Question

prove that
sin ^8 x- cos ^8x = (sin^2 x - cos^2 x) (1 - 2 sin^2x.cos^2x)​

Answer 1

Step-by-step explanation:

L.H.S

sin^8x − cos^8x

⇒(sin⁴x)² − (cos⁴x)²

=>(sin⁴x − cos⁴x)(sin⁴x+cos⁴x) ∵(a²−b²)=(a−b(a+b)

=> [(sin²x)²−(cos²x)²][(sin²x+cos²x)²−2sin² x cos²x) ∵(a+b)²−2ab=a²+b²

=> [(sin²x−cos²x)(sin²x+cos²x)][(sin²x+cos²x)²−2sin²xcos²x)]

=> [(sin²x−cos²x)(1))((1)²−2sin²xcos²x) ∵sin²x+cos²x=1

= (sin²x−cos²x)(1−2sin²xcos²x)

R.H.S

Hence, proved.

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Answer 2

Answer:

Please don't post useless answers....