Let mathrm{f}: mathrm{R} rightarrow mathrm{R} | vedclass

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Question

$f(x)=\cos a x$

$\because f\left(\frac{1}{2}\right)=-1$

So, $-1=\cos \frac{a}{2}$

$\Rightarrow a=2(2 n+1) \pi$

Thus $f(x)=\cos 2(2 n+1) \p

Vedclass · Accepted Answer

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

Vedclass · Answer

$f(x)=\cos a x$

$\because f\left(\frac{1}{2}\right)=-1$

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 $f(k)=1$

$\sum_{k=1}^{20} \frac{1}{\sin k \sin (k+1)}=\frac{1}{\sin 1} \sum_{k=1}^{20}\left[\frac{\sin ((k+1)-k)}{\sin k \cdot \sin (k+1)}\right]$

$=\frac{1}{\sin 1} \sum_{k=1}^{20}(\cot k-\cot (k+1))$

$=\frac{\cot 1-\cot 21}{\sin 1}=\operatorname{cosec}^{2} 1 \cos \operatorname{ec}(21) \cdot \sin 20$

Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as

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

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