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# Prove it

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, 5 months ago

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By 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! · 2 years, 5 months ago

Thanks Good solution · 2 years, 5 months ago

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 :))) ) · 2 years, 3 months ago

thanks for proving, refer this

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For posting question and solutions its not needed as you now @Nicholas Nye · 2 years, 3 months ago

Thanks!! · 2 years, 3 months ago

@Sanjeet Raria You too please · 2 years, 5 months ago

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 · 2 years, 5 months ago

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. · 2 years, 5 months ago

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! · 2 years, 5 months ago