I have to express: cos 6θ and sin 6θ in terms of sinθ and cosθ.
I tried to do it using Demoivre's formula but it looks horribly wrong
I got: (cosθ + sinθ)^6
any assistance would be appreciated.
De Moivre's formula says that $\displaystyle \cos6\theta + i\sin6\theta = (\cos\theta+i\sin\theta)^6$. Expand the right side by the binomial theorem. Then take the real part to get $\displaystyle \cos6\theta = \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta$. Taking the imaginary part yields a similar formula for $\displaystyle \sin6\theta$.