cos ^4 frac{pi}{24} - sin ^4 frac{pi}{24} = | vedclass

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Question

(D) हम सर्वसमिका $a^2 - b^2 = (a - b)(a + b)$ का उपयोग करते हैं।$\cos ^4 \frac{\pi}{24} - \sin ^4 \frac{\pi}{24} = \left(\cos ^2 \frac{\pi}{24}\right)

Vedclass · Accepted Answer

$\frac{\sqrt{2} + \sqrt{6}}{4}$

Vedclass · Answer

(D) हम सर्वसमिका $a^2 - b^2 = (a - b)(a + b)$ का उपयोग करते हैं।$\cos ^4 \frac{\pi}{24} - \sin ^4 \frac{\pi}{24} = \left(\cos ^2 \frac{\pi}{24}ight)^2 - \left(\sin ^2 \frac{\pi}{24}ight)^2$$= \left(\cos ^2 \frac{\pi}{24} + \sin ^2 \frac{\pi}{24}ight) \left(\cos ^2 \frac{\pi}{24} - \sin ^2 \frac{\pi}{24}ight)$चूंकि $\cos ^2 	heta + \sin ^2 	heta = 1$ और $\cos ^2 	heta - \sin ^2 	heta = \cos 2	heta$,इसलिए:$= (1) \cdot \cos \left(2 \cdot \frac{\pi}{24}ight) = \cos \frac{\pi}{12}$$\frac{\pi}{12} = 15^\circ$ का उपयोग करते हुए,$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$$= \left(\frac{1}{\sqrt{2}}ight) \left(\frac{\sqrt{3}}{2}ight) + \left(\frac{1}{\sqrt{2}}ight) \left(\frac{1}{2}ight) = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$

$\cos ^4 \frac{\pi}{24} - \sin ^4 \frac{\pi}{24} = $

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