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\co

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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:$f(	heta) = 2\cos^2 	heta - 1 + 2\cos 	heta = 2(\cos^2 	heta + \cos 	heta) - 1$.Completing the square:$f(	heta) = 2(\cos^2 	heta + \cos 	heta + \frac{1}{4}) - 1 - \frac{1}{2} = 2(\cos 	heta + \frac{1}{2})^2 - \frac{3}{2}$.Since $-1 \le \cos 	heta \le 1$,the range of $(\cos 	heta + \frac{1}{2})$ is $[-\frac{1}{2}, \frac{3}{2}]$.Thus,the range of $(\cos 	heta + \frac{1}{2})^2$ is $[0, \frac{9}{4}]$.Therefore,the range of $2(\cos 	heta + \frac{1}{2})^2 - \frac{3}{2}$ is $[2(0) - \frac{3}{2}, 2(\frac{9}{4}) - \frac{3}{2}] = [-\frac{3}{2}, \frac{9}{2} - \frac{3}{2}] = [-\frac{3}{2}, 3]$.So,the expression is always greater than or equal to $-\frac{3}{2}$ and less than or equal to $3$.

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

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