p.t.: cos³2x +3cos2x= 4(cos^6x-sin^6x)
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Answered by
39
we have to prove that,
cos³(2x) + 3cos(2x) = 4(cos^6x - sin^6x)
LHS = cos³(2x) + 3cos(2x)
we know, cos2x = 2cos²x - 1
then, cos³(2x) + 3cos(2x) = (2cos²x - 1)³ + 3(2cos²x - 1)
= (2cos²x)³ - 3(2cos²x)² + 3(2cos²x).1² - 1 + 6cos²x - 3
= 8cos^6x - 12cos⁴x + 6cos²x + 6cos²x - 4
= 8cos^6x - 12cos⁴x + 12cos²x - 4
= 8cos^6x - 12cos²x(cos²x - 1) - 4
= 8cos^6x - 12(1 - sin²x) sin²x - 4
= 4cos^6x + 4cos^6x - 12sin²x + 12sin⁴x - 4
= 4cos^6x + 4(1 - sin²x)³ - 12sin²x + 12sin⁴x - 4
= 4cos^6x + 4(1 - 3sin⁴x + 3sin²x - sin^6x) - 12sin²x + 12sin⁴x - 4
= 4cos^6x + 4 - 12sin⁴x + 12sin²x - 4sin^6x - 12sin²x + 12sin⁴x - 4
= 4cos^6x - 4sin^6x
= 4(cos^6x - sin^6x) = RHS
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