Trigonometric Arithmetic III

Geometry Level 3

sec 2 x ( cos x + sin x tan x 2 ) = sin ( x 3 0 ) + cos ( 6 0 x ) cos x \sec ^{ 2 }{ x } -\left (\cos { x } +\sin { x } \tan { \frac { x }{ 2 } } \right )=\frac { \sin {(x-30^\circ)} +\cos{(60^\circ-x)} }{ \cos{x} }

If the solution for the trigonometric equation above is x = a ( b n + 1 ) x=a^\circ(bn+1) , for integers a , b , n a,b,n and a , b > 0 a,b > 0 , what is the value of a b \dfrac a b ?


The answer is 20.

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1 solution

Hasan Kassim
Sep 26, 2014

we have:

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

sin ( x 3 0 ) + cos ( 6 0 x ) = 3 2 sin x 1 2 cos x + 1 2 cos x + 3 2 sin x = 3 sin x \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 :

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

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

= > x = 18 0 n \displaystyle => x=180^{\circ}n or x = 6 0 + 18 0 n = 6 0 ( 3 n + 1 ) x= 60^{\circ} + 180^{\circ}n= 60^{\circ}(3n+1) , so a = 60 , b = 3 , a / b = 20 a = 60, b = 3, a/b = 20

Note that if x x is a multiple of 18 0 180^\circ , then tan x 2 \tan \frac x 2 is undefined.

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