Interesting Trigonometry IMO Problem

Geometry Level 4

$\large \frac{1}{\sin (2x)} + \frac{1}{\sin (4x)} + \frac1{\sin(8x) } + \ldots + \frac{1}{\sin {(2^n x)}}$

Let $n$ be a positive integer and $x$ be a real number where $x \ne \dfrac {k \pi}{2^m}$ for non-negative integer $m$ and $k$ is an integer. What is the simplest form of the expression above?

\sec (x) - \sec (2^n x)

\csc (x) - \csc (2^n x)

\tan( x )- \tan (2^n x)

\cot (x) - \cot (2^n x)

1 solution

Sandeep Rathod
Nov 24, 2014

$T_{n} = \frac{sin(2^{n-1}x)}{sin(2^{n-1}x).sin(2^{n}x)}$

$= \frac{sin(2^{n} - 2^{n-1})}{sin(2^{n-1}x).sin(2^{n}x)}$

$= \frac{sin2^{n}cos2^{n-1} - cos2^{n}sin2^{n-1}}{sin(2^{n-1}x).sin(2^{n}x)}$

$T_{n} = cot2^{n - 1}x - cot2^{n}x$

Thus series becomes,

$cotx - cot2x + cot2x - cot4x + ........... + cot2^{n-1}x - cot2^{n}x$

$= cotx - cot2^{n}x$

From JEE point of view, just put $n=1$ and $x=\frac{\pi}{4}$ and match your answer with the options. :) Nice solution !!

Sandeep Bhardwaj - 6 years, 6 months ago

haha didn't think of that :p

Happy Melodies - 6 years, 6 months ago

Nice. I used n=1 & x=pi/3 as there is some restriction on x mentioned which doesn't allow pi/4 .

Deepak Kumar - 6 years, 1 month ago

your second statement is wrong you forgot to write a x there

A Former Brilliant Member - 4 years, 10 months ago

Can you please explain me your second statement .

Sabhrant Sachan - 5 years ago

In the second statement x is missing...

Also don't you think it is an overrated problem?

Rahil Sehgal - 4 years, 2 months ago

nice solution

Harry Jones - 4 years, 5 months ago

@sandeep Rathod That's complete brilliance !!!

Ankit Kumar Jain - 4 years, 3 months ago

Interesting Trigonometry IMO Problem

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