Trigonometric summation

Geometry Level 5

$\displaystyle{f^{ n }\left( \theta \right) =\sum _{ k=0 }^{ n }{ { 4 }^{ k }\sec ^{ 2 }{ \left( { 2 }^{ k }\theta \right) } } }$ Define a function $f$ as above and if $\displaystyle{f^{ 504 }\left( \cfrac { \pi }{ { 2 }^{ 507 } } \right) \equiv { 2 }^{ a }-\csc ^{ 2 }{ \left( \cfrac { \pi }{ { 2 }^{ b } } \right) } }$ for positive integers $a,b$ . Evaluate $a-b$ .

Original

The answer is 504.

1 solution

Deepanshu Gupta
Jun 7, 2015

First start with simple and very elementary telescopic product result :

(Proof is left as an exercise for reader )

$\displaystyle{\prod _{ k=0 }^{ n }{ \cos { \left( { 2 }^{ k }\theta \right) } } =\cfrac { \sin { \left( { 2 }^{ n+1 }\theta \right) } }{ { 2 }^{ n+1 }\sin { (\theta ) } } }$

Now for converting sum into product , take natural logarithm :

$\displaystyle{\sum _{ k=0 }^{ n }{ \ln { \left( \cos { \left( { 2 }^{ k }\theta \right) } \right) } } =\ln { \left( \sin { \left( { 2 }^{ n+1 }\theta \right) } \right) } -\ln { \left( \sin { (\theta ) } \right) } -\ln { \left( { 2 }^{ n+1 } \right) } }$

Now , differentiate above equation w.r.t to $\theta$ We will get :

$\displaystyle{\sum _{ k=0 }^{ n }{ { 2 }^{ k }\tan { \left( { 2 }^{ k }\theta \right) } } =\cot { \theta } -{ 2 }^{ n+1 }\cot { \left( { 2 }^{ n+1 }\theta \right) } }$

Now again differentiate the above equation :

$\displaystyle{f^{ n }\left( \theta \right) \equiv \sum _{ k=0 }^{ n }{ { 4 }^{ k }\sec ^{ 2 }{ \left( { 2 }^{ k }\theta \right) } } ={ 4 }^{ n+1 }\csc ^{ 2 }{ \left( { 2 }^{ n+1 }\theta \right) } -\csc ^{ 2 }{ \theta } }$

Now Put the given values in question and get the answer of $\boxed { 504 }$ .

Moderator note:

A more direct approach would be to verify that the trigonometric identity

$\sec^2 ( \theta ) = 4 \csc^2 ( 2 \theta ) - \csc ^2 \theta$

allows us to perform the telescoping sum.

Of course, it's not easy to find the $u_i$ , especially if you were not given the form of $f^n$ .

I just guessed the answer as 504 , Since It is the special number , as it is total marks of JEE advance 2015 papaer !

I nailed this question from my strong intuntion power .... :D :D

karan shekhawat - 5 years, 12 months ago

My approach was: elaborating the summation and adding 1/sin^2(x)and subtracting we get

4^n+1/sin^2(2^n+1)x - 1/sin^2(x) = above defined function

i have taken theta=x

put the values of n and x we get the answer.

To my extreme carelessness I was unable to locate one expression due to which I typed wrong answer and hence lost points.

Aakash Khandelwal - 5 years, 11 months ago