The least value of (cos^{2}theta - 6sintheta | vedclass

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Question

(C) Let $f(	heta) = \cos^{2}	heta - 6\sin	heta \cos	heta + 3\sin^{2}	heta + 2$.Using the identities $\cos^{2}	heta = \frac{1+\cos 2	heta}{2}$,

Vedclass · Accepted Answer

$4-\sqrt{10}$

Vedclass · Answer

(C) Let $f(	heta) = \cos^{2}	heta - 6\sin	heta \cos	heta + 3\sin^{2}	heta + 2$.Using the identities $\cos^{2}	heta = \frac{1+\cos 2	heta}{2}$, $\sin^{2}	heta = \frac{1-\cos 2	heta}{2}$, and $2\sin	heta \cos	heta = \sin 2	heta$:$f(	heta) = \frac{1+\cos 2	heta}{2} - 3\sin 2	heta + 3(\frac{1-\cos 2	heta}{2}) + 2$$f(	heta) = \frac{1}{2} + \frac{1}{2}\cos 2	heta - 3\sin 2	heta + \frac{3}{2} - \frac{3}{2}\cos 2	heta + 2$$f(	heta) = 4 - 3\sin 2	heta - \cos 2	heta$The expression $a \sin x + b \cos x$ lies in the interval $[-\sqrt{a^{2}+b^{2}}, \sqrt{a^{2}+b^{2}}]$.Here, $a = -3$ and $b = -1$, so $\sqrt{(-3)^{2} + (-1)^{2}} = \sqrt{9+1} = \sqrt{10}$.Thus, $-3\sin 2	heta - \cos 2	heta \in [-\sqrt{10}, \sqrt{10}]$.Therefore, $f(	heta) \in [4-\sqrt{10}, 4+\sqrt{10}]$.The least value is $4-\sqrt{10}$.

The least value of $(\cos^{2}\theta - 6\sin\theta \cos\theta + 3\sin^{2}\theta + 2)$ is:

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