The maximum value of left(2 cos^2 18^{circ} - | vedclass

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(D) Let $E = \left(2 \cos^2 18^{\circ} - \sin 18^{\circ}ight) \left(\cos 	heta + 3 \sqrt{2} \cos \left(	heta + \frac{\pi}{4}ight) + 3ight)$.Fi

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) Let $E = \left(2 \cos^2 18^{\circ} - \sin 18^{\circ}ight) \left(\cos 	heta + 3 \sqrt{2} \cos \left(	heta + \frac{\pi}{4}ight) + 3ight)$.First,simplify the constant term: $2 \cos^2 18^{\circ} - \sin 18^{\circ} = (1 + \cos 36^{\circ}) - \sin 18^{\circ}$.Using $\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$,we get $1 + \frac{\sqrt{5} + 1}{4} - \frac{\sqrt{5} - 1}{4} = 1 + \frac{2}{4} = \frac{3}{2}$.Now,simplify the second term: $\cos 	heta + 3 \sqrt{2} \left(\cos 	heta \cos \frac{\pi}{4} - \sin 	heta \sin \frac{\pi}{4}ight) + 3$.$= \cos 	heta + 3 \sqrt{2} \left(\cos 	heta \cdot \frac{1}{\sqrt{2}} - \sin 	heta \cdot \frac{1}{\sqrt{2}}ight) + 3$.$= \cos 	heta + 3 \cos 	heta - 3 \sin 	heta + 3 = 4 \cos 	heta - 3 \sin 	heta + 3$.Thus,$E = \frac{3}{2} (4 \cos 	heta - 3 \sin 	heta + 3)$.The maximum value of $a \cos 	heta + b \sin 	heta$ is $\sqrt{a^2 + b^2}$.Here,the maximum value of $4 \cos 	heta - 3 \sin 	heta$ is $\sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$.Therefore,the maximum value of $E = \frac{3}{2} (5 + 3) = \frac{3}{2} 	imes 8 = 12$.

The maximum value of $\left(2 \cos^2 18^{\circ} - \sin 18^{\circ}\right) \left(\cos \theta + 3 \sqrt{2} \cos \left(\theta + \frac{\pi}{4}\right) + 3\right)$ is

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