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Prove that $\cos^3A\cos 3A+\sin^3A\sin 3A=\cos^32A$

• $\cos^3A=\large\frac{1}{4}$$(3\cos A+\cos 3A) • \sin^3A=\large\frac{1}{4}$$(3\sin A-\sin 3A)$
$\cos^3A\cos 3A+\sin^3A\sin 3A$
$\large\frac{1}{4}$$(3\cos A+\cos 3A)\cos 3A+\large\frac{1}{4}$$(3\sin A-\sin 3A)\sin 3A$
$\Rightarrow \large\frac{1}{4}$$(3\cos A\cos 3A+\cos^23A+\large\frac{1}{4}$$(3\sin A\sin 3A-\sin ^23A)$
$\Rightarrow \large\frac{1}{4}$$[(3\cos A\cos 3A+\cos^23A+(3\sin A\sin 3A-\sin ^23A)] \Rightarrow \large\frac{3}{4}$$(\cos A\cos 3A+\sin A\sin 3A)+\large\frac{1}{2}$$(\cos^23 A-\sin^2 3A) \Rightarrow \large\frac{3}{4}$$\cos (3A-A)+\large\frac{1}{4}$$(\cos 2.3 A) \cos(A-B)=\cos A.\cos B+\sin A\sin B \Rightarrow \large\frac{3}{4}$$\cos (2A)+\large\frac{1}{4}$$(\cos 6 A) \Rightarrow \large\frac{3}{4}$$\cos (2A)+\large\frac{1}{4}$$(\cos 3.2 A) \Rightarrow \large\frac{3}{4}$$\cos (2A)+\large\frac{1}{4}$$(4\cos^32A-3\cos 2A) \Rightarrow \large\frac{3}{4}$$\cos (2A)+\cos^32A-\large\frac{3}{4}$$\cos 2A)$
$\Rightarrow \cos^32A$=RHS