Question

For any \(\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), the expression
\(3(\cos\theta - \sin\theta)^4 + 6(\sin\theta + \cos\theta)^2 + 4\sin^6\theta\)
equals :

• A. \(13 - 4\cos^2\theta + 6\sin^2\theta\cos^2\theta\)
• B. \(13 - 4\cos^6\theta\)
• C. \(13 - 4\cos^2\theta + 6\cos^2\theta\)
• D. \(13 - 4\cos^4\theta + 2\sin^2\theta\cos^2\theta\)

Hint:

Expand \((\cos\theta - \sin\theta)^4\) using \(s = \sin\theta, c = \cos\theta\) and use \(s^6 + c^6 = 1 - 3s^2c^2\).

Answer 1

The Correct Option is B

Let \(s = \sin\theta,\ c = \cos\theta\), so \(s^2 + c^2 = 1\).
\((c - s)^2 = 1 - 2sc\), so \(3(c - s)^4 = 3(1 - 2sc)^2 = 3 - 12sc + 12s^2c^2\).
\(6(s + c)^2 = 6(1 + 2sc) = 6 + 12sc\).
Sum of the first two terms \(= 9 + 12s^2c^2\).
Now \(s^6 + c^6 = (s^2 + c^2)^3 - 3s^2c^2(s^2 + c^2) = 1 - 3s^2c^2\), so \(4s^6 = 4 - 12s^2c^2 - 4c^6\).
Total \(= 9 + 12s^2c^2 + 4 - 12s^2c^2 - 4c^6 = 13 - 4\cos^6\theta\).
Answer: (2) \(13 - 4\cos^6\theta\)