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

Question

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 $\tan\alpha+\tan\beta\equiv\frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta}$ ?

Answer 1

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.