Question

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

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

Hint:

Use \(\sin^2\theta+\cos^2\theta=1\). Put \(a=\sin^2\theta\), \(b=\cos^2\theta\), then use \(a^3+b^3=(a+b)^3-3ab(a+b)\).

Answer 1

The Correct Option is B

We need to simplify: \[ \sin^6\theta+\cos^6\theta+3\sin^2\theta\cos^2\theta. \] Let \[ a=\sin^2\theta \] and \[ b=\cos^2\theta. \] Then, \[ \sin^6\theta=(\sin^2\theta)^3=a^3 \] and \[ \cos^6\theta=(\cos^2\theta)^3=b^3. \] Also, \[ 3\sin^2\theta\cos^2\theta=3ab. \] So the expression becomes \[ a^3+b^3+3ab. \] Now we know: \[ a+b=\sin^2\theta+\cos^2\theta. \] Using the identity: \[ \sin^2\theta+\cos^2\theta=1. \] Thus, \[ a+b=1. \] Now use the identity: \[ a^3+b^3=(a+b)^3-3ab(a+b). \] Since \(a+b=1\), \[ a^3+b^3=1^3-3ab(1). \] \[ a^3+b^3=1-3ab. \] Now add \(3ab\): \[ a^3+b^3+3ab=(1-3ab)+3ab. \] \[ =1. \] Therefore, \[ \sin^6\theta+\cos^6\theta+3\sin^2\theta\cos^2\theta=1. \]