# Finding the first-quadrant angle $\theta$ satisfying $\cot\theta = \frac{\cos(20^\circ)\sqrt{3}}{2 + \cos(40^\circ) + \cos(10^\circ)\sqrt{3}}$

Find the angle theta in degrees knowing that $$\theta$$ belongs to the first quadrant. $$\cot\theta = \frac{\cos(20^\circ)\sqrt{3}}{2 + \cos(40^\circ) + \cos(10^\circ)\sqrt{3}}$$

Answer: $$70^\circ$$

I try

$$\cos(40^\circ) = 2\cos^2(20^\circ) - 1$$

\begin{align} 2 + \sqrt{3}\cos(10^\circ) + \cos(40^\circ) &= 2 + \sqrt{3}\cos(10^\circ) + 2\cos^2(20^\circ) - 1 \\ &= 1 + \sqrt{3}\cos(10^\circ) + 2\cos^2(20^\circ) \end{align}

\begin{align} \cos(20^\circ) &= \cos(30^\circ - 10^\circ) \\ &= \cos(30^\circ)\cos(10^\circ) + \sin(30^\circ)\sin(10^\circ) \\ &= \frac{\sqrt{3}}{2}\cos(10^\circ) + \frac{1}{2}\sin(10^\circ) \end{align}

$$\therefore =\frac{\sqrt{3}\left(\frac{\sqrt{3}}{2}\cos(10^\circ) + \frac{1}{2}\sin(10^\circ)\right)}{1 + \sqrt{3}\cos(10^\circ) + 2\cos^2(20^\circ)} = \frac{\frac{3}{2}\cos(10^\circ) + \frac{\sqrt{3}}{2}\sin(10^\circ)}{1 + \sqrt{3}\cos(10^\circ) + 2\cos^2(20^\circ)}$$

How do I finish?

• If you set $°= \frac{\pi}{180}$ and use the Moivre formula $\sin \phi = \frac{1}{2i} (q-q^{-1}), \cos \phi = \frac{1}{2} (q+q^{-1}),$ with $q = e^{i \phi}$ you find $e^{2 i \theta} \ = \ (-1)^{\frac{7}{9}}$ Commented Jul 20 at 15:05

$$\dfrac{\cos(20^\circ)\sqrt3}{2+\cos(40^\circ)+\cos(10^\circ)\sqrt3}=$$

$$=\dfrac{2\cos(20^\circ)\cos(30^\circ)}{2+\cos(40^\circ)+2\cos(10^\circ)\cos(30^\circ)}=$$

$$=\dfrac{\cos(50^\circ)+\cos(10^\circ)}{2+\cos(40^\circ)+\cos(40^\circ)+\cos(20^\circ)}=$$

$$=\dfrac{2\cos(50^\circ)+\cos(10^\circ)-\cos(50^\circ)}{2\big[1+\cos(40^\circ)\big]+\cos(20^\circ)}=$$

$$=\dfrac{2\sin(40^\circ)-2\sin(30^\circ)\sin(-20^\circ)}{2\big[1+2\cos^2(20^\circ)-1\big]+\cos(20^\circ)}=$$

$$=\dfrac{4\sin(20^\circ)\cos(20^\circ)+\sin(20^\circ)}{4\cos^2(20^\circ)+\cos(20^\circ)}=$$

$$=\dfrac{\sin(20^\circ)\big[4\cos(20^\circ)+1\big]}{\cos(20^\circ)\big[4\cos(20^\circ)+1\big]}=\tan(20^\circ)=\cot(70^\circ)\,.$$

Method 1

\begin{aligned}&\frac{\sqrt{3}\cos 20^\circ}{2+\cos 40^\circ+\sqrt{3}\cos 10^\circ}\\=&\frac{\sqrt{3}\sin20^\circ\cos 20^\circ}{(2+\cos 40^\circ+\sqrt{3}\cos 10^\circ)\sin20^\circ}\\=&\frac{\sqrt{3}\sin20^\circ\cos 20^\circ }{(2\sin20^\circ+\cos 40^\circ\sin20^\circ+\sqrt{3}\cos 10^\circ\sin20^\circ)}\end{aligned}

Let \begin{aligned} x&=2\sin20^\circ+\cos 40^\circ\sin20^\circ+\sqrt{3}\cos 10^\circ\sin20^\circ,\\y&=2\cos20^\circ+\sin 40^\circ\cos20^\circ+\sqrt{3}\sin 10^\circ\cos20^\circ,\end{aligned} with \begin{aligned} x+y&=4-4+\frac{\sqrt{3}}{2}+\frac{1}{2}\cdot\sqrt{3}=\sqrt{3},\\x-y&=4\cos70^\circ-\sin20^\circ+\sqrt{3}\sin10^\circ\\&=3\sin20^\circ+\sqrt{3}\sin(30-20)^\circ\\&=3\sin20^\circ+\sqrt{3}(\frac{1}{2}\cos20^\circ-\frac{\sqrt{3}}{2}\sin20^\circ)\\&=\sqrt{3}(\frac{1}{2}\cos20^\circ+\frac{\sqrt{3}}{2}\sin20^\circ)\\&=\sqrt{3}\cos40^\circ\\&=2\sqrt{3}\cos^220^\circ-\sqrt{3}.\end{aligned} Then $$x=\sqrt{3}\cos^220^\circ$$, with $$\displaystyle \frac{\sqrt{3}\sin20^\circ\cos 20^\circ }{(2\sin20^\circ+\cos 40^\circ\sin20^\circ+\sqrt{3}\cos 10^\circ\sin20^\circ)}=\tan20^\circ$$.

