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 \(t = \tan \theta\)
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U Z
3 years, 6 months ago
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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!
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Thanks Good solution
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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 :))) )
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thanks for proving, refer this
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Thanks!!
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@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
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@brian charlesworth @Satvik Golechha @Mvs Saketh @Santanu Banerjee @Aman Sharma
you too please help
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@Sanjeet Raria You too please
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@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
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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.
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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!
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