Can anyone help me prove this trig identity:
Cos(4ά)+4cos(2ά)+3=8cos(4ά)
Thanks
Two comments:
1. $\displaystyle \cos(4\alpha)=\cos^{4}(\alpha)-2\sin^{2}(\alpha)\cos^{2}(\alpha)+\sin^{4}(\alpha)-4\sin^{2}(\alpha)\cos^{2}(\alpha).$
You missed some squares in there.
2. In your expression, you still have a $\displaystyle \cos(2\alpha)$ in there. Reduce the angle, and see what you get.