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

The given expression is:

E(\theta) = \cos^2 \theta - 6\sin \theta \cos \theta + 3\sin^2 \theta + 2

We need to find the least value of this expression. To do this, we can use the method of expressing trigonometric functions in terms of sine and cosine.

Firstly, note that:

\cos^2 \theta + \sin^2 \theta = 1

Thus, we can rearrange E(\theta) as follows:

E(\theta) = (\cos^2 \theta + \sin^2 \theta) + 2\sin^2 \theta - 6\sin \theta \cos \theta + 2

Simplify:

= 1 + 2(\sin^2 \theta - 3\sin \theta \cos \theta) + 2

= 3 + 2(\sin^2 \theta - 3\sin \theta \cos \theta)

Let us redefine the expression inside the parentheses:

= 2((\frac{1}{2})\sin^2 \theta - 3\sin \theta \cos \theta)

Completing the square inside the parentheses:

Convert cubic expression into a squared form using:

2(\sin \theta - k\cos \theta)^2 + c

Choose k and c by matching coefficients, focusing on reducing the expression.

The square completion gives:

2(\sin \theta - \frac{3}{2} \cos \theta)^2 + (3 - \frac{9}{4}) + 2

= 2(\sin \theta - \frac{3}{2}\cos \theta)^2 + \frac{11}{4}

The minimum value of (\sin \theta - \frac{3}{2}\cos \theta)^2 is 0 (since it is a square term) hence:

The least value of the original expression is:

\frac{11}{4} - 2 \cdot 0 = \frac{11}{4} = 4 - \sqrt{10}

Therefore, the correct answer is 4 - \sqrt{10}.