Question

proof that sin 6 theta + cos 6 theta = 1 - 3 sin square theta *cos squared theta​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

sin6θ+cos6θ+3sin2θcos2θ=1

Given:

\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1sin6θ+cos6θ+3sin2θcos2θ=1

To prove:

\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1sin6θ+cos6θ+3sin2θcos2θ=1

Proof:  

LHS

\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \thetasin6θ+cos6θ+3sin2θcos2θ

As we see that \sin ^{6} \theta+\cos ^{6} \thetasin6θ+cos6θ  is given so converting it into \left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}(sin2θ)3+(cos2θ)3

Using the value of \left(a^{3}+b^{3}\right)(a3+b3) we get the value as \left(a^{3}+b^{3}\right)=a^{3}+b^{3}+3 a b(a+b)(a3+b3)=a3+b3+3ab(a+b)

Using the value of \sin ^{2} \theta=a, \cos ^{2} \theta=bsin2θ=a,cos2θ=b

We get the solution as \left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{3}(sin2θ+cos2θ)3

Converting the value of \sin ^{2} \theta+\cos ^{2} \theta=1sin2θ+cos2θ=1, we get  

\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{3}=(1)^{3}=1=R H S(sin2θ+cos2θ)3=(1)3=1=RHS

\bold{\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1}sin6θ+cos6θ+3sin2θcos2θ=1

Therefore LHS=RHSLHS=RHS

Hence. Proved.