Trigonometry Tricks

Date: 11/14/2002 at 11:09:31
From: Tanya Riston
Subject: Calculus 1

To find how (1/4)cos^4x-(1/6)cos^6x = -(1/4)sin^4x+(1/6)sin^6x, I've 
tried to change all the cos to sin by using sin^2x+cos^2x = 1 and then 
using double angle formulas. I am trying to prove that they are equal 
to each other.

Date: 11/15/2002 at 10:47:17
From: Doctor Ian
Subject: Re: Calculus 1

Hi Tanya,

The first thing that occurs to me is that you can get rid of the
fractions:

  (1/4)cos^4x - (1/6)cos^6x = -(1/4)sin^4x + (1/6)sin^6x

       cos^4x - (4/6)cos^6x = -sin^4x + (4/6)sin^6x

          6cos^4x - 4cos^6x = -6sin^4x + 4sin^6x

Now I'd look for symmetries that I can exploit. For example, can I get 
the same exponents together? That would let me use the distributive 
property. (At this point, I honestly don't know if this will work. I'm 
just trying to let you see how I'm thinking about it.)

          6cos^4x + 6sin^4x = 4cos^6x + 4sin^6x

         6(cos^4x + sin^4x) = 4(cos^6x + sin^6x)

Now at this point, I don't see anything obvious, so I'll go look (in
our FAQ) for some identities that I might be able to use. These look
promising:

   sin^4(x) = (3 - 4 cos[2x] + cos[4x])/8,
   cos^4(x) = (3 + 4 cos[2x] + cos[4x])/8,

   sin^6(x) = (10 - 15 cos[2x] + 6 cos[4x] - cos[6x])/32,
   cos^6(x) = (10 + 15 cos[2x] + 6 cos[4x] + cos[6x])/32,

Now, instead of substituting into the equation, which would give me a
huge mess, I'll just try adding one pair together:

    sin^4(x) + cos^4(x) 

  = (3 - 4 cos[2x] + cos[4x])/8 + (3 + 4 cos[2x] + cos[4x])/8 

  = (1/8)(3 - 4cos[2x] + cos[4x] + 3 + 4cos[2x] + cos[4x])

  = (1/8)(6 + 2cos[4x])

So that doesn't look to bad!  So the left side of my equation becomes

   6(1/8)(6 + 2cos^4x) = 4(cos^6x + sin^6x)

Just looking at the other pair, I can see that the [2x] terms and [6x]
terms are going to kill each other, leaving me with only the constant
and [4x] terms.  So it looks like smooth sailing from here. Can you
finish up? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/