Solving trigonometric equation

[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}) ?

 

[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

 

[thereddevils]

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

got it thanks !