1. Trigonometry as you’ve already seen it before
We have defined sine and cosine for right triangles. What is we would like to
consider sine and cosine for arbitrary real numbers? Consider a circle of radius r
centered at the origin. Draw a line from the point (1, 0) to the origen. Pick any
point P = (x, y) in the first quadrant (x > 0, y > 0). Consider Figure 1 below.
y
1
1
2
−1
− 21
− 12
−1
The angle θ is 30○ in the example (π/6 in radians). The sine of
θ, which is the height of the red
line, is
P
sin θ = 1/2.
sin θ By the Pythagorean Thoerem,
2
tan θ =
2
sin θ
cos θ we have cos θ + sin θ=1. So
the length of the blue line, which
θ
x
is cosine of θ, is
cos θ
1
√
√
3
cos θ = 1 − 1/4 =
.
2
This gives us that tan θ, which is
the height of the orange lline, it
1
sin θ
=√
tan θ =
cos θ
3
Figure 1
opp
We already know that for right trangles, sin θ = hyp
. Since our hypotenuse is 1,
adj
we have that sin θ = y. The same goes for cos θ. cos θ = hyp
= x. Of course, for
circles with radius r, we have that x = r cos θ and y = r sin θ.
This is all well and good, but we would like to extend our notion of sine and
cosine to arbitrary real numbers rather than restricting to 0 < θ < π/2. First, let
θ ≥ 0. Imagine tracing a curve from the point (1, 0) in a counter-clockwise direction
for a distance of θ. If θ ≥ 2π, we just keep tracing around the circle as necessary.
Place a point P where you stop tracing the curve. For example, for θ = π, our
point P is (−1, 0). For θ = 5π
2 , our point is (0, 1) (which happens to be the same
as with θ = π2 ).
For an arbitrary point P = (x, y), let’s define cos θ = x and sin θ = y. So for θ = 3π
2 ,
3π
3π
we have that P = (0, −1). So cos 2 = 0 and sin 2 = −1. Take any real number
1
2
θ ≥ 0. Imagine winding counterclockwise around the circle of radius one, starting
at (1, 0), for exactly a distance of θ. The end point will be a point P = (x, y).
Then cos θ = x and sin θ = y.
But what about for negative real numbers? As you may imagine, we handle
”negative angles” by thinking in reverse. For an arbitrary negative number θ,
imagine winding clockwise, again starting at (1, 0), for a distance of θ. The endpoint will be a point P = (x, y). Again, define cos θ = x and sin θ = y. For example,
if θ = − 5π
2 , our point will be (0, 1).
5π
So cos − 5π
2 = 0 and sin − 2 = 1. Now we have a notion for sin θ and cos θ that
makes since for any real value of θ. Of course, when thinking about circles with
radius r, we have r cos θ = x and r sin θ = y.
2. Some common equalities
Consider figure 1 again. Since x = cos θ and y = sin θ, by the pythagorean
theorem, cos2 θ + sin2 θ = 11.
Recall that sin 0 = sin 2π and cos 0 = cos 2π. So for an arbitrary angle θ,
sin θ = sin(θ + 2π) and cos θ = cos(θ + 2π)
2π).
Note that for any angle θ, θ + π will give a point directly across the unit circle.
For example, if θ = π6 , then θ + π = 7π
6 . Notice how these are directly across the unit
circle from each other. It should be easy to convince yourself that for any angle θ,
sin(θ + π) = − sin θ and cos(θ + π) = − cos θθ.
You should also convince yourself that sin(−θ) = − sin θ and cos(−θ) = cos θθ.
3. Tangent
In chapter 1, we defined tangent as the quotient of sine and cosine. That is,
sin θ
tan θ = cos
θ . Notice that whenever θ is such that cos θ = 0, tan θ is undefined. (For
θ = π/2 + kπ, for any integer k.) Also note that tan(−θ) = − tan θ.