Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$
I have tried $\sin^3\alpha(3\sin\alpha – 4 \sin^3\alpha) = 3\sin^4\alpha – 4\sin^6\alpha$ and $\cos^3\alpha(4\cos^3\alpha – 3\cos\alpha) = 4\cos^6\alpha – 3\cos^4\alpha$ to give
$$\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = 3\sin^4\alpha – 4\sin^6\alpha + 4\cos^6\alpha – 3\cos^4\alpha$$
I can't work out how to simplify this to $\cos^32\alpha$.
I also noticed that the LHS of the question resembles $\cos(A-B)$, but I can't figure a way of making that useful.
Best Answer
Let $\cos^2\alpha=a,\sin^2\alpha=b\implies a+b=1, a-b=\cos2\alpha$
$ \sin3\alpha\sin^3\alpha + \cos3\alpha\cos^3\alpha = $
$ = 3\sin^4\alpha - 4\sin^6\alpha + 4\cos^6\alpha - 3\cos^4\alpha = $
$ = 3b^2-4b^3+4a^3-3a^2=4(a^3-b^3)-3(a^2-b^2)=(a-b)\{4(a^2+ab+b^2)-3(a+b)\} $
Now, $4(a^2+ab+b^2)-3(a+b)=4\{(a+b)^2-ab\}-3=4(1-ab)-3=1-4ab$ $=1-4\cos^2\alpha\sin^2\alpha=1-(2\sin\alpha\cos\alpha)^2=1-\sin^22\alpha=\cos^22\alpha$