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माना $f : R \rightarrow R$,$f(x+y)+f(x-y)=2 f(x) f(y), f\left(\frac{1}{2}\right)=-1$ द्वारा परिभाषित है। तो $\sum_{ k =1}^{20} \frac{1}{\sin ( k ) \sin ( k + f ( k ))}$ बराबर है
$\operatorname{cosec}^{2}(1) \operatorname{cosec}(21) \sin (20)$
$\sec ^{2}(1) \sec (21) \cos (20)$
$\operatorname{cosec}^{2}(21) \cos (20) \cos (2)$
$\sec ^{2}(21) \sin (20) \sin (2)$
Solution
$f(x)=\cos a x$
$\because f\left(\frac{1}{2}\right)=-1$
So, $-1=\cos \frac{a}{2}$
$\Rightarrow a=2(2 n+1) \pi$
Thus $f(x)=\cos 2(2 n+1) \pi x$
Now k is natural number
Thus $f(k)=1$
$\sum_{k=1}^{20} \frac{1}{\sin k \sin (k+1)}=\frac{1}{\sin 1} \sum_{k=1}^{20}\left[\frac{\sin ((k+1)-k)}{\sin k \cdot \sin (k+1)}\right]$
$=\frac{1}{\sin 1} \sum_{k=1}^{20}(\cot k-\cot (k+1))$
$=\frac{\cot 1-\cot 21}{\sin 1}=\operatorname{cosec}^{2} 1 \cos \operatorname{ec}(21) \cdot \sin 20$
