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)?
 A: $\textbf{Hint:}$
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, :)
A: 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
