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I just learned this cool trick but I can't figure out why it works, obviously I know how it's done:

1 + 2 + 3 + ... + n ( I know the no of terms have to be even for it to work in pairs)

$\sin$(1)+ $\sin$(2) + $\sin$(3)+... = [$\sin$(1)+ $\sin$(n)] + [$\sin$(2) + $\sin$(n - 1)] + ...

upto halfway through the series and stop since it repeats the pairs in reverse basically doubling the sum.

then use sum-product identities,

$\sin\alpha+\sin\beta\equiv2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$\cos\alpha+\cos\beta\equiv2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$\sum_1^6\sin x \equiv2\sin\frac{7}{2}(\cos\frac{5}{2}+\cos\frac{3}{2}+\cos\frac{1}{2})$

or

$\sum_1^4\cos x \equiv2\cos\frac{5}{2}(\cos\frac{3}{2}+\cos\frac{1}{2})$

But why in $2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ does

when n = 6 , $\sum_1^6\sin x$

$\sin\frac{\alpha+\beta}{2}$ takes the first (1 + n) while $\cos\frac{\alpha-\beta}{2}$ does (1 - n) + (2 - (n - 1)) + (3 -(n - 2))

Why $\sum_x^n\tan x$ doesn't work using $\tan\alpha+\tan\beta\equiv\frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta}$ ?

Anthony
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  • I don't understand what you are asking. $\alpha +\beta$ is constant, and $\alpha-\beta$ is not. Why is that a problem? – saulspatz Feb 25 '20 at 16:17
  • @saulspatz I guess I don't understand why the formula involving cos(A - B) extends like (1 - n) + (2 - (n - 1)) + (3 -(n - 2)) I guess? And also why tanx doesn't work the same as sinx and cosx? I don't understand any of it, I can only calculate it but don't know why it works? – Anthony Feb 25 '20 at 16:28
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    I am not sure if it answers your question, but it is possible to get a much nicer compact form that applies for all $n$, odd or even. Namely, $\displaystyle \sum_{k=1}^n \sin k = \frac{\sin\frac n 2 \sin \frac{n+1}2}{\sin \frac 1 2}$. If this helps, and you'd like to see how it's done, let me know. – Deepak Feb 25 '20 at 16:29
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    The easiest way to show the above is to observe that the sum of sines with arguments in arithmetic progression is actually the imaginary part of a geometric series involving complex numbers. Then just a bit of fiddling with the product-sum identities you already know gets you that nice form. – Deepak Feb 25 '20 at 16:36
  • @Deepak I'd like to see how you'd arrive to this formula, post it man. Also why there isn't any for cosine or tangent? – Anthony Feb 25 '20 at 16:53
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    You can use the same method to derive one for the sum of cosines. It is unlikely you will find something so nice for the sum of tangents, as it is just different from sines and cosines (the fact that it's a ratio of a sine and a cosine is neither here nor there). There is no reason to suppose that the same methods will work for tangent. I can post an outline of the comment in the next method, but I am limited in the number of characters. – Deepak Feb 25 '20 at 16:57
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    Briefly, you start by observing that $\sin k = \Im(e^{ik})$, so $\displaystyle \sum_{k=1}^n \sin k = \Im \sum_{k=1}^n e^{ik} = \Im \frac{e^i(e^{in}-1)}{e^i-1}$, where I just used the geometric series sum formula. You have to simplify from that step on - make the denominator a real number, separate out the imaginary part and apply the sum-product and half angle identity. – Deepak Feb 25 '20 at 17:01

1 Answers1

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For the first question, you just need to look at the values of $\alpha$ and $\beta$. As you have noted, they are paired so the sum is always $n+1$. The first few $(\alpha,\beta)$ pairs are $$(n,1)\\(n-1,2)\\(n-2,3)\\\vdots$$ The sums are always $n+1$ and the differences are $$n-1\\n-3\\n-5\\\vdots$$ Symbolically, we have $$\alpha=n-k\\\beta=k+1\\\alpha+\beta=n+1\\\alpha-\beta=n-2k-1$$

As to why a similar formula doesn't work for tangent, it's because you don't have a common denominator.

saulspatz
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