Cos4x-4cos2x+3
Lets break it into parts:
1) Cos4x:
= cos (2x+2x)
Cos2xcos2x-sin2xsin2x
(cos^2x-sin^2x)^2 – (2sinxcosx)^2
(1-2sin^2 x)^2 - 4sin^2 x cos^2 x
(1-2sin^2 x)(1-2sin^2 x) – 4sin^2 x(1-sin^2 x)
1-4sin^2 x+ 4sin^4 x-4sin^2 x + 4sin^4 x
=8sin^4 x-8sin^2 x +1
2) -4cos2x:
-4(cos^2 x –sin^2 x)
-4(1-2sin^2 x)
=-4+8sin^2 x
3) +3
Therfore, combining all parts gives us:
8sin^4 x-8sin^2x+1-4+8sin^2x+3
Which equals to: 8sin^4 x..
Got it?
