Prove it

Prove that

tan⁡nθ=(n1)t−(n3)t3+(n5)t5−...........1−(n2)t2+(n4)t4−.........................\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⁡θt = \tan \theta

#Algebra

Note by U Z
4 years, 9 months ago

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By De Moivre's theorem

(cosθ+isinθ)n=cos(nθ)+isin(nθ)(cos\theta + i sin\theta)^{n} = cos(n\theta) + i sin(n\theta)

Writting the binomial expression of

(cosθ+isinθ)n(cos\theta + i sin\theta)^{n}

Now

Equating real part to cos(nθ)cos(n\theta)

And imaginary part to sin(nθ)sin(n\theta)

We get(let cos = c, sin = s)

c(nθ)=cn−C2ncn−2s2+…\displaystyle c(n\theta) = c^{n} - C_2^{n} c^{n -2} s^{2} +\ldots

s(nθ)=C1ncn−1s−C3ncn−3s3+…\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

t(nθ)=C1ncn−1s−C3ncn−3s3+…cn−C2ncn−2s2+…\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⁡nθ\cos^{n} \theta in numerator and denominator to get the required expression.

Hence Proved!

Krishna Sharma - 4 years, 8 months ago

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Thanks Good solution

U Z - 4 years, 8 months ago

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@brian charlesworth @Satvik Golechha @Mvs Saketh @Santanu Banerjee @Aman Sharma

you too please help

U Z - 4 years, 8 months ago

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@Sanjeet Raria You too please

U Z - 4 years, 8 months ago

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@U Z – @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

U Z - 4 years, 8 months ago

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@U Z – 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 - 4 years, 8 months ago

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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 :))) )

Nicholas Nye - 4 years, 6 months ago

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thanks for proving, refer this

For posting problems with mathematical expressions refer this

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U Z - 4 years, 6 months ago

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Thanks!!

Nicholas Nye - 4 years, 6 months ago

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tan (A+B+C+......) = S1−S3+S5−.....1−S2+S4−......\frac{S_{1}-S_{3}+S_{5}-.....}{1-S_{2}+S_{4}-......}

Where SiS_{i} is sum of products of i terms taken at a time.

For A=B=C...=θ\theta, You will get desired result!

Pranjal Jain - 4 years, 8 months ago

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