have a square-root: √ and try using the X2 tag just above the Reply box )

Hint: C = 1/2 sin-1 20/29 is the same as sin2C = 20/29

3. Apr 16, 2010

thereddevils

Re: trigonometry

got it thanks !