cos frac{pi}{12} = ? | vedclass

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(D) We know that $\cos 	heta = \cos(45^{\circ} - 30^{\circ})$ where $	heta = \frac{\pi}{12} = 15^{\circ}$.Using the formula $\cos(A - B) = \cos A \c

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) We know that $\cos 	heta = \cos(45^{\circ} - 30^{\circ})$ where $	heta = \frac{\pi}{12} = 15^{\circ}$.Using the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$:$\cos(45^{\circ} - 30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ}$$= (\frac{1}{\sqrt{2}} 	imes \frac{\sqrt{3}}{2}) + (\frac{1}{\sqrt{2}} 	imes \frac{1}{2})$$= \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$$= \frac{\sqrt{3} + 1}{2\sqrt{2}}$Rationalizing the denominator by multiplying by $\frac{\sqrt{2}}{\sqrt{2}}$:$= \frac{\sqrt{2}(\sqrt{3} + 1)}{2 	imes 2} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

$\cos \frac{\pi}{12} = ?$

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