The value of 5 cos theta + 3 cos left(theta + | vedclass

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(D) Let $f(	heta) = 5 \cos 	heta + 3 \cos \left(	heta + \frac{\pi}{3}ight) + 3$. Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:

Vedclass · Accepted Answer

$-4$ and $10$

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(D) Let $f(	heta) = 5 \cos 	heta + 3 \cos \left(	heta + \frac{\pi}{3}ight) + 3$. Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$: $f(	heta) = 5 \cos 	heta + 3 \left(\cos 	heta \cos \frac{\pi}{3} - \sin 	heta \sin \frac{\pi}{3}ight) + 3$. $f(	heta) = 5 \cos 	heta + 3 \left(\frac{1}{2} \cos 	heta - \frac{\sqrt{3}}{2} \sin 	hetaight) + 3$. $f(	heta) = \left(5 + \frac{3}{2}ight) \cos 	heta - \frac{3\sqrt{3}}{2} \sin 	heta + 3 = \frac{13}{2} \cos 	heta - \frac{3\sqrt{3}}{2} \sin 	heta + 3$. The expression $a \cos 	heta + b \sin 	heta$ ranges between $-\sqrt{a^2 + b^2}$ and $\sqrt{a^2 + b^2}$. Here,$a = \frac{13}{2}$ and $b = -\frac{3\sqrt{3}}{2}$. $\sqrt{a^2 + b^2} = \sqrt{\frac{169}{4} + \frac{27}{4}} = \sqrt{\frac{196}{4}} = \sqrt{49} = 7$. Thus,$-7 \leq \frac{13}{2} \cos 	heta - \frac{3\sqrt{3}}{2} \sin 	heta \leq 7$. Adding $3$ to all parts: $-7 + 3 \leq f(	heta) \leq 7 + 3$. $-4 \leq f(	heta) \leq 10$.

The value of $5 \cos \theta + 3 \cos \left(\theta + \frac{\pi}{3}\right) + 3$ lies between

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