The least value of ( cos 2 - 6 sin cos + 3 sin 2 + 2 ) is [2026]

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Solution

Q.
The least value of $(\cos^{2} θ - 6 \sin θ \cos θ + 3 \sin^{2} θ + 2)$ is [2026]

1 $4 - \sqrt{10}$

2 $4 + \sqrt{10}$

3 1

4 -1

Ans.
(1)

$f (θ) = \frac{1 + \cos 2 θ}{2} - 3 \sin 2 θ + 3 (\frac{1 - \cos 2 θ}{2}) + 2$

$f (θ) = 4 - 3 \sin 2 θ - \cos 2 θ$

$f (θ) \in [4 - \sqrt{10}, 4 + \sqrt{10}]$

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