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

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) $$\left( \begin{array}{cc} k & R_k \\ 5 & \frac{1}{\sqrt{3}} \\ 7 & \sqrt{3} \\ 19 & \sqrt{3} \\ 20 & 2-\frac{1}{\sqrt{3}} \\ 25 & \frac{2}{\sqrt{3}} \\ 30 & 1+\frac{1}{\sqrt{3}} \\ 31 & \sqrt{3} \\ 35 & \frac{5}{\sqrt{3}} \\ 43 & \sqrt{3} \\ 50 & -2-\frac{1}{\sqrt{3}} \\ 55 & -\frac{2}{\sqrt{3}} \\ 60 & -1+\frac{1}{\sqrt{3}} \end{array} \right)$$


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

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 $4s^2+2s-1 = 0\quad → \large{2s \over 1-4s^2} = 1$

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

Tags:

Trigonometry

Calculus

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