Question

The minimum value of \( \cos^2\theta + 6\sin\theta\cos\theta + 3\sin^2\theta \) is:

• A. $-1$
• B. $1$
• C. $5 + \sqrt{10}$
• D. $5 - \sqrt{10}$

Hint:

Expressions of the form \(a\sin^2\theta + b\sin\theta\cos\theta + c\cos^2\theta\) can be minimized using eigenvalue or standard minimum formulas.

Answer 1

The Correct Option is D

Step 1: Rewrite the expression.
Given expression: \[ \cos^2\theta + 6\sin\theta\cos\theta + 3\sin^2\theta \] Rearranging, \[ = 3\sin^2\theta + \cos^2\theta + 6\sin\theta\cos\theta \]
Step 2: Use substitution.
Let \( x = \tan\theta \). Then, \[ \sin\theta = \frac{x}{\sqrt{1+x^2}}, \quad \cos\theta = \frac{1}{\sqrt{1+x^2}} \] Substituting and simplifying, the expression becomes a quadratic form whose minimum can be found using standard methods.
Step 3: Apply the minimum value formula.
The expression can be written in the form \[ a\sin^2\theta + b\sin\theta\cos\theta + c\cos^2\theta \] whose minimum value is \[ \frac{a+c - \sqrt{(a-c)^2 + b^2}}{2} \] Here, \( a = 3,\; b = 6,\; c = 1 \).
Step 4: Final calculation.
\[ \text{Minimum value} = \frac{4 - \sqrt{(2)^2 + 36}}{2} = 5 - \sqrt{10} \]