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 :
$$4 - \sqrt {10} $$
Solution :
The expression $$ = 2{\sin ^2}\theta - 3\sin 2\theta + 3 = 4 - \left( {\cos 2\theta + 3\sin 2\theta } \right)$$
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 - \sqrt {{1^2} + {3^2}} \left( {\frac{1}{{\sqrt {10} }}\cos 2\theta + \frac{3}{{\sqrt {10} }}\sin 2\theta } \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 - \sqrt {10} \cos \left( {2\theta - \alpha } \right),\,{\text{where}}\,\,\cos \alpha = \frac{1}{{\sqrt {10} }}. \cr} $$
∴ the least value $$ = 4 - \sqrt {10} \times $$ the greatest value of $$\cos \left( {2\theta - \alpha } \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 - \sqrt {10} \times 1 = 4 - \sqrt {10} .$$