# How can one simplify $\sum_{n=1}^8\frac{\sin10n^\circ}{\cos5^\circ\cos10^\circ\cos20^\circ}$?

I have trouble solving the following question:

Consider the following expression: $$\sum_{n=1}^8\frac{\sin10n^\circ}{\cos5^\circ\cos10^\circ\cos20^\circ}$$ The value of the above expression can be fully simplified into the form $$a\sqrt b$$. What is $$a+b$$?

I know that $$\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}2\right)\cos\left(\frac{\alpha-\beta}2\right)$$ but I'm not too sure what pairs of $$\sin$$ I should apply it on, or if this is even the right approach to this problem. Could someone give me hints for solving this problem(Am I on the right track)?

• The index $n$ is not used in your summation. Is it supposed to be used? Apr 15, 2021 at 4:26
• Do you mean $n$ where you have $x?$ Apr 15, 2021 at 4:30
• Judging by the tags used and your profile page, this may not be appropriate, but my first instinct is that you should use complex numbers. The geometric series $\sum_{n=1}^8 e^{in\pi/18}$ has real part $\sum_{i=1}^8 \sin(10n^\circ)$. Apr 15, 2021 at 4:33

$$\textbf{Hint:}$$ Use the formula that, $$\sum_{r=0}^{n-1}\sin(a+rh) = \frac{\sin(\frac{nh}{2})\sin(a+(n-1)\frac{h}{2})}{\sin(\frac{h}{2})}$$

so, Let $$P = \sum_{n=1}^{8}\frac{\sin(10n^\circ)}{\cos(5^\circ)\cos(10^\circ)\cos(20^\circ)}$$ $$=\frac{1}{\cos(5^\circ)\cos(10^\circ)\cos(20^\circ)}\sum_{n=1}^{8}\sin(10n^\circ)$$ $$=\frac{1}{\cos(5^\circ)\cos(10^\circ)\cos(20^\circ)}\cdot\frac{\sin(\frac{8\cdot10}{2})\sin(10+7\cdot\frac{10}{2})}{\sin(\frac{10}{2})}$$ $$=\frac{\sin(40^\circ)\sin(45^\circ)}{\cos(5^\circ)\cos(10^\circ)\cos(20^\circ)\sin(5^\circ)}$$ $$=2\cdot\frac{\sin(40^\circ)\sin(45^\circ)}{2\cos(5^\circ)\sin(5^\circ)\cos(10^\circ)\cos(20^\circ)}$$ $$=4\cdot\frac{\sin(40^\circ)\sin(45^\circ)}{2\sin(10^\circ)\cos(10^\circ)\cos(20^\circ)}$$ $$=8\cdot\frac{\sin(40^\circ)\sin(45^\circ)}{2\sin(20^\circ)\cos(20^\circ)}$$ $$=8\frac{\sin(40^\circ)\sin(45^\circ)}{\sin(40^\circ)}$$ $$P=8\sin(45^\circ)$$

So, $$P=4\cdot\sqrt{2}$$, therefore $$a = 4$$ & $$b=2$$, so $$a+b=6$$

I hope this was helpful, :)

• This was helpful but, because I'm not too experienced in trigonometry, I have no idea how I would reduce something like $8\cos20^\circ$ into the form $a\sqrt b$. Could you help me on that as well? Thanks. Apr 15, 2021 at 5:49
• @AidenChow, There were some mistakes, now iIve corrected it. Apr 15, 2021 at 6:25
• Thanks for your help! My answer key is saying the answer is $6$, but it's only because $\frac 8{\sqrt2}=4\sqrt2\implies a+b=4+2=6$. Apr 15, 2021 at 6:37
• @AidenChow, Yes that is right, it will be 6 Apr 15, 2021 at 6:41
• Also near the end, I think you meant to write $8\frac{\sin40^\circ\sin45^\circ}{\color{red}{\sin40^\circ}}$ instead of $8\frac{\sin40^\circ\sin45^\circ}{\cos40^\circ}$ if I'm not mistaken. Apr 15, 2021 at 6:51

Proof:

\begin{align} S&= \sum_{n=1}^{n}\sin(\alpha+(n-1)\beta) \\S\cdot2\sin\frac\beta2&=\sum_{n=1}^{n}\sin(\alpha+(n-1)\beta)\cdot2\sin\frac\beta2 \\S\cdot2\sin\frac\beta2&=\sum_{n=1}^{n}\cos\left(\alpha+\left(n-\frac{1}{2}\right)\beta\right)-\sum_{n=1}^{n}\cos\left(\alpha+\left(n-\frac{3}{2}\right)\beta\right) \\S\cdot2\sin\frac\beta2&=\cos\left(\alpha+(n-1)\frac{\beta}{2}\right)-\cos\left(\alpha-\frac{\beta}{2}\right) \\S\cdot2\sin\frac\beta2&=2\sin\left(\alpha+\frac{(n-1)\beta}{2}\right)\cdot\sin\left(\frac{n\cdot\beta}{2}\right) \end{align}

Hope this proof is understandable to you. Thanks

• This is brilliant, very good, well done. +1 One very nice and small thing, you can do \sin to get $\sin$ instead of $sin$, the same with $\cos$ and so on. Apr 15, 2021 at 9:37
• Just clarifying, are the periods supposed to indicate multiplication? If so, you can use \cdot. Apr 15, 2021 at 9:41
• Ok, next time I will include both things.
– lee
Apr 15, 2021 at 9:45