Question

If tan⁡ x=√3, and 180°<x<270°, what is sin⁡(x−210°)?

still don't quite get it...

Answer 1

tan60° = √3
so, in QIII, tan(180+60) = tan240 = √3
240-210 = 30
tan30 = 1/√3
or, more generally, since tan210 = tan30 = 1/√3,
tan(x-210) = (tanx - tan210)/(1 + tanx tan210)
= (√3 - 1/√3)/(1 + √3 * 1/√3)
= (√3 - 1/√3)/2
= (3-1)/2√3
= 1/√3

Answer 2

oops. misread the problem.
sin(x-210) = sin30 = 1/2
or
in QIII, tanx = √3
so sec^2x = 1+tan^2x = 1+3 = 4
and cosx = -1/2, sinx = -√3/2
sin210 = sin(180+30) = -sin30 = -1/2
cos210 = -√3/2
sin(x-210) = sinx cos210 - cosx sin210
= (-√3/2)(-√3/2) - (-1/2)(-1/2)
= 3/4 - 1/4 = 1/2

Answer 3

given: tan x = √3 , and given: x is in III
the angle in standard position is 60°

so x = (180+60)° = 240°
then sin (240-210) = sin 30° = 1/2

Answer 4

thank u!!