Let f : R [] R be defined as [f(x + y) + f(x - y) = 2f(x)f(y),f( 1 2 ) = - 1].

Question

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 :

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

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

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

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

Solutions · 1