Using the identity that: #sin(A+B)-=sinAcosB+cosAsinB#
We can make #sin(3x)=sin(x+2x)=sinxcos2x+cosxsin2x#
Using the double angle identities:
#sin2A-=2sinAcosA#
#cos2A=cos^2A-sin^2A#
We can expand this further by writing #sinx(cos^2x-sin^2x)+cosx(2sinxcosx)=sinx(cos^2x-sin^2x)+2sinxcos^2x#
#(cancel(sinx)(cos^2x-sin^2x)+2cancel(sinx)cos^2x)/cancel(sinx)=cos^2x-sin^2x+2cos^2x#
The identity #cos^2x+sin^2x-=1# gives us #sin^2x-=1-cos^2x#
So, #cos^2x-(1-cos^2x)+2cos^2x=cos^2x-1+cos^2x+2cos^2x=4cos^2x-1#
Proof:
#(sin(3(78)))/sin(78)~~-0.83#
#2cos^2(78)~~0.0865#
#4cos^2(78)-1~~-0.83#
#4cos^2x-1#:
graph{4cos(x)cos(x)-1 [-20, 20, -10.42, 10.42]}
#(sin3x)/sinx#:
graph{sin(3x)/sin(x) [-20, 20, -10.42, 10.42]}
#2cos^2x#:
graph{2cos(x)*cos(x) [-20, 20, -10.42, 10.42]}
