Solve $\tan x =\sec 42^\circ +\sqrt{3}$

Question

Question: Solve the trigonometric equation

$$\tan x =\sec 42^\circ+\sqrt{3}$$

with $0^\circ<x<180^\circ$.

I tried to find for an unknown angle $x$ in a geometry problem via a trigonometric approach. I ended up with the trig equation above. Without hesitation, I reached for my calculator, entering the right-hand-side and arctan-ing it for $x$. To my surprise, the angle $x$ jumps out to be exactly $72 ^\circ$. I did not expect such a pleasant outcome.

Then, I thought I should be able solve the equation for the whole-degree angle by hand. I spent a good amount of time already and was not able to derive it yet.

Either the equation is not as innocent as it looks, or a straightforward method just eludes me.

Answer 1

$\begin{align} \cos(42°) &= \cos(60°-18°) \cr &= \cos(60°)\cos(18°) + \sin(60°)\sin(18°) \cr &= {1\over2} (\cos(18°) + \sqrt3 \sin(18°)) \cr \sec(42°) &= \left({2 \over \cos(18°) + \sqrt3 \sin(18°)}\right) \left({\cos(18°) - \sqrt3 \sin(18°) \over \cos(18°) - \sqrt3 \sin(18°)}\right) \cr &= {2(\cos(18°) - \sqrt3 \sin(18°)) \over \cos^2(18°) - 3\sin^2(18°)} \cr &= \left({2\sin(18°) \over 1 -4 \sin^2(18°)}\right) (\cot(18°) - \sqrt3) \cr \end{align}$

Let $s=\sin(18°)$, using multiple angles formula

$\sin(90°) = \sin(5 \times 18°) = 16s^5 - 20s^3 + 5s = 1$

$16s^5 - 20s^3 + 5s - 1 = 0$
$(s-1)(4s^2+2s-1)^2 = 0$

Since $s≠1$, we get $4s^2+2s-1 = 0\quad → \large{2s \over 1-4s^2} = 1$

$\tan(x) = \sec(42°) + \sqrt3 = (\cot(18°) - \sqrt3) + \sqrt3 = \tan(72°)$

Answer 2

Sorry, but I am unable to work with degrees.

If you look here

$$\sec \left(\frac{7 \pi }{30}\right)=\sqrt{8+2 \sqrt{5}-2 \sqrt{15+6 \sqrt{5}}}$$ and here $$\tan \left(\frac{2 \pi }{5}\right)=\sqrt{5+2 \sqrt{5}}$$ Simplify $$\left(\sqrt{8+2 \sqrt{5}-2 \sqrt{15+6 \sqrt{5}}}+\sqrt 3\right)^2=5+2 \sqrt{5}$$

I understand your surprise.

Edit

Thinking that this could not be the only one, I computed $$R_k=\tan \left(\frac{(k+5) \pi}{30} \right)-\sec \left(\frac{k\pi }{30}\right)$$ for $k=1,\cdots,60$.

Here are the "funny" results (I hope I did not miss any) $$\left( \begin{array}{cc} k & R_k \\ 5 & \frac{1}{\sqrt{3}} \\ 7 & \sqrt{3} \\ 19 & \sqrt{3} \\ 20 & 2-\frac{1}{\sqrt{3}} \\ 25 & \frac{2}{\sqrt{3}} \\ 30 & 1+\frac{1}{\sqrt{3}} \\ 31 & \sqrt{3} \\ 35 & \frac{5}{\sqrt{3}} \\ 43 & \sqrt{3} \\ 50 & -2-\frac{1}{\sqrt{3}} \\ 55 & -\frac{2}{\sqrt{3}} \\ 60 & -1+\frac{1}{\sqrt{3}} \end{array} \right)$$

