Question

\[ \sin^6\theta+\cos^6\theta+3\sin^2\theta\cos^2\theta= \]

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(-1\)

Hint:

Use \(\sin^2\theta+\cos^2\theta=1\) and convert higher powers into algebraic identities.

Answer 1

The Correct Option is B

Step 1: Let \[ a=\sin^2\theta,\qquad b=\cos^2\theta \] Then, \[ a+b=\sin^2\theta+\cos^2\theta=1 \]

Step 2: The given expression becomes: \[ a^3+b^3+3ab \]

Step 3: Use identity: \[ a^3+b^3=(a+b)^3-3ab(a+b) \]

Step 4: Since \(a+b=1\), \[ a^3+b^3=1^3-3ab(1)=1-3ab \]

Step 5: Therefore, \[ a^3+b^3+3ab=1-3ab+3ab=1 \] \[ \boxed{1} \]