Question

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

• A. $-1$
• B. $4 + \sqrt{10}$
• C. $4 - \sqrt{10}$
• D. 1

Hint:

Minimum value of $a\cos x + b\sin x + c$ is $c - \sqrt{a^2+b^2}$.

Answer 1

The Correct Option is C

We convert squared terms to double angles: $\cos^2\theta = \frac{1+\cos2\theta}{2}$, $\sin^2\theta = \frac{1-\cos2\theta}{2}$.
Also, $2\sin\theta\cos\theta = \sin2\theta$.
Let $f(\theta) = \frac{1+\cos2\theta}{2} - 3\sin2\theta + 3\left(\frac{1-\cos2\theta}{2}\right) + 2$.
$f(\theta) = \frac{1}{2} + \frac{1}{2}\cos2\theta - 3\sin2\theta + \frac{3}{2} - \frac{3}{2}\cos2\theta + 2$.
Combine constants: $\frac{1}{2} + \frac{3}{2} + 2 = 4$.
Combine cosine coefficients: $\frac{1}{2} - \frac{3}{2} = -1$.
$f(\theta) = 4 - \cos2\theta - 3\sin2\theta$.
The range of $-\cos2\theta - 3\sin2\theta$ is $[-\sqrt{(-1)^2+(-3)^2}, \sqrt{(-1)^2+(-3)^2}] = [-\sqrt{10}, \sqrt{10}]$.
The minimum value is $4 - \sqrt{10}$.