Question 1110588
We like to "reference" an acute (or right) angle {{{theta}}} , such that {{{0<=theta<=pi/2}}} ,
because we can visualize such an angle in a right triangle.
So, when we find an angle that does not fit in that first quadrant interval,
we find a "symmetrical" reference angle in the first quadrant, because we know that, give or not a minus sign,
the trigonometric function values of both angles will be the same.
 
{{{theta=38pi/11=(33+5)pi/11=33pi/11+5pi/11=3pi+5pi/11}}}
{{{5pi/11}}} is between {{{0}}} and {{{5.5pi/11=pi/2}}} ,
so it is less than a {{{pi/2}}} right angle,
but {{{3pi}}} is {{{1&1/2}}} counterclockwise turns,
or {{{6}}} quadrants.
The symmetrical reference angle, in the first quadrant is
{{{(3pi+5pi/11)-2pi-pi=highlight(5pi/11)}}}
{{{drawing(400,400,-1.2,1.2,-1.2,1.2,grid(0),
arrow(0,0,-0.171,-1.188),red(circle(0,0,1)),
arrow(0,0,0.171,1.188),circle(0.142,0.99,0.02),
circle(-0.142,-0.99,0.02),locate(0.142,0.97,Q),
locate(-0.14,-1,P),arc(0,0,0.4,0.4,-81.8,0),
arc(0,0,0.4,0.4,98.18,180),
locate(0.16,0.25,5pi/2),green(arc(0,0,1.8,1.8,270,0)),
green(arc(0,0,1.7,1.8,180,270)),green(arc(0,0,1.7,1.7,90,180)),
green(arc(0,0,1.6,1.7,0,90)),green(arc(0,0,1.6,1.6,270,0)),
green(arc(0,0,1.5,1.6,180,270)),green(arc(0,0,1.5,1.5,98.18,180)),
green(triangle(-0.107,-0.742,-0.14,-0.7,-0.14,-0.76)),
locate(-0.53,-0.33,green(38pi/11))
)}}}
 
The coordinates of {{{Q(cos(5pi/11),sin(5pi/11))}}} and {{{P(-cos(5pi/11),-sin(5pi/11))}}} have the same absolute value,
but with {{{P}}} being in the third quadrant,
both of the coordinates of {{{P}}} are negative.
 
{{{pi=about3.14}}} and
{{{3pi/2=about 4.71}}} ,
so {{{theta=4}}} is in the third quadrant, with {{{pi<theta<3pi/2}}} .
Just as before, for the third quadrant we add or subtract {{{pi}}} (a half turn),
as many times as needed,
until we get to the first quadrant.
The reference angle is {{{4-pi}}}
{{{drawing(400,400,-1.2,1.2,-1.2,1.2,grid(0),
arrow(0,0,-0.98,-1.135),red(circle(0,0,1)),
arrow(0,0,0.98,1.135),
arc(0,0,0.6,0.6,-49.18,0),
arc(0,0,0.5,0.5,130.82,180),
red(arc(0,0,0.5,0.5,180,360)),
green(arc(0,0,1.1,1.1,130.82,360)),
green(arc(0,0,1,1,130.82,360)),
locate(0.28,0.15,4-pi),locate(-0.177,0.177,red(pi)),
locate(-0.23,0,4-pi),locate(-0.353,0.354,green(4))
)}}}
 
NOTE:
For second quadrant, and fourt angles,
the first quadrant reference angle can be found,
by adding and/or subtracting {{{pi}}} as many times as needed
to get to an angle between {{{pi/2}}} and {{{pi}}}
(in the second quadrant, but less than one whole turn),
and then subtracting that angle from {{{pi}}} to get the supplementary angle.
For example, {{{64pi/11=55pi/11+9pi/11=5pi+9pi/11}}} is
{{{2&1/2)(2pi)}}} (or {{{2&1/2}}} counterclockwise turns, or {{{5}}} half turns)
plus {{{9pi/11}}} .
Subtracting those {{{5half-turns=5pi}}} , we get {{{9pi/11}}} ,
with {{{pi/2=5.5pi/11<9pi/11<11pi/11=pi}}} .
Then, {{{pi-9pi/11=2pi/11}}} is the first quadrant reference angle.