Question

# For non-negative integers n, let f(n)=n∑k=0sin(k+1n+2π)sin(k+2n+2π)n∑k=0sin2(k+1n+2π) Assuming cos−1x takes values in [0,π], which of the following options is/are correct?f(4)=√32limn→∞f(n)=12If α=tan(cos−1f(6)),  then α2+2α−1=0 sin(7cos−1f(5))=0

Solution

## The correct options are A f(4)=√32 C If α=tan(cos−1f(6)),  then α2+2α−1=0  D sin(7cos−1f(5))=0f(n)=n∑k=0[cos(πn+2)−cos(2k+3n+2π)]n∑k=01−cos2(k+1n+2π) ⇒f(n)=(n+1)cos(πn+2)−n∑k=0cos(2k+3n+2π)(n+1)−n∑k=0cos2(k+1n+2π) ⇒f(n)=(n+1)cos(πn+2)−[cos3πn+2+cos5πn+2+...+cos2n+3n+2π](n+1)−[cos2πn+2+cos4πn+2+...+cos2n+1n+2π] ⇒f(n)=(n+1)cos(πn+2)−⎡⎢ ⎢ ⎢ ⎢⎣sin(n+1)πn+2sinπn+2⎤⎥ ⎥ ⎥ ⎥⎦cos(n+3)π(n+2)(n+1)−⎡⎢ ⎢ ⎢ ⎢⎣sin(n+1)πn+2sinπn+2⎤⎥ ⎥ ⎥ ⎥⎦cos2(n+2)π2(n+2) ⇒f(n)=(n+1)cos(πn+2)−⎡⎢ ⎢ ⎢ ⎢⎣sin(π−πn+2)sinπn+2⎤⎥ ⎥ ⎥ ⎥⎦cos(π+π(n+2))(n+1)−⎡⎢ ⎢ ⎢ ⎢⎣sin(π−πn+2)sinπn+2⎤⎥ ⎥ ⎥ ⎥⎦cosπ ⇒f(n)=(n+1)cos(πn+2)+cosπn+2(n+1)−cosπ f(n)=(n+2)cos(πn+2)(n+2)=cos(πn+2) Option 1: f(4)=cosπ6=√32 Option 2: limn→∞f(n)=limn→∞cosπn+2=cos(0)≠12 Option 3:  If α=tan(cos−1cosπ8)=tanπ8=√2−1 then α2+2α−1=(√2−1)2+2(√2−1)−1=0  Option 4: sin(7cos−1cosπ7)=sinπ=0

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