How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$

Question

How to prove : $\cos^32\theta + 3\cos2\theta = 4(\cos^6 \theta -\sin^6 \theta)$

Answer 1

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 $$4\left(\frac{(1+x)^3}8-\frac{(1-x)^3}8\right)=\frac12(1+3x+3x^2+x^3-(1-3x+3x^2-x^3))=x^3+3x=LHS$$

Answer 2

$4(\cos^6\theta-\sin^6\theta)$

$=4((\cos^2\theta)^3-(\sin^2\theta)^3)$

$=4(\cos^2\theta-\sin^2\theta)(\cos^4\theta+\sin^4\theta+\cos^2\theta\sin^2\theta)$

$=4\cos 2\theta[\{(\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta\}+\cos^2\theta\sin^2\theta]$

$=4\cos 2\theta[\{1-2\cos^2\theta\sin^2\theta\}+\cos^2\theta\sin^2\theta]$

$=4\cos 2\theta(1-\cos^2\theta\sin^2\theta)$

$=4\cos2\theta-\cos2\theta\sin^22\theta$

$=4\cos2\theta-\cos2\theta(1-\cos^22\theta)$

$=\cos^32\theta+3\cos2\theta$

Answer 3

$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 $

Answer 4

Writing $\cos^2\theta=c,\sin^2\theta=s\implies c+s=1$

$\cos^32\theta + 3\cos2\theta$

$=(c-s)^3+3(c-s)$

$=c^3-s^3-3cs(c-s)+3(c-s)$

$=c^3-s^3+3(c-s)(1-cs)$

$=c^3-s^3+3(c-s)((c+s)^2-cs)$

$=c^3-s^3+3(c^3-s^3)=?$