Trig or Treat

...In response to a student's e-mail...
>
> Dr. Ritter, I know that when you have sin^2(x), that it is equivalent to 1 -
> cos (2x) / 2, but what happens with the odd powers?  For example, 9.1D #3,
> sin^3(x) is equivalent to 3sin(x) - sin(3x) / 4 according to the solutions
> manual, but where is this coming from?  Taken at face value, this doesn't do
> me a whole lot of good because the very next problem is cos^3(x).  I need to
> know either the trig identity that this involves, or the process needed to
> obtain this result.  Thank you.
> 


Trig or Treat Sir,

It helps to learn how to _play_ with trig identities. [Incidentally, the
term is _equal_.]  I'll show you where the sin^3 originates and by doing
similar things you can derive an identity for cos^3.  [You may need to
break out a trig text to follow.]  I'll just use identities that should be
in your back pocket.

sin^3(t) = sin(t)sin^2(t)

	 = sin(t)[(1/2)(1 - cos(2t))]  //Replaced sin^2//

	 = (1/2)sin(t) - (1/2)sin(t)cos(2t) //Distribute ...//
	 
	 = (1/2)sin(t) - (1/2)[(1/2)[sin(3t) + sin(-t)]]
	
			//Use identity for sin(x)cos(y) that's easy to
	obtain, real time, from sin(x+y) and sin(x-y) identities.//

	 = (1/2)sin(t) - (1/4)sin(3t) + (1/4)sin(t) 

			//sin's odd, remember.//
	
	 = [3sin(t) - sin(3t)]/4  //Voila!//

You can perform similar magic to deal with cos^3 using perhaps
cos(x+y) and cos(x-y) to deal with cos(x)cos(y) and sin(x)sin(y).

Time for you to play...