Question

prove that sin power 6 theta + cos 6 theta is equal to 1 - 3 sin squared theta cos squared theta

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\sin^{6}\theta+\cos^{6}\theta=1-3\sin^{2}\theta\cos^{2}\theta$, proved.

Step-by-step explanation:

Prove that, $\sin ^{6} \theta + \cos ^{6} \theta =1-3\sin ^{2} \theta \cos ^{2} \theta$

We have,

$(\sin^{2}\theta+\cos^{2}\theta )^{3}$

Using identity,

$(a+b)^{3} =a^{2}+b^{3}+3ab(a+b)$

$(\sin^{2}\theta+\cos^{2}\theta )^{3} =(\sin^{2}\theta)^{3}+(\cos^{2}\theta)^{3}+3\sin^{2}\theta\cos^{2}\theta(\sin^{2}\theta+\cos^{2}\theta)$

⇒ $(1)^{3} =\sin^{6}\theta+\cos^{6}\theta+3\sin^{2}\theta\cos^{2}\theta(1)$

⇒ $1 =\sin^{6}\theta+\cos^{6}\theta+3\sin^{2}\theta\cos^{2}\theta$

⇒ $\sin^{6}\theta+\cos^{6}\theta=1-3\sin^{2}\theta\cos^{2}\theta$, proved.