Question

Prove That

Cos³2A+3Cos2A=4(cos^6 A-sin^6A)

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Answer 1

Answer:

Making assumptions about where parentheses should be,

cos^3(2A) + 3cos(2A) =

cos(2A)(cos^2(2A) + 3) =

(cos^2A - sin^2A)((cos^2A - sin^2A)^2 + 3) =

(cos^2A - sin^2A)(cos^4A - 2cos^2Asin^2A + sin^4A + 3) =

cos^6A - 2cos^4Asin^2A + cos^2Asin^4A + 3cos^2A - cos^4Asin^2A + 2cos^2Asin^4A - sin^6A - 3sin^2A =

cos^6A - 3cos^4Asin^2A + 3cos^2Asin^4A + 3cos^2A - 3sin^2A - sin^6A =

cos^6A - 3cos^4Asin^2A - 3sin^2A + 3cos^2Asin^4A + 3cos^2A - sin^6A =

cos^6A - 3sin^2A(cos^4A + 1) + 3cos^2A(sin^4A + 1) - sin^6A =

cos^6A - 3(1 - cos^2A)(cos^4A + 1) + 3(1 - sin^2A)(sin^4A + 1) - sin^6A =

cos^6A - 3(cos^4A + 1 - cos^6A - cos^2A) + 3(sin^4A + 1 - sin^6A - sin^2A) - sin^6A =

cos^6A - 3cos^4A - 3 + 3cos^6A + 3cos^2A + 3sin^4A + 3 - 3sin^6A - 3sin^2A - sin^6A =

4cos^6A - 3cos^4A + 3sin^4A + 3cos^2A - 3sin^2A - 4sin^6A =

4cos^6A - 3(cos^4A - sin^4A) + 3cos^2A - 3sin^2A - 4sin^6A =

4cos^6A - 3(cos^2A + sin^2A)(cos^2A - sin^2A) + 3(cos^2A - sin^2A) - 4sin^6A =

4cos^6A - 3(cos^2A - sin^2A) + 3(cos^2A - sin^2A) - 4sin^6A =

4cos^6A - 4sin^6A