JEE Advance - Mathematics (1990 - No. 1)
Prove that for any positive integer $$k$$,
$${{\sin 2kx} \over {\sin x}} = 2\left[ {\cos x + \cos 3x + ......... + \cos \left( {2k - 1} \right)x} \right]$$
Hence prove that $$\int\limits_0^{\pi /2} {\sin 2kx\,\cot \,x\,dx = {\pi \over 2}} $$
$${{\sin 2kx} \over {\sin x}} = 2\left[ {\cos x + \cos 3x + ......... + \cos \left( {2k - 1} \right)x} \right]$$
Hence prove that $$\int\limits_0^{\pi /2} {\sin 2kx\,\cot \,x\,dx = {\pi \over 2}} $$
This question requires trigonometric identities and integration techniques. It's beyond the scope of basic high school mathematics.
The first part involves mathematical induction and trigonometric sum-to-product formulas, while the second part requires integration of trigonometric functions.
This problem can be solved using complex numbers and De Moivre's theorem.
The identity can be proven using the properties of geometric series and Euler's formula.
It tests knowledge of trigonometric series and their convergence properties.