If A = frac{pi}{24},then frac{cos A + cos 3A | vedclass

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Question

(A) Let $I = \frac{\cos A + \cos 3A + \cos 5A + \cos 7A}{\sin A + \sin 3A + \sin 5A + \sin 7A}$.Grouping the terms in the numerator and denominator:$I

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(A) Let $I = \frac{\cos A + \cos 3A + \cos 5A + \cos 7A}{\sin A + \sin 3A + \sin 5A + \sin 7A}$.Grouping the terms in the numerator and denominator:$I = \frac{(\cos 7A + \cos A) + (\cos 5A + \cos 3A)}{(\sin 7A + \sin A) + (\sin 5A + \sin 3A)}$.Using the sum-to-product formulas $\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$ and $\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$:$I = \frac{2\cos 4A \cos 3A + 2\cos 4A \cos A}{2\sin 4A \cos 3A + 2\sin 4A \cos A}$.Factoring out common terms:$I = \frac{2\cos 4A (\cos 3A + \cos A)}{2\sin 4A (\cos 3A + \cos A)}$.$I = \frac{\cos 4A}{\sin 4A} = \cot 4A$.Given $A = \frac{\pi}{24}$,then $4A = 4 	imes \frac{\pi}{24} = \frac{\pi}{6}$.$I = \cot\frac{\pi}{6} = \sqrt{3}$.

If $A = \frac{\pi}{24}$,then $\frac{\cos A + \cos 3A + \cos 5A + \cos 7A}{\sin A + \sin 3A + \sin 5A + \sin 7A} = $

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