The value of cos^{3}left(frac{pi}{8}right) cd | vedclass

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(C) We know that $\cos\left(\frac{3\pi}{8}\right) = \sin\left(\frac{\pi}{2} - \frac{3\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right)$ and $\sin\left(\

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

Vedclass · Answer

(C) We know that $\cos\left(\frac{3\pi}{8}ight) = \sin\left(\frac{\pi}{2} - \frac{3\pi}{8}ight) = \sin\left(\frac{\pi}{8}ight)$ and $\sin\left(\frac{3\pi}{8}ight) = \cos\left(\frac{\pi}{2} - \frac{3\pi}{8}ight) = \cos\left(\frac{\pi}{8}ight)$.Substituting these into the expression:$= \cos^{3}\left(\frac{\pi}{8}ight) \cdot \sin\left(\frac{\pi}{8}ight) + \sin^{3}\left(\frac{\pi}{8}ight) \cdot \cos\left(\frac{\pi}{8}ight)$$= \sin\left(\frac{\pi}{8}ight) \cos\left(\frac{\pi}{8}ight) \left[ \cos^{2}\left(\frac{\pi}{8}ight) + \sin^{2}\left(\frac{\pi}{8}ight) ight]$Since $\sin^{2}	heta + \cos^{2}	heta = 1$,the expression simplifies to:$= \sin\left(\frac{\pi}{8}ight) \cos\left(\frac{\pi}{8}ight) \cdot 1$$= \frac{1}{2} \left( 2 \sin\left(\frac{\pi}{8}ight) \cos\left(\frac{\pi}{8}ight) ight)$$= \frac{1}{2} \sin\left(2 \cdot \frac{\pi}{8}ight) = \frac{1}{2} \sin\left(\frac{\pi}{4}ight)$$= \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.

The value of $\cos^{3}\left(\frac{\pi}{8}\right) \cdot \cos\left(\frac{3\pi}{8}\right) + \sin^{3}\left(\frac{\pi}{8}\right) \cdot \sin\left(\frac{3\pi}{8}\right)$ is:

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