sin left(frac{5 pi}{24}right) cdot cos left(f | vedclass

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Question

(A) $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$ सूत्र का उपयोग करते हुए: $\sin \left(\frac{5 \pi}{24}\right) \cos \left(\frac{\pi}{24}\right) = \frac{1}

Vedclass · Accepted Answer

$\frac{1+\sqrt{2}}{4}$

Vedclass · Answer

(A) $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$ सूत्र का उपयोग करते हुए: $\sin \left(\frac{5 \pi}{24}\right) \cos \left(\frac{\pi}{24}\right) = \frac{1}{2} \left[ \sin \left(\frac{5 \pi}{24} + \frac{\pi}{24}\right) + \sin \left(\frac{5 \pi}{24} - \frac{\pi}{24}\right) \right]$ $= \frac{1}{2} \left[ \sin \left(\frac{6 \pi}{24}\right) + \sin \left(\frac{4 \pi}{24}\right) \right]$ $= \frac{1}{2} \left[ \sin \left(\frac{\pi}{4}\right) + \sin \left(\frac{\pi}{6}\right) \right]$ $= \frac{1}{2} \left[ \frac{1}{\sqrt{2}} + \frac{1}{2} \right] = \frac{1}{2} \left[ \frac{\sqrt{2}+1}{2} \right] = \frac{\sqrt{2}+1}{4}$

$\sin \left(\frac{5 \pi}{24}\right) \cdot \cos \left(\frac{\pi}{24}\right)$ का मान है

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