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

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Accepted Answer

"f(x) = cos$\lambda$x

$\because$ $f\left( {{1 \over 2}} \right)$ = $-$1

So, $-$1 = cos${\lambda \over 2}$

$\Rightarrow$ $\lambda$ = 2$\pi$

Thus f(x) = cos2$\pi$x

Now k is natural number

Thus f(k) = 1

$$\sum\limits_{k = 1}^{20} {{1 \over {\sin k\sin (k + 1)}} = {1 \over {\sin 1}}\sum\limits_{k = 1}^{20} {\left[ {{{\sin \left( {(k + 1) - k} \right)} \over {\sin k.\sin (k + 1)}}} \right]} } $$

$= {1 \over {\sin 1}}\sum\limits_{k = 1}^{20} {(\cot k - \cot (k + 1)}$)

$= {{\cot 1 - \cot 21} \over {\sin 1}} = \cos e{c^2}1\cos ec(21).\sin 20$"

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

Solution

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

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