Question : If $\sin \theta \cos \theta=\frac{\sqrt{2}}{3}$,then the value of $\left(\sin ^6 \theta+\cos ^6 \theta\right)$ is:
Option 1: $\frac{1}{3}$
Option 2: $\frac{4}{3}$
Option 3: $\frac{2}{3}$
Option 4: $\frac{5}{3}$
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Correct Answer: $\frac{1}{3}$
Solution :
Given: $\sin \theta \cos \theta=\frac{\sqrt{2}}{3}$
$\left(\sin ^6 \theta+\cos ^6 \theta\right)=(\sin^2 \theta +\cos^2 \theta)(\sin^4 \theta + \cos^4 \theta-\sin^2 \theta\cos^2 \theta)$
⇒ $\left(\sin ^6 \theta+\cos ^6 \theta\right)=(\sin^2 \theta +\cos^2 \theta)((\sin^2 \theta + \cos^2 \theta)^2-3\sin^2 \theta\cos^2 \theta)$
⇒ $\left(\sin ^6 \theta+\cos ^6 \theta\right)=(1)((1)^2-3(\frac{\sqrt{2}}{3})^2)$
⇒ $\left(\sin ^6 \theta+\cos ^6 \theta\right)=(1-\frac{{2}}{3})$
⇒ $\left(\sin ^6 \theta+\cos ^6 \theta\right)=\frac{1}{3}$
Hence, the correct answer is $\frac{1}{3}$.
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