cos^3 frac{pi}{8} cos frac{3pi}{8} + sin^3 fr | vedclass

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Question

(A) Let $x = \frac{\pi}{8}$. Then $\frac{3\pi}{8} = 3x$. The expression is $\cos^3 x \cos 3x + \sin^3 x \sin 3x$. We know that $\cos 3x = 4\cos^3 x -

Vedclass · Accepted Answer

$\frac{1}{2\sqrt{2}}$

Vedclass · Answer

(A) Let $x = \frac{\pi}{8}$. Then $\frac{3\pi}{8} = 3x$. The expression is $\cos^3 x \cos 3x + \sin^3 x \sin 3x$. We know that $\cos 3x = 4\cos^3 x - 3\cos x$ and $\sin 3x = 3\sin x - 4\sin^3 x$. Substituting these: $\cos^3 x (4\cos^3 x - 3\cos x) + \sin^3 x (3\sin x - 4\sin^3 x) = 4\cos^6 x - 3\cos^4 x + 3\sin^4 x - 4\sin^6 x$. $= 4(\cos^6 x - \sin^6 x) - 3(\cos^4 x - \sin^4 x)$. Using $a^2 - b^2 = (a-b)(a+b)$ and $a^3 - b^3 = (a-b)(a^2+ab+b^2)$: $= 4(\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) - 3(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)$. Since $\cos^2 x + \sin^2 x = 1$,$\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$. The expression becomes $(\cos^2 x - \sin^2 x) [4(1 - 2\sin^2 x \cos^2 x + \sin^2 x \cos^2 x) - 3]$. $= \cos 2x [4(1 - \sin^2 x \cos^2 x) - 3] = \cos 2x [1 - 4\sin^2 x \cos^2 x] = \cos 2x [1 - (2\sin x \cos x)^2]$. $= \cos 2x [1 - \sin^2 2x] = \cos 2x \cdot \cos^2 2x = \cos^3 2x$. Since $x = \frac{\pi}{8}$,$2x = \frac{\pi}{4}$. $\cos^3(\frac{\pi}{4}) = (\frac{1}{\sqrt{2}})^3 = \frac{1}{2\sqrt{2}}$.

$\cos^3 \frac{\pi}{8} \cos \frac{3\pi}{8} + \sin^3 \frac{\pi}{8} \sin \frac{3\pi}{8} = $

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