Question

sin8A-cos8A=(sin4A+cos4A)(1-2cos2A)

Answer 1

Answer:

We have to prove the relation:

\sin^8\theta-\cos^8\theta=(\sin^2\theta-\cos^2\theta)(1-2\sin^2\theta\cos^2\theta)

We will evaluate the left hand side term to obtain the right hand side term.

Step-by-step explanation:

On taking Left hand side:

\sin^8\theta-\cos^8\theta=(\sin^4\theta)^2-(\cos^4\theta)^2\\\\=(\sin^4\theta-\cos^4\theta)(\sin^4\theta+\cos^4\theta)

Since, we know that:

a^2-b^2=(a+b)(a-b)

=[(\sin^2\theta)^2-(\cos^2\theta)^2][(\sin^2\theta)^2+(\cos^2\theta)^2]\\\\=(\sin^2\theta-\cos^2\theta)(\sin^2\theta+\cos^2\theta)[(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta]

Also we know that:

\sin^2\theta+\cos^2\theta=1

and

a^2+b^2=(a+b)^2-2ab )

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

Hence, the expression is proved.

Hence,  we get

\sin^8\theta-\cos^8\theta=(\sin^2\theta-\cos^2\theta)(1-2\sin^2\theta\cos^2\theta)