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

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Question

(D) આપણે નિત્યસમ $a^2 - b^2 = (a - b)(a + b)$ નો ઉપયોગ કરીએ છીએ.$\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) આપણે નિત્યસમ $a^2 - b^2 = (a - b)(a + b)$ નો ઉપયોગ કરીએ છીએ.$\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$$= \left(\cos ^2 \frac{\pi}{24} + \sin ^2 \frac{\pi}{24}ight) \left(\cos ^2 \frac{\pi}{24} - \sin ^2 \frac{\pi}{24}ight)$$\cos ^2 	heta + \sin ^2 	heta = 1$ અને $\cos ^2 	heta - \sin ^2 	heta = \cos 2	heta$ હોવાથી:$= (1) \cdot \cos \left(2 \cdot \frac{\pi}{24}ight) = \cos \frac{\pi}{12}$$\frac{\pi}{12} = 15^\circ$ હોવાથી,$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$$= \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} = $

Similar Questions

$\cos \alpha \sin (\beta - \gamma ) + \cos \beta \sin (\gamma - \alpha ) + \cos \gamma \sin (\alpha - \beta ) = $

જો $\cos (\alpha+\beta)=\frac{4}{5}$,$\sin (\alpha-\beta)=\frac{5}{13}$ અને $\alpha, \beta$ એ $0$ અને $\frac{\pi}{4}$ ની વચ્ચે હોય,તો $\tan 2 \alpha$ ની કિંમત શોધો.

જો $(1+\tan \alpha)(1+\tan 4 \alpha)=2$ અને $\alpha \in \left(0, \frac{\pi}{16}\right)$ હોય,તો $\alpha$ ની કિંમત શોધો.

કિંમત શોધો: $\sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) + \sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma )$

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