Trigonometric Arithmetic III

we have:

$\displaystyle *\cos x +\sin x \tan \frac{x}{2} = \cos x + \sin x (\frac{1-\cos x}{\sin x} )= 1$

$\displaystyle * \sin (x-30^{\circ}) +\cos(60^{\circ} - x) = \frac{\sqrt{3}}{2}\sin x - \frac{1}{2} \cos x + \frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x = \sqrt{3}\sin x$

Therefore our equation can be written as :

$\displaystyle \sec^2x -1=\frac{ \sqrt{3}\sin x}{\cos x}$

$\displaystyle <=> \tan x (\tan x -\sqrt{3})=0$

$\displaystyle => x=180^{\circ}n$ or $x= 60^{\circ} + 180^{\circ}n= 60^{\circ}(3n+1)$ , so $a = 60, b = 3, a/b = 20$

Note that if $x$ is a multiple of $180^\circ$ , then $\tan \frac x 2$ is undefined.