Sine possibility

Geometry Level 4

( 1 + 1 sin x ) cos x + ( 1 + 1 cos x ) sin x = 2 + 2 \left(1+\frac{1}{\sin x}\right)\cos x+\left(1+\frac{1}{\cos x}\right)\sin x=2+\sqrt2

Find the number of possible values of sin x \sin x ?


The answer is 3.

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1 solution

Chew-Seong Cheong
Nov 26, 2020

( 1 + 1 sin x ) cos x + ( 1 + 1 cos x ) sin x = 2 + 2 cos x + cot x + sin x + tan x = 2 + 2 sin x + cos x 2 = 2 tan x cot x 2 sin ( x + π 4 ) 2 = 2 ( tan x + cot x ) \begin{aligned} \left(1+\frac 1{\sin x}\right)\cos x + \left(1+\frac 1{\cos x}\right)\sin x & = 2 + \sqrt 2 \\ \cos x + \cot x + \sin x + \tan x & = 2 + \sqrt 2 \\ \sin x + \cos x - \sqrt 2 & = 2 - \tan x - \cot x \\ \sqrt 2 \sin \left(x + \frac \pi 4 \right) - \sqrt 2 & = 2 - (\tan x + \cot x) \end{aligned}

Note that 2 2 sin ( x + π 4 ) 2 -\sqrt 2 \le \sqrt 2 \sin \left(x + \frac \pi 4 \right) \le \sqrt 2 then the left-hand side is 2 2 L H S 0 \rm -2\sqrt 2 \le LHS \le 0 . For 0 x < 2 π 0 \le x < 2\pi L H S = 0 \rm LHS = 0 , when x = π 4 x = \frac \pi 4 .

Considering tan x + cot x > 0 \tan x + \cot x > 0 , by AM-GM inequality tan x + cot x 2 tan x cot x = 2 \tan x + \cot x \ge 2 \sqrt{\tan x \cot x} = 2 . Then the right-hand side, R H S 0 \rm RHS \le 0 . For 0 x < 2 π 0\le x < 2\pi , R H S = 0 \rm RHS = 0 , when x = π 4 x = \frac \pi 4 and x = 5 4 π x= \frac 54 \pi . Therefore, L H S = R H S \rm LHS = RHS , when x = π 4 x = \frac \pi 4 and a value of sin x = sin π 4 = 1 2 \sin x = \sin \frac \pi 4 = \frac 1{\sqrt 2} . Since the R H S = 2 2 \rm RHS = -2\sqrt 2 , when x 0.214 , 1.357 , 3.355 , 4.499 x \approx 0.214, 1.357, 3.355, 4.499 , there are two other points when L H S = R H S \rm LHS = RHS . Therefore there are 3 \boxed 3 possible values of sin x \sin x .

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