Question 1077833
1) If you know (or figure out, or look up)
the values of the main trigonometric functions
(cosine, sine, and tangent)
for {{{pi/3}}} and {{{pi/4}}} ,
you are 1/3 of the way to the answer.
{{{matrix(3,4,theta,cosine, sine, tangent,
pi/4,sqrt(2)/2,sqrt(2)/2,1,
pi/3,1/2,sqrt(3)/2,sqrt(3))}}}
 
2) You need to know some trigonometric identities that are really definitions:
{{{cot(theta)=1/tan(theta)}}} , and {{{sec(theta)=1/cos(theta)}}} .
 
3) You also need to understand the relation between
angles in other quadrants and
their symmetrical first quadrant reference angles.
{{{drawing(300,300,-1.2,1.2,-1.2,1.2,grid(0),
red(circle(0,0,1)),arrow(0,0,1.2,0.8),
arrow(0,0,-1.2,0.8),arrow(0,0,-1.2,-0.8),
arrow(0,0,1.2,-0.8),arc(0,0,1,1,-33.7,0),
locate(0.36,0.2,theta),arc(0,0,1,1,180,213.7),
locate(-0.4,0.2,theta),arc(0,0,1.2,1.2,0,33.7),
locate(0.45,-0.1,theta),arc(0,0,1.2,1.2,146.3,180),
locate(-0.5,-0.1,theta),arc(0,0,1.8,1.8,180,360),
locate(-0.63,0.63,pi),triangle(-0.9,0,-0.93,0.05,-0.87,0.05),
red(arc(0,0,1.4,1.4,213.7,360)),locate(0.07,0.63,red(pi-theta)),
red(triangle(-0.58,0.39,-0.54,0.39,-0.58,0.43))
)}}}
Foe example (and relevant to the problem)
the diagram shows you that
{{{matrix(5,4,Angle,"cosine =", "sine =", "tangent =",
pi-theta,-cos(theta), sin(theta), -tan(theta),
pi+theta,-cos(theta), -sin(theta), tan(theta),
2pi-theta,cos(theta), -sin(theta), -tan(theta),
-theta,cos(theta), -sin(theta), -tan(theta))}}}
 
There are diagrams, tables and charts that can help you.
Your textbook probably has then.
Your teacher probably handed out that kind of help.
You can find them online searching for "trig identities."
Wikipedia has a nice diagram, and a table.
However, you can figure out the answers by yourself,
with the Pythagorean theorem.
For {{{pi/3}}} think of half of an equilateral triangle.
{{{drawing(300,300,-1.1,1.1,-0.2,1.8,
green(triangle(0,0,1,0,0,1.732)),
green(rectangle(0,0,0.05,0.05)),
triangle(-1,0,1,0,0,1.732),locate(0.02,0.8,h),
locate(0.35,0.1,"1 / 2"),locate(0.45,0.88,1),
red(arc(0,1.732,0.6,0.6,60,90)),
red(arc(1,0,0.65,0.65,180,240)),
locate(0.05,1.45,red(pi/6)),
locate(0.75,0.23,red(pi/3))
)}}}
The Pythagorean theorem lets you figure out
the length of the long leg as {{{h=sqrt(3)/2}}} .
So, for the {{{pi/3=60^o}}} , we have the trigonometric ratios
{{{cos(pi/3)=1/2}}} , {{{sin(pi/3)=sqrt(3)/2}}} , and {{{tan(pi/6)=(sqrt(3)/2)/(1/2)=sqrt(3)}}} .
For {{{pi/4}}} think of half of a square.