Let f: R rightarrow R be defined as f(x+y)+f( | vedclass

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(A) The given functional equation is $f(x+y)+f(x-y)=2 f(x) f(y)$.This is a well-known Cauchy-type functional equation whose solution is $f(x) = \cos(a

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$\operatorname{cosec}^{2}(1) \operatorname{cosec}(21) \sin (20)$

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(A) The given functional equation is $f(x+y)+f(x-y)=2 f(x) f(y)$.This is a well-known Cauchy-type functional equation whose solution is $f(x) = \cos(ax)$.Given $f\left(\frac{1}{2}\right) = -1$,we have $\cos\left(\frac{a}{2}\right) = -1$,which implies $\frac{a}{2} = (2n+1)\pi$,so $a = 2(2n+1)\pi$.For any integer $k$,$f(k) = \cos(2(2n+1)\pi k) = \cos(2m\pi) = 1$ where $m$ is an integer.Thus,the expression becomes $\sum_{k=1}^{20} \frac{1}{\sin(k) \sin(k+1)}$.Using the identity $\frac{1}{\sin(k) \sin(k+1)} = \frac{1}{\sin(1)} \left( \frac{\sin((k+1)-k)}{\sin(k) \sin(k+1)} \right) = \frac{1}{\sin(1)} (\cot(k) - \cot(k+1))$.Summing from $k=1$ to $20$,we get $\frac{1}{\sin(1)} (\cot(1) - \cot(21))$.$= \frac{1}{\sin(1)} \left( \frac{\cos(1)}{\sin(1)} - \frac{\cos(21)}{\sin(21)} \right) = \frac{1}{\sin(1)} \left( \frac{\cos(1)\sin(21) - \sin(1)\cos(21)}{\sin(1)\sin(21)} \right)$.$= \frac{\sin(21-1)}{\sin^2(1) \sin(21)} = \frac{\sin(20)}{\sin^2(1) \sin(21)} = \operatorname{cosec}^2(1) \operatorname{cosec}(21) \sin(20)$.

Let $f: R \rightarrow R$ be defined as $f(x+y)+f(x-y)=2 f(x) f(y)$ and $f\left(\frac{1}{2}\right)=-1$. Then,the value of $\sum_{k=1}^{20} \frac{1}{\sin (k) \sin (k+f(k))}$ is equal to:

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