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]}