Answer 3

$$\require{cancel}\begin{align}\sec42^\circ+\sqrt3&\overset{(1)}{=}\csc48^\circ+\csc24^\circ-\tan36^\circ\\&=\frac{\cancel2\sin36^\circ\cancel{\cos12^\circ}}{\cancel{\sin24^\circ}^{\sin12^\circ}\sin48^\circ}-\frac{\sin36^\circ}{\cos36^\circ}\\&=\sin36^\circ\left(\frac2{\cos36^\circ-\cos60^\circ}-\frac1{\cos36^\circ}\right)\\&=\frac{\sin36^\circ(\cos36^\circ+\cos60^\circ)}{2\sin12^\circ\sin48^\circ\cos36^\circ}\\&\overset{(2)}{=}\cancel{\tan36^\circ}\bcancel{\cot48^\circ\cot12^\circ\cot72^\circ}\tan72^\circ\\&=\boxed{\tan72^\circ}\end{align}$$

$(1):$ A proof of $\csc24^\circ-\tan36^\circ=\sqrt3$ is given in this post.

$(2):\cot(60^\circ-\theta)\cot\theta\cot(60^\circ+\theta)=\cot3\theta$ which can proved by using the identities $\cos(60^\circ-\theta)\cos\theta\cos(60^\circ+\theta)=\frac14\cos3\theta$ and $\sin(60^\circ-\theta)\sin\theta\sin(60^\circ+\theta)=\frac14\sin3\theta$.

Answer 4

Following are the most of the identities of the form $$\tan x+\tan y=\sec z$$

The current one has been derived at the last :) (See $I7$)

$$\sec2x+\tan2x=\tan(45^\circ+x)\ \ \ \ (1)$$

$$x\to-x\implies\sec2x-\tan2x=\tan(45^\circ-x)\ \ \ \ (2)$$

$$2x\to180^\circ-2x\implies(1)$$

$$2x\to180^\circ+2x\implies(2)$$

$$\tan2x+\tan2y=\sec2(x+y)\ \ \ \ (3A)$$ $$\iff\sin(4x+4y)=2\cos2x\cos2y\ \ \ \ (3B)$$

If $2x\to180^\circ-2x, y\to-y\implies\ \ \ \ (3A)$

So, if we replace $y$ with $-y,2x$ will be replaced by $180^\circ-2x$

So, we can safely avoid $y<0$

Case $\#1:$ If $2\cos2y=1,2y=\pm60^\circ$

$'+'\implies\cos2x=\sin(4x+120^\circ)=\cos(4x+30^\circ)$

$$4x+30^\circ=360^\circ n\pm2x$$

$'+'\implies2x\equiv-30^\circ\pmod{360^\circ}\implies\tan(-30^\circ)+\tan(60^\circ)=\sec(-30^\circ)$

which is a special case of $(2)$ with $2x=30^\circ$

$'-'\implies2x=120^\circ n-10^\circ$ $$\tan(120^\circ n-10^\circ)+\tan60^\circ=\sec(120^\circ n+50^\circ)$$

$$n=0\implies\tan60^\circ-\tan10^\circ=\sec50^\circ\ \ \ \ (I1)$$

$$n=1\implies\tan110^\circ+\tan60^\circ=\sec170^\circ\iff\tan70^\circ-\tan60^\circ=\sec10^\circ\ \ \ \ (I2)$$

$$n=2\implies\tan230^\circ+\tan60^\circ=\sec290^\circ\iff\tan50^\circ+\tan60^\circ=\sec70^\circ\ \ \ \ (I3)$$

Case $\#2:$ If $2\cos2y=-1,2y=\pm120^\circ$

$2y\to180^\circ-2y,x\to-x\implies (3A)$

Case $\#3:$ If $\cos2y=0,2y=180^\circ n+90^\circ$

$(3B)$ becomes $$\sin(4x+360^\circ n+180^\circ)=0$$

$4x+360^\circ n+180^\circ=180^\circ m\iff2x=90^\circ(m-2n-1)$

$$\tan2x+\tan60^\circ=\sec2y\ \ \ \ (4)$$

$y\to-y\implies(4)$

$$\cos2x=2\sin(2x+60^\circ)\cos2y=2\cos(2x-30^\circ)\cos2y$$

$$\iff\cos(2x-2y-30^\circ)+\cos(2x+2y-30^\circ)+\cos(180^\circ-2x)=0$$

Now from Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$,

$$\cos36^\circ+\cos108^\circ+\cos120^\circ=0$$

Case $\#4A:$

If $\cos(180^\circ-2x)=\cos36^\circ,180^\circ-2x=\pm36^\circ$

Case $\#4A1:$ $+\implies2x=144^\circ,$

$2x-2y-30^\circ,2x+2y-30^\circ$ becomes $\{114^\circ-2y,114^\circ+2y\}$

which need to be $\equiv\{\pm108^\circ,\pm120^\circ\} $

which is satisfied by $2y=\pm6^\circ$

$$\implies\tan144^\circ+\tan60^\circ=\sec6^\circ\iff-\tan36^\circ+\tan60^\circ=\sec6^\circ\ \ \ \ (I4)$$

