<s>[INST] Calculate $\sin{55}-\sin{19}+\sin{53}-\sin{17}$ without calculator So here is a trigonometric series.

$$\sin{55^\mathrm{o}}-\sin{19^\mathrm{o}}+\sin{53^\mathrm{o}}-\sin{17^\mathrm{o}}$$

Strange isn't it,  and I have to calculate the total result of the series (without calculator). I don't think Maclaurin series will help me any way. Further I tried almost all trigonometric identities (as per my knowledge) but so far I had no clue. Probably I am missing some kind of identity. Anyone can help me in this?

Note: It's not a homework question.

 [/INST] You can use the following product-sum trig identity:
$$\sin (a+b) - \sin (a-b) = 2 \cos a \sin b$$
$$\sin (a+b) + \sin (a-b) = 2 \sin a \cos b$$
So, $$(\sin 55^\circ - \sin 17^\circ) + (\sin 53^\circ - \sin 19^\circ)$$
$$(2 \cos 36^\circ \sin 19^\circ) + (2 \cos 36^\circ \sin 17^\circ)$$
$$2 \cos 36^\circ (\sin 19^\circ + \sin 17^\circ)$$
$$4 \cos 36^\circ \sin 18^\circ \cos 1^\circ$$

We can find $\sin 18^\circ$ using the following method, and use $\cos 2x = 1 - 2 \sin^2 x$ to find $\cos 36^\circ$.
$\sin 72^\circ = 2 \sin 36^\circ \cos 36^\circ$
$\cos 18^\circ = 4 \sin 18^\circ \cos 18^\circ (1 - 2 \sin^2 18^\circ)$
$1 = 4x(1 - 2x^2)$ where $x = \sin 18^\circ$
$8x^3 - 4x + 1 = 0$
We can factor using a few methods, I think the easiest is to observe that $x = 1/2$ is a root and do polynomial long division of $(2x-1)$:
$(2x - 1)(4x^2 + 2x - 1) = 0$ We can't have $x = \frac{1}{2}$ because that's $\sin 30^\circ$, not $\sin 18^\circ$.
Quadratic formula: $x = \frac{-2 \pm \sqrt{20}}{8} = \frac{1}{4} (-1 \pm \sqrt{5})$
The root is positive, so it's $\frac{1}{4} (-1 + \sqrt{5})$. From this follows $\cos 36^\circ = \frac{1}{4}(1 + \sqrt{5})$

So, the above answer reduces to just $\cos 1^\circ$. This doesn't have a nice formula, but it is algebraic!
</s>