# Trigonometry Olympiad problem: Evaluate $1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$

Find the value of $$1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$$

My attempt

I converted the $\sin$ functions which have arguments greater than $90^\circ$ to $\cos$ but I have gone no where with it!

I also tried using double angle formula for the angles which are even.

• Shouldn't the first term be $\sin 2^\circ$? Jul 23, 2016 at 16:17
• I also thinks first will be that $\sin2°$ Jul 23, 2016 at 16:23
• Yes you are right...I will fix it Jul 23, 2016 at 16:27

An approach. One may write \begin{align} \sum_{k=1}^nk\sin(ka)&=\sum_{k=1}^n\frac{d}{da}(-\cos(ka)) \\\\&=-\frac{d}{da}\sum_{k=0}^n\cos(ka) \\\\&=-\frac{d}{da}\left(\frac{1}{2}+\frac{\sin\left[(n+\frac12)a\right]}{2\sin \frac a2} \right) \\\\&=\frac{(n+1) \sin(na)-n \sin((n+1)a)}{4\sin^2 \frac a2} \end{align} then one may take $a:=2^{\circ}, \, n:=90$.

• Smart, simple and beautiful, for sure ! (+1) Jul 23, 2016 at 16:33
• @Oliver Oloa can you explain how you got from the second line to the third line. Aug 10, 2016 at 3:50
• @James Dickens Sure. Please have a look here: math.stackexchange.com/questions/933402/… Aug 10, 2016 at 7:18

Hints: Your sum is related with: $$S=\sum_{n=1}^{90} n \sin\left(\frac{\pi n}{90}\right) = \sum_{n=0}^{89}(90-n)\sin\left(\frac{\pi n}{90}\right)\tag{1}$$ that fulfills: $$\color{red}{2\,S} = 90\sum_{n=1}^{89}\sin\left(\frac{\pi n}{90}\right) = \color{red}{90\cdot\cot\left(\frac{\pi}{180}\right)}.\tag{2}$$

Use $\sin(\theta)=\sin(180^\circ-\theta)$. Let the summation as $S$. Then \begin{align} 2S&=\sum_{k=1}^{90}(k\sin(2k^\circ)+(90-k)\sin((180-2k)^\circ)) \\ &=90\sum_{k=1}^{90}\sin(2k^{\circ}) \end{align} Now use $-2\sin(2k^\circ)\sin(1^\circ)=\cos((2k+1)^\circ)-\cos((2k-1)^\circ)$, then \begin{align} S=45\sum_{k=1}^{90}\sin(2k^{\circ})=-\frac{45}{2\sin(1^\circ)}\sum_{k=1}^{90}(\cos((2k+1)^\circ)-\cos((2k-1)^\circ)=\frac{45\cos(1^\circ)}{\sin(1^\circ)}=45\cot(1^\circ) \end{align}

• $\sin x \geq 0, \; \forall x\in [0^{\circ},180^{\circ}]$ Jul 23, 2016 at 16:25
• @NgChungTak I edited. Thanks. Jul 23, 2016 at 16:53

$$\sin\theta=\sin(\pi-\theta)$$

$$\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}\cdots 88\sin4^{\circ}+89\sin 2^{\circ}+ 90\sin 0^{\circ}$$

$$90(\sin 2^{\circ} +\sin 4^{\circ} + \sin6 ^{\circ}\cdots \sin 90^{\circ}) -45\sin90^{\circ}$$

Now it is easy to evaluate this sum .

$$\sin \alpha +\sin (\alpha+\beta)+\sin (\alpha+2\beta) +\cdots n$$ terms
$$=\frac{\sin{\frac{n\beta}{2}}}{\sin{\frac{\beta}{2}}}{\sin\left[ {\alpha + \frac{\beta}{2}{(n-1)}}\right]}$$

$$90\sin 45^{\circ}\left[\frac{\sin 46^{\circ}}{\sin 1^{\circ}}\right] -90\sin 45^{\circ}.\cos45^{\circ}$$ $$90\sin 45^{\circ}\left[\frac{\sin 46^{\circ}}{\sin 1^{\circ}} -\cos45^{\circ}\right]$$

Now it is almost done simplify it and get $$45\cot 1^{\circ}$$

• @sidt36: how can the sum of positive numbers be zero? Jul 23, 2016 at 16:41

As a complement to other answers, you should know that the following identities are valid

\begin{align} 2 \sin \frac{1}{2} x \sum_{k=1}^{n} \cos kx &= +\sin(n+\frac{1}{2})x - \sin\frac{1}{2}x \\ 2\sin \frac{1}{2} x \sum_{k=1}^{n} \sin kx &= -\cos(n+\frac{1}{2})x + \cos\frac{1}{2}x \end{align}

Thsese formulas can easily be proved by moving $\sin \frac{1}{2} x$ into the summation, using the identities

\begin{align} 2 \sin \frac{1}{2} x \cos kx &= + \sin(k + \frac{1}{2})x - \sin(k - \frac{1}{2})x \\ 2 \sin \frac{1}{2} x \sin kx &= - \cos(k + \frac{1}{2})x + \cos(k - \frac{1}{2})x \end{align}

and the telescoping property of partial sums. In your example, we have

$$x=2\frac{\pi}{180}=\frac{\pi}{90}$$