Maximum value of 5 cos theta + 3 cos left( th | vedclass

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(B) Given expression: $5 \cos 	heta + 3 \cos \left( 	heta + \frac{\pi}{3} ight) - 1$Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin

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$6$

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(B) Given expression: $5 \cos 	heta + 3 \cos \left( 	heta + \frac{\pi}{3} ight) - 1$Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:$= 5 \cos 	heta + 3 \left( \cos 	heta \cdot \cos \frac{\pi}{3} - \sin 	heta \cdot \sin \frac{\pi}{3} ight) - 1$Substituting $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$:$= 5 \cos 	heta + 3 \left( \frac{1}{2} \cos 	heta - \frac{\sqrt{3}}{2} \sin 	heta ight) - 1$$= 5 \cos 	heta + \frac{3}{2} \cos 	heta - \frac{3\sqrt{3}}{2} \sin 	heta - 1$$= \frac{13}{2} \cos 	heta - \frac{3\sqrt{3}}{2} \sin 	heta - 1$The expression is of the form $a \cos 	heta + b \sin 	heta + c$,where the maximum value is $\sqrt{a^2 + b^2} + c$.Here,$a = \frac{13}{2}$,$b = -\frac{3\sqrt{3}}{2}$,and $c = -1$.Maximum value $= \sqrt{\left( \frac{13}{2} ight)^2 + \left( -\frac{3\sqrt{3}}{2} ight)^2} - 1$$= \sqrt{\frac{169}{4} + \frac{27}{4}} - 1$$= \sqrt{\frac{196}{4}} - 1$$= \sqrt{49} - 1 = 7 - 1 = 6$.

Maximum value of $5 \cos \theta + 3 \cos \left( \theta + \frac{\pi}{3} \right) - 1$ is

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