Recent content by desertwonder

The 3/4 turned into a 3 after I multiplied everything by 4. Afterwards, the 4sin and 4cos on the left became 3sin and 3cose when moving the stuff on the right to the left. I then divided everything by 3.

Hmm...after fiddling around some more following your suggestions I have: (sin(x))^4+(cos(x))^4 = = -2(sin(x))^2(cos(x))^2

Hmm...this was all I could narrow it down to: cos(4x)= (cos(x))^4 + (sin(x))^4 - 6(sin(x))^2(cos(x))^2

Hmmm...in that case...I'm not really sure how to go about reducing cos(4x) to a single angle using the double-angle cosine formula. :confused: EDIT: Ahh...I now see that cos(4x)=cos(2x+2x)=cos(2x)cos(2x)-sin(2x)sin(2x)... more work underway! EDIT2: Ohhh, here we go: cos(4x)=((cos(x))^2-...

Cos(4x)=Cos(2(2x))= ...So does that mean Cos(4x)=2((cosx)^2-(sinx)^2)?

I've wracked my brains out trying to prove this identity. If anyone could offer some suggestions, I'd greatly appreciate it! (sin(x))^4+(cos(x))^4 = 0.25∙cos(4x)+0.75