Question
Prove the following identities:
$\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$
$\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$
✓
Answer
To prove $\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$
We know, $\text{a}^3+\text{b}^3=(\text{a}+\text{b})^3-3\text{ab}(\text{a}+\text{b})$
Put $\text{a}=\sin^2\theta,\text{b}=\cos^2\theta$
$\therefore\sin^6\theta+\cos^6\theta=\big(\sin^2\theta+\cos^2\theta\big)^3\\ \ \ -3\sin^2\theta\cos^2\theta\times\big(\sin^2\theta+\cos^2\theta\big)$
$=1-3\sin^2\theta\cos^2\theta$
$=\text{R.H.S}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
We know, $\text{a}^3+\text{b}^3=(\text{a}+\text{b})^3-3\text{ab}(\text{a}+\text{b})$
Put $\text{a}=\sin^2\theta,\text{b}=\cos^2\theta$
$\therefore\sin^6\theta+\cos^6\theta=\big(\sin^2\theta+\cos^2\theta\big)^3\\ \ \ -3\sin^2\theta\cos^2\theta\times\big(\sin^2\theta+\cos^2\theta\big)$
$=1-3\sin^2\theta\cos^2\theta$
$=\text{R.H.S}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$
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