How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$ How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta  -\sin^6 \theta)$
 A: Write $\cos2\theta=x$. We have
$$x=2\cos^2\theta-1=1-2\sin^2\theta$$
$$\frac{1+x}2=\cos^2\theta$$
$$\frac{1-x}2=\sin^2\theta$$
Thus the RHS becomes
$$4\left(\frac{(1+x)^3}8-\frac{(1-x)^3}8\right)=\frac12(1+3x+3x^2+x^3-(1-3x+3x^2-x^3))=x^3+3x=LHS$$
A: $4(\cos^6\theta-\sin^6\theta)$
$=4((\cos^2\theta)^3-(\sin^2\theta)^3)$
$=4(\cos^2\theta-\sin^2\theta)(\cos^4\theta+\sin^4\theta+\cos^2\theta\sin^2\theta)$
$=4\cos 2\theta[\{(\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta\}+\cos^2\theta\sin^2\theta]$
$=4\cos 2\theta[\{1-2\cos^2\theta\sin^2\theta\}+\cos^2\theta\sin^2\theta]$
$=4\cos 2\theta(1-\cos^2\theta\sin^2\theta)$
$=4\cos2\theta-\cos2\theta\sin^22\theta$
$=4\cos2\theta-\cos2\theta(1-\cos^22\theta)$
$=\cos^32\theta+3\cos2\theta$
A: $4(\cos^6 \theta  -\sin^6 \theta) = 4(\cos^2 \theta  -\sin^2 \theta)(\cos^4 \theta + \cos^2 \theta\sin^2 \theta+\sin^4 \theta) = 4(\cos^2 \theta  -\sin^2 \theta)(  (\cos^2 \theta  +\sin^2 \theta)^2 - 2\cos^2 \theta\sin^2 \theta + \cos^2 \theta\sin^2 \theta) = 4(\cos^2 \theta  -\sin^2 \theta)(  1 - \cos^2 \theta\sin^2 \theta ) = 4\cos2\theta(1- \dfrac{\sin^2 2\theta}{4})  = 4\cos2\theta - \sin^2 2\theta\cos2\theta = 4\cos2\theta - (1-\cos^22\theta)\cos2\theta =  3\cos2\theta +\cos^3\theta $
A: Writing $\cos^2\theta=c,\sin^2\theta=s\implies c+s=1$
$\cos^32\theta + 3\cos2\theta$
$=(c-s)^3+3(c-s)$
$=c^3-s^3-3cs(c-s)+3(c-s)$
$=c^3-s^3+3(c-s)(1-cs)$
$=c^3-s^3+3(c-s)((c+s)^2-cs)$
$=c^3-s^3+3(c^3-s^3)=?$
