f4(x)=41(sin4x+cos4x)
=41[(sin2x+cos2x)2−2sin2xcos2x]
=41[1−2sin2xcos2x]
f6(x)=61[sin6x+cos6x]
=61[(sin2x+cos2x)3−3sin2xcos2x(sin2x+cos2x)]
=61[1−3sin2xcos2x(1)]
f4(x)−f6(x)=41−61
=121
Let fk(x)=k1(sinkx+coskx) where x∈R and k≥1. Then f4(x)−f6(x) equals
Held on 6 Apr 2014 · Verified 6 Jul 2026.
41
121
61
31
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