JEE Main 2019TrigonometryTrigonometric Ratio and IdentitesEasyQuestionLet fk(x)=1k(sinkx+coskx)f_k\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)fk(x)=k1(sinkx+coskx) where x∈Rx \in Rx∈R and k≥ 1.k \ge \,1.k≥1. Then f4(x)−f6(x) {f_4}\left( x \right) - {f_6}\left( x \right)\,\,f4(x)−f6(x) equals :OptionsA14{1 \over 4}41B112{1 \over 12}121C16{1 \over 6}61D13{1 \over 3}31Check AnswerHide SolutionSolutionLet fk(x)=1k(sinkx+cosk.x){f_k}\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}.x} \right)fk(x)=k1(sinkx+cosk.x) Consider f4(x)−f6(x){f_4}\left( x \right) - {f_6}\left( x \right) f4(x)−f6(x) === 14(sin4x+cos4x)−16(sin6x+cos6x){1 \over 4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {1 \over 6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right)41(sin4x+cos4x)−61(sin6x+cos6x) =14[1−2sin2xcos2x]−16[1−3sin2xcos2x] = {1 \over 4}\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - {1 \over 6}\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right]=41[1−2sin2xcos2x]−61[1−3sin2xcos2x] =14−16=112 = {1 \over 4} - {1 \over 6} = {1 \over {12}}=41−61=121