When dealing with things like cos3xcos3x, it helps to simplify it to trigonometric functions of a unit xx; i.e. something like cosxcosx or cos^3xcos3x. We can use the sum rule for cosine to accomplish this:
cos(alpha+beta)=cosalphacosbeta-sinalphasinbetacos(α+β)=cosαcosβ−sinαsinβ
So, since cos3x=cos(2x+x)cos3x=cos(2x+x), we have:
cos(2x+x)=cos2xcosx-sin2xsinxcos(2x+x)=cos2xcosx−sin2xsinx
=(cos^2x-sin^2x)(cosx)-(2sinxcosx)(sinx)=(cos2x−sin2x)(cosx)−(2sinxcosx)(sinx)
Now we can replace cos3xcos3x with the above expression:
(cos3x)/cosx=1-4sin^2xcos3xcosx=1−4sin2x
((cos^2x-sin^2x)(cosx)-(2sinxcosx)(sinx))/cosx=1-4sin^2x(cos2x−sin2x)(cosx)−(2sinxcosx)(sinx)cosx=1−4sin2x
We can split this larger fraction up into two smaller fractions:
((cos^2x-sin^2x)(cosx))/cosx-((2sinxcosx)(sinx))/cosx=1-4sin^2x(cos2x−sin2x)(cosx)cosx−(2sinxcosx)(sinx)cosx=1−4sin2x
Note how the cosines cancel:
((cos^2x-sin^2x)cancel(cosx))/cancel(cosx)-((2sinxcancel(cosx))(sinx))/cancelcosx=1-4sin^2x
->cos^2x-sin^2x-2sin^2x=1-4sin^2x
Now add a sin^2x-sin^2x into the left side of the equation (which is the same thing as adding 0). The reasoning behind this will become clear in a minute:
cos^2x-sin^2x-2sin^2x+(sin^2x-sin^2x)=1-4sin^2x
Rearrange terms:
cos^2x+sin^2x-(sin^2x+sin^2x+2sin^2x)=1-4sin^2x
Use the Pythagorean Identity sin^2x+cos^2x=1 and combine the sin^2xs in the parentheses:
1-(4sin^2x)=1-4sin^2x
You can see that our little trick of adding sin^2x-sin^2x has allowed us to use the Pythagorean Identity and collect the sin^2x terms.
And voila:
1-4sin^2x=1-4sin^2x
Q.E.D.
