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Find $x$ in degrees if $\cot(x)=\csc(12^{\circ})-\sqrt{3}$

My attempt: $$\cot (x)=\frac{1}{\sin (12^{\circ})}-2 \sin \left(60^{\circ}\right)$$ $$\Rightarrow \cot x=\frac{1-2 \sin (12^{\circ}) \sin (60^{\circ})}{\sin \left(12^{\circ}\right)}$$

$$\Rightarrow \cot x=\frac{1-\cos 48^{\circ}+\cos 72^{\circ}}{\sin \left(12^{\circ}\right)}$$

Now let $, \theta=12^{\circ},s=\sin(\theta)$, then we get $$\cot x=\frac{1-\cos (4 \theta)+\cos (6\theta)}{\sin (\theta)}$$

Converting to rational function in $s$, we get $$\cot x=\frac{-32 s^{6}+40 s^{4}-10 s^{2}+1}{s}$$

Umesh shankar
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    Merely on your attempt and nothing more : I feel that the jump into powers of $s$ is too much : I'd rather look to try and manipulate things so that I get a single trigonometric function on the RHS, although at the moment it's very hard for me to see what's going on. I'll see if I can get something to work. – Sarvesh Ravichandran Iyer Jun 08 '21 at 04:00
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    I would advise you to calculate values – Lalit Tolani Jun 08 '21 at 04:05

1 Answers1

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This is a very similar yet a little different question compared to this problem, and the answer by @albert chan paves the way for a similar solution to this problem.

$$ \sin(12°) = \sin(30°-18°) = \sin(30°)\cos(18°) - \cos(30°)\sin(18°)$$ $$= {1\over2} (\cos(18°) - \sqrt3 \sin(18°))$$

Converting $\sin$ to $\csc$ and rationalizing the denominator:

$$\csc(12°) = \left({2 \over \cos(18°) - \sqrt3 \sin(18°)}\right) \left({\cos(18°) + \sqrt3 \sin(18°) \over \cos(18°) + \sqrt3 \sin(18°)}\right)$$ $$= {2(\cos(18°) + \sqrt3 \sin(18°)) \over \cos^2(18°) - 3\sin^2(18°)} $$

Using $\cot(x)\sin(x)=\cos(x)$ for numerator and Pythagorean trig identity for denominator: $$= \left({2\sin(18°) \over 1 -4 \sin^2(18°)}\right) (\cot(18°) + \sqrt3) $$

Let $s=\sin(18°)$, using multiple angles formula and the fact that $s≠1$:

$$\sin(90°) = \sin(5 \times 18°) = 16s^5 - 20s^3 + 5s -1 = (s-1)(4s^2+2s-1)^2 = 0 $$

$$\implies 4s^2+2s-1 = 0 \implies {2s \over 1-4s^2} = 1$$

$$\fbox{$ \cot(x) = (\cot(18°) + \sqrt3) - \sqrt3 = \cot(18°)$}$$

krazy-8
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