Question

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

• A. $4+\sqrt{10}$
• B. $4-\sqrt{10}$
• C. $0$
• D. none of these.

Answer 1

The correct option is B $4-\sqrt{10}$

$={\cos }^2\theta -6\sin \theta \cos \theta +3{\sin }^2\theta +2$
$=-6\sin \theta \cos \theta +2{\sin }^2\theta +3$
$f\left(x\right)=-6\sin \theta \cos \theta +2{\sin }^2\theta +3$
$f^{\prime }\left(x\right)=-6({\cos }^2\theta -{\sin }^2\theta )+4\sin \theta \cos \theta =0$
$\cos 2\theta =\frac{\sin 2\theta }{3}$
$\tan 2\theta =3$
$\Rightarrow \sin 2\theta =\frac{3}{\sqrt{10}}$
$\Rightarrow \cos 2\theta =\frac{1}{\sqrt{10}}$
these are the minimum values
putting it back in $f\left(x\right)$,
$f\left(x\right)=\frac{-9-1}{\sqrt{10}}+4=4-\sqrt{10}$