MathematicsLet f: R→R be defined as f(x+y)+f(x−y)= 2f(x)f(y), f(12)=−1. Then, the value of ∑k=1201sin(k)sin(k+f(k))is equal to:

Letf:RRbedefinedasf(x+y)+f(xy)=2f(x)f(y),f(12)=1.Then,thevalueofk=1201sin(k)sin(k+f(k))isequalto:

  1. A

    sec2(21)sin(20)sin(2)

  2. B

    cosec2(1)cosec(21)sin(20)

  3. C

    sec2(1)sec(21)cos(20)

  4. D

    cosec2(21)cos(20)cos(2)

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    Solution:

    f(x)=cos2πx f(k)=cos 2=1
    k=1201sin(k) sin(k+f(k))) =k=1201sin 1sin(k+1-k)sin(k) sin(k+f(k) =k=1201sin 1sin(k+1) cosk-cos (k+1) sinksin(k) sin(k+1) =k=1201sin 1cot k-cot (k+1) =1sin 1cot 1-cot 21 =1sin 1sin 20sin1 sin 21 cosec21 cosec 21sin20

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