cos^8a-sin^8a=1/8cos2a(3+cos4a)
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Question: cos^8a-sin^8a=1/4 cos2a(3+cos4a)
Answer: LHS= (cos^4A)^2 - (sin^4A)^2
=(cos^4A+sin^4A) (cos^4A-sin^4A)
=[(cos^2A)^2 + (sin^2A) ] [(cos^2A)^2 - (sin^2A)]
=[(cos^2A +sin^2A)^2 - 2cos^2Asin^2A] [(cos^2A-sin^2A) (cos^2A +sin^2A)]
=[1 - 2 cos^2A sin^2A] cos2A
=1/4 x 4 [1 - 2 cos^2A sin^2A] cos2A
=1/4 [4 - 2.(2 cosA sinA)^2]cos2A
=1/4 [4-2sin^2 2A] cos2A
=1/4 [4-2(1-cos^2 A)]cos2A
=1/4 [4-2+2cos^2 A]cos2A
= 1/4 [2+2cos^2 2A]cos2A
=1/4 [2+(cos4A +1)]cos2A
=1/4 [2+1+ cos 4A]cos2A
=1/4[3+cos4A]cos2A
=1/4 cos2A [3+cos4A]
=RHS
Hence, RHS =LHS
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