Question

Sin3A - Cos3a = [ Sin2a – Cos2a ] [ 1 – 2 Sin2a Cos2a​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

You probably typed the question wrong. It's $sin^8-cos^8$

$= (sin^4 A + cos ^4 A)(sin^4 A - cos^4 A)$

$=(sin^4 A + cos ^4 A)(sin^2 A + cos^2 A)(sin^2 A -cos^2 A)$

As we know, sin^2 A +cos^2 A =1

$=(Sin^4 A + cos^4 A)(1)(sin^2 A -cos^2 A)$

We know that sin^2 A + cos^2 A =1

$= (Sin^2 A + cos^2 A)^2=1^2$

$= Sin^4 A + cos^4 A + 2 sin^2 A × cos^2 A = 1$

Putting sin^4 A + cos^4 A = 1 - 2 sin^2 A cos^2 A

$=(1 - 2sin^2 A cos^2 A)(sin^2 A - cos^2 A)$