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

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

Vedclass · Accepted Answer

$4-\sqrt{10}$

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(C) Let $f(	heta) = cos^{2}	heta - 6sin	heta cos	heta + 3sin^{2}	heta + 2$.Using the identities $cos^{2}	heta = \frac{1+cos 2	heta}{2}$,$sin^{2}	heta = \frac{1-cos 2	heta}{2}$,and $2sin	heta cos	heta = sin 2	heta$:$f(	heta) = \frac{1+cos 2	heta}{2} - 3sin 2	heta + 3(\frac{1-cos 2	heta}{2}) + 2$$f(	heta) = \frac{1}{2} + \frac{1}{2}cos 2	heta - 3sin 2	heta + \frac{3}{2} - \frac{3}{2}cos 2	heta + 2$$f(	heta) = 4 - 3sin 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,$-3sin 2	heta - cos 2	heta \in [-\sqrt{10}, \sqrt{10}]$.Therefore,$f(	heta) \in [4-\sqrt{10}, 4+\sqrt{10}]$.The least value is $4-\sqrt{10}$.

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

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