cos 2theta + 2cos theta is always | vedclass

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(C) Let $f(	heta) = \cos 2	heta + 2\cos 	heta$. Using the identity $\cos 2	heta = 2\cos^2 	heta - 1$,we get: $f(	heta) = 2\cos^2 	heta - 1 + 2\

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Greater than or equal to $-\frac{3}{2}$ and less than or equal to $3$

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(C) Let $f(	heta) = \cos 2	heta + 2\cos 	heta$. Using the identity $\cos 2	heta = 2\cos^2 	heta - 1$,we get: $f(	heta) = 2\cos^2 	heta - 1 + 2\cos 	heta$ $f(	heta) = 2(\cos^2 	heta + \cos 	heta) - 1$ Completing the square: $f(	heta) = 2(\cos^2 	heta + \cos 	heta + \frac{1}{4}) - 1 - \frac{1}{2}$ $f(	heta) = 2(\cos 	heta + \frac{1}{2})^2 - \frac{3}{2}$ Since $-1 \le \cos 	heta \le 1$,the minimum value occurs when $\cos 	heta = -\frac{1}{2}$,giving $f(	heta) = -\frac{3}{2}$. The maximum value occurs at the boundaries of $\cos 	heta$,i.e.,$\cos 	heta = 1$,giving $f(	heta) = 2(1 + \frac{1}{2})^2 - \frac{3}{2} = 2(\frac{9}{4}) - \frac{3}{2} = \frac{9}{2} - \frac{3}{2} = 3$. Thus,$-\frac{3}{2} \le \cos 2	heta + 2\cos 	heta \le 3$.

$\cos 2\theta + 2\cos \theta$ is always

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