Question

(sinx)^6+(cosx)^6

Answer 1

s^6+c^6=(s^4)-( sc )^2+c^4 =(1-2( sc )^2) - (sc)^2 =1-3(sc)^2 consider cos4x=cos( 2(2x) )=1-2( sin2x )^2=1-2( 2sc )^2=1-8(sc)^2 8(sc)^2 = 1-cos4x (sc)^2 =(1-cos4x)/8 Hence s^6+c^6=1-3(sc)^2 =1-3(1-cos4x)/8 = (5+3cos4x)/8
cos4x =2cos^2(2x)-1 =2[[2cos^4x-1]^2]-1 =8cos^4x-8cos^2x+1 [(sin^2x)+(cos^2x)]^3 =sin^6x+3sin^4xcos^2x+3cos^4xsin^2x+cos^6x So (sinx)^6+(cosx)^6 =1-3sin^4xcos^2x+3cos^4xsin^2x =1-3sin^2xcos^2x =1-3(1-cos^2x)cos^2x =1-3cos^2x+3cos^4x =5/8+3/8(8cos^4x-8cos^2x+1) =(5/8)+(3/8)cos4x