Case $\#4A2:$ $-\implies2x=216^\circ,$

$2x-2y-30^\circ,2x+2y-30^\circ$ becomes $\{186^\circ-2y,186^\circ+2y\}$

which is satisfied by $2y=\pm66^\circ$

$$\implies\tan216^\circ+\tan60^\circ=\sec66^\circ\iff\tan36^\circ+\tan60^\circ=\sec66^\circ\ \ \ \ (I5)$$

Case $\#4B:$

If $\cos(180^\circ-2x)=\cos108^\circ,180^\circ-2x=\pm108^\circ$

Case $\#4B1:$ $+\implies2x=72^\circ,$

$2x-2y-30^\circ,2x+2y-30^\circ$ becomes $\{42^\circ-2y,42^\circ+2y\}$

which is satisfied by $2y=78^\circ$

$$\implies\tan72^\circ+\tan60^\circ=\sec78^\circ\ \ \ \ (I6)$$

Case $\#4B2:$ $-\implies2x=288^\circ,$

$2x-2y-30^\circ,2x+2y-30^\circ$ becomes $\{258^\circ-2y,258^\circ+2y\}$

which is satisfied by $2y=138^\circ$

$$\implies\tan288^\circ+\tan60^\circ=\sec138^\circ\iff-\tan72^\circ+\tan60^\circ=-\sec42^\circ\ \ \ \ (I7)$$

Answer 5

A very nice euclidean approach.

Starting with a regular pentagon $ABCDE$ of side $2$, draw the equilateral triangles $\triangle ABF$ and $\triangle EBG$, as shown below. Produce $DF$ to $H \in AB$.

1. $\overline{FH} = \sqrt 3$.
2. $\overline{DH} = \tan 72^\circ$.
3. $\triangle DGB$ is isosceles, with $\measuredangle DBG = 24^\circ$, hence $\measuredangle BDG = 78^\circ$, and $\measuredangle GDC = 42^\circ$.
4. Since $\triangle GDC$ is isosceles with base $\overline{CD} = 2$, $\overline{DG} = \sec 42^\circ$.
5. By SAS criterion $\triangle DAG \cong \triangle ADF$, and, therefore $DF \cong DG$.
6. By 1., 2., 4., and 5. we have $$\boxed{\tan 72^\circ = \sec 42^\circ + \sqrt 3}.$$

Answer 6

Golden triangles, I use these mnemonic:

$\;\displaystyle \cos 36° = \frac{\sqrt{2+\frac{1}{ϕ}}}{2} = \frac{ϕ}{2}\;,\; \sin 36° = \frac{\sqrt{2-\frac{1}{ϕ}}}{2}$

$\;\displaystyle \cos 18° = \frac{\sqrt{2+ϕ}}{2} \qquad\;\;,\; \sin 18° = \frac{\sqrt{2-ϕ}}{2} = \frac{1}{2ϕ}$

---

$\displaystyle \cos (60°-18°) = \frac{1}{2} × \frac{\sqrt{2+ϕ}}{2} \;+\; \frac{\sqrt{3}}{2} × \frac{1}{2ϕ}$

$\displaystyle \sec 42° = \frac{4ϕ}{ϕ\sqrt{2+ϕ} \,+ \sqrt{3}} = \frac{4ϕ\,(ϕ\sqrt{2+ϕ} \,- \sqrt{3})}{ϕ^2\,(2+ϕ)-3} = ϕ\sqrt{2+ϕ} \,- \sqrt{3}$

First term: $\;ϕ\sqrt{2+ϕ} \,= \cot 18° = \tan 72°$