[a] sin θ=1/2, θ in Q1; cos θ
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:
Since it's Q1 we know that x is positive, so we put √3 on the adjacent x:
So now we know that
[b] cos θ=3/5, θ in Q4; csc θ
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:
Since it's Q4 we know that y is negative, so we put -4 on the opposite y:
So now we know that
See if you can do the other two by yourself. [c] is in Q2 and [d] is in Q1
Edwin