Writing sins 3s in terms of sin s In the reduction below, I do not understand line 4 and 6.  What identities were applied to line 3 and 5 to reach those conclusions?  How were those identities introduced?
1 ) $\sin  3a $
2 )  $= \sin(2a +s)$
3 )  $= \sin2a ·\cos a + \cos 2a·\sin a$
4 )  $=(2\sin a·\cos a)\cos a + (\cos^2a - \sin^2a)\sin a$ 
5 )  $= 2\sin a·\cos²a + \cos²a·\sin a - \sin³a$
6 ) $ = 2\sin a(1-\sin²a) + (1 - \sin²a)\sin a - \sin³a$ 
7 ) $ = 2\sin a - 2\sin³a + \sin a - \sin³a - \sin³a$
8 ) $= 3\sin a - 4\sin³a$
This is important rewriting all forms of $\sin$ $n·s$ in terms  of $\sin$ $s$. All help is greatly appreciated

Following are the formula's used above -
1.) $\sin(a + b) = \sin a \cos b + \cos a \sin b$
2.) $\sin 2a = 2 \sin a \cos a$
3.) $\cos 2a = \cos^2 a - \sin^2 a$
4.) $\cos^2 a = 1 - \sin^2 a$
