2
$\begingroup$

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)?

$\endgroup$
3
  • 2
    $\begingroup$ The index $n$ is not used in your summation. Is it supposed to be used? $\endgroup$
    – HiMatt
    Apr 15, 2021 at 4:26
  • 2
    $\begingroup$ Do you mean $n$ where you have $x?$ $\endgroup$ Apr 15, 2021 at 4:30
  • $\begingroup$ 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 $\sum_{i=1}^8 \sin(10n^\circ)$. $\endgroup$ Apr 15, 2021 at 4:33

2 Answers 2

2
$\begingroup$

$\textbf{Hint:}$ 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, :)

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – Aiden Chow
    Apr 15, 2021 at 5:49
  • $\begingroup$ @AidenChow, There were some mistakes, now iIve corrected it. $\endgroup$
    – prAnjal
    Apr 15, 2021 at 6:25
  • $\begingroup$ Thanks for your help! My answer key is saying the answer is $6$, but it's only because $\frac 8{\sqrt2}=4\sqrt2\implies a+b=4+2=6$. $\endgroup$
    – Aiden Chow
    Apr 15, 2021 at 6:37
  • $\begingroup$ @AidenChow, Yes that is right, it will be 6 $\endgroup$
    – prAnjal
    Apr 15, 2021 at 6:41
  • 1
    $\begingroup$ 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. $\endgroup$
    – Aiden Chow
    Apr 15, 2021 at 6:51
2
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Apr 15, 2021 at 9:37
  • 1
    $\begingroup$ Just clarifying, are the periods supposed to indicate multiplication? If so, you can use \cdot. $\endgroup$
    – Aiden Chow
    Apr 15, 2021 at 9:41
  • $\begingroup$ Ok, next time I will include both things. $\endgroup$
    – lee
    Apr 15, 2021 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.