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If f(x)=sinx-cosx is written as f1(x)+f2(x) where f1(x) is even and f2(x) is odd then

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a
f1(x)=cosx
b
f1(x)=-cosx
c
f2(x)=-sinx+cosx
d
f2(x)=sin(2π-x)

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detailed solution

Correct option is B

f1(x)=f(x)+f(-x)2=12Sinx-cosx-sinx-cosx=-cosx f2(x)=f(x)-f(-x)2=12sinx-cosx+sinx+cosx=sinx


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