Question

sin raise to 8 theta minus Cos raise to 8 theta equal to 1 - 2 cos square theta into 1 minus 2 sin square theta cos square theta​

Answer 1

Answer:

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

Step-by-step explanation:

$Formula\:used:\\\\a^2-b^2=(a+b)(a-b)$

Now,

$sin^8\theta-cos^8\theta\\\\=(sin^4\theta)^2-(cos^4\theta)^2\\\\=(sin^4\theta+cos^4\theta)(sin^4\theta-cos^4\theta)\\\\=[(sin^2\theta)^2+(cos^2\theta)^2][(sin^2\theta)^2-(cos^2\theta)^2]\\\\=[(sin^2\theta+cos^2\theta)^2-2.sin^2\theta.cos^2\theta](sin^2\theta+cos^2\theta)(sin^2\theta-cos^2\theta)\\\\=[(1)^2-2.sin^2\theta.cos^2\theta](1)(sin^2\theta-cos^2\theta)\\\\=(1-2.sin^2\theta.cos^2\theta)(1-cos^2\theta-cos^2\theta)\\\\=(1-2.sin^2\theta.cos^2\theta)(1-2cos^2\theta)$

Answer 2

Answer:

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Step-by-step explanation:

sin8θ+cos8θ

=(sin4θ+cos4θ)2-2sin4θcos4θ

=((sin2θ+cos2θ)2-2sin2θcos2θ)2-2sin4θcos4θ

=(1-2sin2θcos2θ)2-2sin4θcos4θ

=1+4sin4θcos4θ-4sin2θcos2θ-2sin4θcos4θ

=2sin4θcos4θ-4sin2θcos2θ+1

=2sin2θcos2θ(sin2θcos2θ-2)+1.