When dealing with things like #cos3x#, it helps to simplify it to trigonometric functions of a unit #x#; i.e. something like #cosx# or #cos^3x#. We can use the sum rule for cosine to accomplish this:
#cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta#
So, since #cos3x=cos(2x+x)#, we have:
#cos(2x+x)=cos2xcosx-sin2xsinx#
#=(cos^2x-sin^2x)(cosx)-(2sinxcosx)(sinx)#
Now we can replace #cos3x# with the above expression:
#(cos3x)/cosx=1-4sin^2x#
#((cos^2x-sin^2x)(cosx)-(2sinxcosx)(sinx))/cosx=1-4sin^2x#
We can split this larger fraction up into two smaller fractions:
#((cos^2x-sin^2x)(cosx))/cosx-((2sinxcosx)(sinx))/cosx=1-4sin^2x#
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^2x#s 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.
