Question

Minimum value of the expression ${\cos }^2\theta -\left(6\sin \theta \cos \theta \right)+3{\sin }^2\theta +2,$ is

• A. $4+\sqrt{10}$
• B. $4-\sqrt{10}$
• C. $0$
• D. $4$

Answer 1

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

${\cos }^2\theta -\left(6\sin \theta \cos \theta \right)+3{\sin }^2\theta +2$

$1+2{\sin }^2\theta -6\sin \theta \cos \theta +2$

$\therefore$ $\left[\begin{array}{c}{\sin }^2\theta =\left(\frac{1-\cos 20}{2}\right)\\ 2\sin \theta \cos \theta =\sin 2\theta \end{array}\right]$

$\frac{(1-\cos 2\theta )\times 2}{2}-3\left(\sin 2\theta \right)+2$

$4-(\cos 2\theta +2\sin 2\theta )$

$\sqrt{(1-1{)}^2+(3{)}^2}\le \cos 2\theta +2\sin 2\theta \le \sqrt{(a{)}^2+(3{)}^2}$

$-\sqrt{10}\le \cos 2\theta +3\sin 2\theta \le \sqrt{10}$

$a\le x\le b$

$-b\le -x\le -a$

$4-\sqrt{10}\le 4-(\cos 2\theta +3\sin 2\theta )\le 4+\sqrt{10}$

So, $4-\sqrt{10}$ is minimum value.