Trigonometry

Question

Prove that cos⁶x + sin⁶x = (5+3cos4x)/8

 

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Answer 1

cos⁶x + sin⁶x = (cos²x)³ + sin⁶x

                    = (1-sin²x)³ + sin⁶x            note: (a-b)³ = a³ - 3a²b +3ab² - b³, here a=1 and b=sin²x

                    = 1 - 3sin²x + 3sin⁴x - sin⁶x + sin⁶x

                    = 1 - 3sin²x(1 - sin²x)

                    = 1 - 3sin²xcos²x                                                   sin2a = 2sinacosa

                    = 1 - 3(sin2x/2)²

                    = 1 - 3sin²2x/4                                                     sin²a = (1-cos2a)/2

                    = 1 - 3((1 - cos4x)/2)/4

                    = 1 - 3/8 + 3cos4x/8

                    =5/8 +3cos4x/8

cos⁶x + sin⁶x = (5 + 3cos4x)/8