De moivre's theorem (1 Viewer)

[Joshmosh2]

Show that
sin^3x(cos^2x) = 1/16 (2sinx +sin3x -sin5x)

(using the fact that z + 1/z = 2cosx, and z - 1/z = 2isinx)


My approach was to expand LHS into sin^3x (1-sin^2x)
= sin^3x - sin^5x
= (z - 1/z)^3 - (z - 1/z)^5
(then expand and simplify)

Is this a viable method?
The suggested solutions suggest otherwise.

 

[Carrotsticks]

That approach is fine. What do the suggested solutions suggest?

 

[Joshmosh2]

That approach is fine. What do the suggested solutions suggest?

Starts it off differently with:
http://imgur.com/xkKGwfx (it continues based off this)


I don't really like this approach because of the manipulation from step 4 to 5

 

[Carrotsticks]

How do they get from here to the final result? Do they use a trig identity that may look somewhat unfamiliar to you?