Question 1174577
[a] sin θ=1/2, θ in Q1; cos θ<pre>
We draw a right triangle with an angle in Q1, and remember that the sine is the
opposite over the hypotenuse (or the y over the r). So we put the numerator of
1/2, which is 1, on the opposite side (or the y) and the denominator of 1/2,
which is 2, on the hypotenuse, (or the r). Then we calculate the adjacent using
the Pythagorean theorem:

{{{matrix(5,1,
x^2+y^2=r^2,
x^2+1^2=2^2,
x^2+1=4,
x^2=3,
x="" +- sqrt(3))}}}

Since it's Q1 we know that x is positive, so we put √3 on the adjacent x:

{{{drawing(300,200,-1,1.5,-1,1,line(-2,0,2,0), line(0,-2,0,2),
green(line(sqrt(3)/2,1/2,sqrt(3)/2,0)),
locate(.3,-.02,x=sqrt(3)),locate(.9,.3,y=1),locate(.2,.4,r=2),
line(0,0,sqrt(3)/2,1/2)  )}}}

So now we know that

{{{cos(theta)=adjacent/hypotenuse = x/r=sqrt(3)/2}}}
</pre>[b] cos θ=3/5, θ in Q4; csc θ<pre>
We draw a right triangle with an angle in Q4, and remember that the cosine is
the adjacent over the hypotenuse (or the x over the r). So we put the numerator
of 3/5, which is 3, on the adjacent side (or the x) and the denominator of 3/5,
which is 5, on the hypotenuse, (or the r). Then we calculate the opposite using
the Pythagorean theorem:

{{{matrix(5,1,
x^2+y^2=r^2,
3^2+y^2=5^2,
9+y^2=25,
y^2=16,
y="" +- 4)}}}

Since it's Q4 we know that y is negative, so we put -4 on the opposite y:

{{{drawing(300,200,-1,1.5,-1,1,line(-2,0,2,0), line(0,-2,0,2),
green(line(3/5,-4/5,3/5,0)),
locate(.2,.2,x=3),locate(.6,-.3,y=-4),locate(.06,-.4,r=5),
line(0,0,3/5,-4/5)  )}}}

So now we know that

{{{csc(theta)=hypotenuse/opposite = r/y=5/(-4)=-5/4}}}

See if you can do the other two by yourself. [c] is in Q2 and [d] is in Q1

Edwin</pre>