JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 3)
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 :
cosec2(21) cos(20) cos(2)
sec2(1) sec(21) cos(20)
cosec2(1) cosec(21) sin(20)
sec2(21) sin(20) sin(2)
Explanation
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
$$\sum\limits_{k = 1}^{20} {{1 \over {\sin k\sin (k + 1)}} = {1 \over {\sin 1}}\sum\limits_{k = 1}^{20} {\left[ {{{\sin \left( {(k + 1) - k} \right)} \over {\sin k.\sin (k + 1)}}} \right]} } $$
$$ = {1 \over {\sin 1}}\sum\limits_{k = 1}^{20} {(\cot k - \cot (k + 1)} $$)
$$ = {{\cot 1 - \cot 21} \over {\sin 1}} = \cos e{c^2}1\cos ec(21).\sin 20$$
$$\because$$ $$f\left( {{1 \over 2}} \right)$$ = $$-$$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
$$\sum\limits_{k = 1}^{20} {{1 \over {\sin k\sin (k + 1)}} = {1 \over {\sin 1}}\sum\limits_{k = 1}^{20} {\left[ {{{\sin \left( {(k + 1) - k} \right)} \over {\sin k.\sin (k + 1)}}} \right]} } $$
$$ = {1 \over {\sin 1}}\sum\limits_{k = 1}^{20} {(\cot k - \cot (k + 1)} $$)
$$ = {{\cot 1 - \cot 21} \over {\sin 1}} = \cos e{c^2}1\cos ec(21).\sin 20$$
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