Prove that
\(\tan n\theta =\large{ \frac{ \binom{n}{1}t - \binom{n}{3}t^{3} + \binom{n}{5}t^{5} - ...........}{ 1 - \binom{n}{2}t^{2} + \binom{n}{4}t^{4} - .........................}}\)
where \(t = \tan \theta\)
Prove that
\(\tan n\theta =\large{ \frac{ \binom{n}{1}t - \binom{n}{3}t^{3} + \binom{n}{5}t^{5} - ...........}{ 1 - \binom{n}{2}t^{2} + \binom{n}{4}t^{4} - .........................}}\)
where \(t = \tan \theta\)
Note by
Megh Choksi
2 years, 9 months ago
Problem Loading...
Note Loading...
Set Loading...
Comments
Sort by:
Top NewestBy De Moivre's theorem
\((cos\theta + i sin\theta)^{n} = cos(n\theta) + i sin(n\theta)\)
Writting the binomial expression of
\((cos\theta + i sin\theta)^{n}\)
Now
Equating real part to \(cos(n\theta)\)
And imaginary part to \(sin(n\theta)\)
We get(let cos = c, sin = s)
\(\displaystyle c(n\theta) = c^{n} - C_2^{n} c^{n -2} s^{2} +\ldots\)
\(\displaystyle s(n\theta) = C_1^{n} c^{n-1} s - C_3^{n}c^{n-3} s^{3} + \ldots\)
Dividing equation 2 by 1 we get
\(\displaystyle t(n\theta) = \frac{C_1^{n} c^{n-1}s - C_3^{n} c^{n-3}s^{3} + \ldots}{c^{n} - C_2^{n}c^{n-2}s^{2} + \ldots}\)
Now divide by \(\cos^{n} \theta\) in numerator and denominator to get the required expression.
Hence Proved! – Krishna Sharma · 2 years, 9 months ago
Log in to reply
Thanks Good solution – Megh Choksi · 2 years, 9 months ago
Log in to reply
I would solve this one by using De Moiver's identity (cosθ+jsinθ)^n=cosnθ+jsinnθ. Then tannθ=sinnθ/cosnθ. By expanding into powers we have (cosθ+jsinθ)^n=Sum(j^k *sin(θ)^k * cos(θ)^(n-k) * (n per k) )=Sum(j^k *tan(θ)^k * cos(θ)^n * (n per k) ). By grouping into real and imaginary parts we can get the result given. (I should learn to write in these posts in a more beautiful way :))) ) – Nicholas Nye · 2 years, 7 months ago
Log in to reply
thanks for proving, refer this
For posting problems with mathematical expressions refer this
For a complete guide for higher mathematical expressions refer This Wikibook
An example -
Writing this we get,
You can refer others too ,
Right Click a problem and select open in a new window
Now just copy it your job becomes easier
For posting images in a comment write
! [Description or leave it blank] (Url of the image) ( don't leave any gap here)
For posting question and solutions its not needed as you now @Nicholas Nye – Megh Choksi · 2 years, 7 months ago
Log in to reply
Thanks!! – Nicholas Nye · 2 years, 7 months ago
Log in to reply
@Calvin Lin @DEEPANSHU GUPTA @Kostub Deshmukh @Agnishom Chattopadhyay @Pratik Shastri @Pranshu Gaba @Pranav Arora @Pranjal Jain @Victor Loh @Vinay Sipani @Aditya Raut @Christopher Boo @Krishna Ar @Krishna Sharma
Please help – Megh Choksi · 2 years, 9 months ago
Log in to reply
@brian charlesworth @Satvik Golechha @Mvs Saketh @Santanu Banerjee @Aman Sharma
you too please help – Megh Choksi · 2 years, 9 months ago
Log in to reply
@Sanjeet Raria You too please – Megh Choksi · 2 years, 9 months ago
Log in to reply
@Mursalin Habib @John Muradeli @Cody Johnson You too please help
To Brilliant.org ,
Here after a comment is posted and when we click the edit button and want mention someone the menu is not coming . we have to make a new comment for this
For example firstly I forget about the people in the second, third and four comment and when i clicked the edit and type their name the menu was not displayed – Megh Choksi · 2 years, 9 months ago
Log in to reply
Yes that's an issue - what you need to do, as far as I'm as a non-moderator am concerned, is mention all the names first, in the order you want, and only then type your text. If text needs to be between the names, space the names appropriately.
This does need to be fixed, though. I thought I was the only one, but apparently not. – John Muradeli · 2 years, 9 months ago
Log in to reply
tan (A+B+C+......) = \(\frac{S_{1}-S_{3}+S_{5}-.....}{1-S_{2}+S_{4}-......}\)
Where \(S_{i}\) is sum of products of i terms taken at a time.
For A=B=C...=\(\theta\), You will get desired result! – Pranjal Jain · 2 years, 9 months ago
Log in to reply