Question

sin6q +cos6q=1-3sin2qcos2q

Answer 1

Step-by-step explanation:

hope you will understand

Answer 2

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

Step-by-step explanation:

Step-by-step explanation:

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

L.H.S. = $(\sin^2 \theta)^3+(\cos^2\theta)^3$

Using the identity,

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

⇒ $a^{3} +b^{3}=(a+b)^{3} -3ab(a+b)$

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

Using the trigonometric identity,

$\sin^2 A+\cos^2 A=1$

$=(1)^3 -3\sin^2\theta\cos^2\theta(1)$

$=1-3\sin^2 \theta \cos^2 \theta$

= L.H.S., proved.

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