# If $\sin 5°+\sin 10°+\sin15°+\cdots+\sin 40°=a$, then $\sin 5°+\sin 10°+\sin15°+\cdots+\sin 175°=?$

I'm stuck in this question

If $$\sin 5°+\sin 10°+\sin15°+\cdots+\sin 40°=a$$

$$\sin 5°+\sin 10°+\sin15°+\cdots+\sin 175°=?$$

I know that, (I asked before) $$\sin 5°+\sin 10°+\sin15°+\cdots+\sin 175°=\tan\frac{175}{2}$$

But, I didn't catch a hint here.

• $\sin(\pi/4+x)=\frac{1}{\sqrt 2}(\sin x+\cos x)$,$\sin(\pi/2+x)=\cos x$,$\sin(3\pi/4+x)=\frac{1}{\sqrt 2}(\sin x-\cos x)$-These identities may be helpful. – Thomas Shelby Jan 4 '19 at 12:20
• You can link to your question more easier than anybody. – kelalaka Jan 4 '19 at 15:02
• math.stackexchange.com/questions/17966/… – lab bhattacharjee Jan 4 '19 at 19:47

Let $$a = \sin 5°+\sin 10°+\sin15°+\cdots+\sin 40°= \\ \operatorname{Im} (\sum_{n=0}^{8}\exp(i n 5 \pi/180))$$ and $$b = \sin 5°+\sin 10°+\sin15°+\cdots+\sin 175°= \\ \operatorname{Im} \left((1 + \exp(i 9\cdot 5 \pi/180)+ \exp(i 2\cdot 9\cdot 5 \pi/180)+ \exp(i 3\cdot9\cdot 5 \pi/180))\cdot \sum_{n=0}^{8}\exp(i n 5 \pi/180)\right) =\\ \operatorname{Im} \left((1 + \exp(i \pi/4)+ \exp(i 2 \pi/4)+ \exp(i 3 \pi/4))\cdot \sum_{n=0}^{8}\exp(i n 5 \pi/180)\right) =\\ \operatorname{Im} \left((1 + i(1 +\sqrt 2))\cdot \sum_{n=0}^{8}\exp(i n 5 \pi/180)\right) =\\ a + (1 +\sqrt 2)\operatorname{Re} \left( \sum_{n=0}^{8}\exp(i n 5 \pi/180)\right) = a + (1 +\sqrt 2)\sum_{n=0}^{8}\cos( n 5 \pi/180)\\ = a + (1 +\sqrt 2)\sum_{n=0}^{8}\sin((90 - 5 n) \pi/180)$$ Denote $$c = \sum_{n=0}^{8}\sin((90 - 5 n) \pi/180)$$. Then we have $$b = 2 a + 2c +\frac{2}{\sqrt 2} -1$$ The last two terms are $$\sin 45° = \frac{1}{\sqrt 2}$$, $$\sin 135° = \frac{1}{\sqrt 2}$$. And $$\sin 90° = 1$$ has to be subtracted because it was counted twice in $$2c$$.

Solving $$b = 2 a + 2c +\frac{2}{\sqrt 2} -1 \\ b = a + (1 +\sqrt 2) c$$

gives $$b = \frac{\left( \sqrt{2}+2\right) \, \left( 4 a+\sqrt{2}\right) }{2}$$

• (+1) Many thanks. I hope we can complete. – Elementary Jan 4 '19 at 12:19
• $\frac{\left( \sqrt{2}+2\right) \, \left( 4 a+\sqrt{2}\right) }{2}$ – Aleksas Domarkas Jan 4 '19 at 12:43
• @Beginner I completed that approach. – Andreas Jan 4 '19 at 12:59

$$2a\cdot\sin2.5^\circ=\cos2.5^\circ-\cos(45^\circ-2.5^\circ)$$

$$\iff\left(2a+\dfrac1{\sqrt2}\right)\sin2.5^\circ=\left(1-\dfrac1{\sqrt2}\right)\cos2.5^\circ\ \ \ \ (1)$$

If $$b=\sin 5^\circ+\sin 10^\circ+\sin15^\circ+\cdots+\sin 175^\circ,$$

$$2b\cdot\sin2.5^\circ=\cos2.5^\circ-\cos(180^\circ-2.5^\circ)=2\cos2.5^\circ\ \ \ \ (2)$$

Divide $$(2)$$ by $$(1)$$ to find $$\dfrac{2b}{\left(2a+\dfrac1{\sqrt2}\right)}=\dfrac2{\left(1-\dfrac1{\sqrt2}\right)}$$ as $$\sin2.5^\circ\cdot\cos2.5^\circ\ne0$$

• Thank you for answer. – Elementary Jan 5 '19 at 14:03