Question 1182618
Evaluate the six trigonometric function of: s=2π/3. Show your solution.<pre>
<font size=4>
{{{2pi/3}}} is 120<sup>o</sup>, has a referent angle of 60<sup>o</sup> and is in the second quadrant, 
so we draw a 30-60-90 triangle with hypotenuse=r=2, adjacent=x=1, and
opposite=y=√3

However, x goes left in QII, so we must make x negative, so we have x=-1
instead of 1 in the graph below.  y is positive because it goes upward.
r, the hypotenuse or radius vector is always taken positive in all quadrants.
<pre>
{{{drawing(400,400,-2,2,-2,2, graph(400,400,-2,2,-2,2),
red(locate(.12,.56,s=2pi/3=120^o)),
line(-1,sqrt(3),-1,0), line(0,0,-1,sqrt(3)),locate(-3.2,.5,y=sqrt(3)),
red(arc(0,0,.7,-.7,0,120)),locate(-1.38,1,y=sqrt(3)),
locate(-.8,0.2,x=-1), locate(-.5,1,r=2) )}}}

adjacent = x = -1,  opposite = y = √3, hypotenuse = r = 2

{{{sin(s) = opposite/hypotenuse = y/r = sqrt(3)/2}}}
{{{cos(s) = adjacent/hypotenuse = x/r = (-1)/2 =-1/2}}}
{{{tan(s) = opposite/adjacent = y/x = sqrt(3)/(-1) = -sqrt(3)}}}
{{{sec(s) = hypotenuse/adjacent = r/x = 2/(-1)=-2}}}
{{{csc(s) = hypotenuse/opposite = r/y = 2/(sqrt(3)) = 2sqrt(3)/3}}} 
{{{cot(s) = adjacent/opposite = x/y = (-1)/sqrt(3) = -sqrt(3)/3}}} 
 
Edwin</font></pre>