frac{sqrt{2} cos 45^{circ}+cos 56^{circ}+cos | vedclass

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Question

(A) Let $A = \frac{\sqrt{2} \cos 45^{\circ} + \cos 56^{\circ} + \cos 58^{\circ} - \cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) Let $A = \frac{\sqrt{2} \cos 45^{\circ} + \cos 56^{\circ} + \cos 58^{\circ} - \cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}}$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the numerator becomes $1 + \cos 56^{\circ} + \cos 58^{\circ} - \cos 66^{\circ}$.Using $\cos 58^{\circ} - \cos 66^{\circ} = 2 \sin 62^{\circ} \sin 4^{\circ}$ is not helpful,so we rewrite the expression as $(1 - \cos 66^{\circ}) + (\cos 56^{\circ} + \cos 58^{\circ})$.Using $1 - \cos 66^{\circ} = 2 \sin^2 33^{\circ}$ and $\cos 56^{\circ} + \cos 58^{\circ} = 2 \cos 57^{\circ} \cos 1^{\circ}$.Note that $\sin 33^{\circ} = \cos 57^{\circ}$.So,the numerator is $2 \sin^2 33^{\circ} + 2 \cos 57^{\circ} \cos 1^{\circ} = 2 \cos 57^{\circ} (\sin 33^{\circ} + \cos 1^{\circ})$.Since $\sin 33^{\circ} = \cos 57^{\circ}$,this is $2 \cos 57^{\circ} (\cos 57^{\circ} + \cos 1^{\circ})$.Using $2 \cos 28^{\circ} \cos 29^{\circ} = \cos 57^{\circ} + \cos 1^{\circ}$.The expression becomes $A = \frac{2 \cos 57^{\circ} (\cos 57^{\circ} + \cos 1^{\circ})}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}}$.Since $\sin 33^{\circ} = \cos 57^{\circ}$,we have $A = \frac{2 (\cos 57^{\circ} + \cos 1^{\circ})}{\sqrt{2} (\cos 57^{\circ} + \cos 1^{\circ})} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

$\frac{\sqrt{2} \cos 45^{\circ}+\cos 56^{\circ}+\cos 58^{\circ}-\cos 66^{\circ}}{\sqrt{2} \cos 28^{\circ} \cos 29^{\circ} \sin 33^{\circ}} = $

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