prove that
sin^6 thita + cos^6 thita + 3 sin^2 ×cos^2 = 1
Answers
Step-by-step explanation:
sin
6
θ+cos
6
θ+3sin
2
θcos
2
θ=1
Given:
\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1sin
6
θ+cos
6
θ+3sin
2
θcos
2
θ=1
To prove:
\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1sin
6
θ+cos
6
θ+3sin
2
θcos
2
θ=1
Proof:
LHS
\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \thetasin
6
θ+cos
6
θ+3sin
2
θcos
2
θ
As we see that \sin ^{6} \theta+\cos ^{6} \thetasin
6
θ+cos
6
θ is given so converting it into \left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}(sin
2
θ)
3
+(cos
2
θ)
3
Using the value of \left(a^{3}+b^{3}\right)(a
3
+b
3
) we get the value as \left(a^{3}+b^{3}\right)=a^{3}+b^{3}+3 a b(a+b)(a
3
+b
3
)=a
3
+b
3
+3ab(a+b)
Using the value of \sin ^{2} \theta=a, \cos ^{2} \theta=bsin
2
θ=a,cos
2
θ=b
We get the solution as \left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{3}(sin
2
θ+cos
2
θ)
3
Converting the value of \sin ^{2} \theta+\cos ^{2} \theta=1sin
2
θ+cos
2
θ=1 , we get
\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{3}=(1)^{3}=1=R H S(sin
2
θ+cos
2
θ)
3
=(1)
3
=1=RHS
\bold{\sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1}sin
6
θ+cos
6
θ+3sin
2
θcos
2
θ=1
Therefore LHS=RHSLHS=RHS
Hence. Proved.
