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JEE Main 2019
Trigonometry
Trigonometric Ratio and Identites
Easy

Question

Let fk(x)=1k(sinkx+coskx)f_k\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right) where xRx \in R and k1.k \ge \,1. Then f4(x)f6(x){f_4}\left( x \right) - {f_6}\left( x \right)\,\, equals :

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Solution

Let fk(x)=1k(sinkx+cosk.x){f_k}\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}.x} \right) Consider f4(x)f6(x){f_4}\left( x \right) - {f_6}\left( x \right) == 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) =14[12sin2xcos2x]16[13sin2xcos2x] = {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] =1416=112 = {1 \over 4} - {1 \over 6} = {1 \over {12}}

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