Trigonometric Identities

[jazz98]

Hello! everyone

Show that sin 3A = 3 sin A – 4 sin3 A
Solution
Sin 3A= sin(2A + A)
= sin 2A cos A + cos 2A sin A
= (2sin A cos A) cos A + (1 – 2 sin2 A) sin A
= 2sin A cos2 A + sin A – 2sin3 A
= 2sin A(1 – sin2 A) + sin A – 2sin3 A
= 3sin A – 4sin3 A
Someone please let me know why + cos 2A sin A become + (1 – 2 sin2 A) sin A

[midas]

it's one of the double-angle identities

http://www.purplemath.com/modules/idents.htm#double

cos(2x)=1-2sin^2(x)