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

Question

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?

Answer 1

We will use product-to-sum identities, sum-to-product identities and double-angle formulas.

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

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

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

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

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

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

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

Answer 2

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.