Method 2

To prove $$\displaystyle \frac{\sqrt{3}\cos 20^\circ}{2+\cos 40^\circ+\sqrt{3}\cos 10^\circ}=\cot70^\circ=\frac{\cos70^\circ}{\sin70^\circ},$$ we need only to prove $$\sqrt{3}\cos 20^\circ\sin70^\circ=\cos70^\circ(2+\cos 40^\circ+\sqrt{3}\cos 10^\circ).$$

Now use $$\displaystyle \cos20^\circ=\sin70^\circ=\frac{1}{2}\cos40^\circ+\frac{\sqrt{3}}{2}\sin40^\circ$$, $$\displaystyle \cos 10^\circ=\sin80^\circ=2\sin40^\circ\cos40^\circ$$, $$\displaystyle \cos70^\circ=\frac{\sqrt{3}}{2}\cos40^\circ-\frac{1}{2}\sin40^\circ$$, and we then only need to prove \begin{aligned}&\sqrt{3}\cos 20^\circ\sin70^\circ-\cos70^\circ(2+\cos 40^\circ+\sqrt{3}\cos 10^\circ)\\=&\sqrt{3}\big(\frac{1}{2}\cos40^\circ+\frac{\sqrt{3}}{2}\sin40^\circ\big)^2-\big(\frac{\sqrt{3}}{2}\cos40^\circ-\frac{1}{2}\sin40^\circ\big)\big(2+\cos40^\circ+2\sqrt{3}\sin40^\circ\cos40^\circ\big)\\=&2\bigg(\big(\frac{1}{2}\sin40^\circ-\frac{\sqrt{3}}{2}\cos40^\circ\big)+\sqrt{3}\cos40^\circ\sin40^\circ\big(\frac{1}{2}\sin40^\circ-\frac{\sqrt{3}}{2}\cos40^\circ\big)\\+&\frac{\sqrt{3}}{2}\sin^240^\circ+\cos40^\circ\sin40^\circ-\frac{\sqrt{3}}{8}\bigg)\\=&2\bigg(-\sin20^\circ-\frac{\sqrt{3}}{2}\sin20^\circ\sin80^\circ+\frac{\sqrt{3}}{4}\big(2\sin^240^\circ-1+1\big)+\frac{1}{2}\sin80^\circ-\frac{\sqrt{3}}{8}\bigg)\\=&2\bigg(-\sin20^\circ-\frac{\sqrt{3}}{2}\sin20^\circ\sin80^\circ-\frac{\sqrt{3}}{4}\cos80^\circ+\frac{1}{2}\sin80^\circ+\frac{\sqrt{3}}{8}\bigg)\\=&2\bigg(-\sin20^\circ+\frac{1}{4}\sin80^\circ-\frac{\sqrt{3}}{2}\sin20^\circ\sin80^\circ+\big(-\frac{\sqrt{3}}{4}\cos80^\circ+\frac{1}{4}\sin80^\circ\big)+\frac{\sqrt{3}}{8}\bigg)\\=&2\bigg(-\sin20^\circ+\frac{1}{4}\sin80^\circ-\frac{\sqrt{3}}{2}\sin20^\circ\sin80^\circ+\frac{1}{2}\sin20^\circ+\frac{\sqrt{3}}{8}\bigg)\\=&2\bigg(-\frac{1}{2}\sin20^\circ+\frac{1}{4}\sin80^\circ-\frac{\sqrt{3}}{2}\sin20^\circ\sin80^\circ+\frac{\sqrt{3}}{8}\bigg)\\=&2\bigg(-\frac{1}{2}\big(2\cos^235^\circ-1\big)+\frac{1}{4}\big(2\cos^25^\circ-1\big)-\frac{\sqrt{3}}{2}\sin20^\circ\sin80^\circ+\frac{\sqrt{3}}{8}\bigg)\\=&-2\cos^235^\circ+\cos^25^\circ-\sqrt{3}\sin20^\circ\sin80^\circ+\frac{1}{2}+\frac{\sqrt{3}}{4}\\=&0.\end{aligned}

To prove this, just let \begin{aligned}&x=-2\cos^235^\circ+\cos^25^\circ-\sqrt{3}\sin20^\circ\sin80^\circ+\frac{1}{2}+\frac{\sqrt{3}}{4},\\&y=-2\sin^235^\circ+\sin^25^\circ-\sqrt{3}\cos20^\circ\cos80^\circ+\frac{1}{2}+\frac{\sqrt{3}}{4},\end{aligned} with \begin{aligned}x+y&=-2+1-\sqrt{3}\cdot\frac{1}{2}+1+\frac{\sqrt{3}}{2}=0,\\x-y&=-2\cos70^\circ+\cos10^\circ+\sqrt{3}\cos100^\circ\\&=-2\cos70^\circ+\cos10^\circ-\sqrt{3}\sin10^\circ\\&=-2\cos70^\circ+2\sin20^\circ\\&=0.\end{aligned}

Then we know that $$x=0$$, and we're done.

• You can't start from the solution: At first we don't know the value of the cot $\theta$ Commented Jul 20 at 18:48
• Check method 1. I've updated the solution. Commented Jul 20 at 20:34