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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.

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

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  • $\begingroup$ (+1) Many thanks. I hope we can complete. $\endgroup$ – Elementary Jan 4 '19 at 12:19
  • $\begingroup$ $\frac{\left( \sqrt{2}+2\right) \, \left( 4 a+\sqrt{2}\right) }{2}$ $\endgroup$ – Aleksas Domarkas Jan 4 '19 at 12:43
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    $\begingroup$ @Beginner I completed that approach. $\endgroup$ – Andreas Jan 4 '19 at 12:59
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Using How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?,

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

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

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 $\sin2.5^\circ\cdot\cos2.5^\circ\ne0$

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

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