Prove the following identities (1-16)
Question:

Prove the following identities (1-16)

$\sin ^{6} x+\cos ^{6} x=1-3 \sin ^{2} x \cos ^{2} x$

Solution:

$\mathrm{LHS}=\sin ^{6} x+\cos ^{6} x$

$=\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}$

$=\left(\sin ^{2} x+\cos ^{2} x\right)\left[\left(\sin ^{2} x\right)^{2}+\left(\cos ^{2} x\right)^{2}-\sin ^{2} x \cos ^{2} x\right]$

$\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$

$=1 \times\left[\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x-\sin ^{2} x \cos ^{2} x\right]$

$\left[\because \sin ^{2} x+\cos ^{2} x=1\right.$ and $\left.a^{2}+b^{2}=(a+b)^{2}-2 a b\right]$

$=1^{2}-3 \sin ^{2} x \cos ^{2} x$

$=1-3 \sin ^{2} x \cos ^{2} x$

$=\mathrm{RHS}$

Hence proved.

 

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