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Trigonometric Functions Review
Circular Representation. Recall that angles of any size can be represented in a unit circle (circle of
radius 1). More specifically, we wind the horizontal radius OA (fixed at O) starting from the horizontal
axis through the size of the angle we want to depict. For example, to represent the angle θ = 3π/4 we
rotate the radius OA counter-clockwise till we capture the size of the angle θ. See Figure 1 below. To
represent a negative angle rotate the radius OA clockwise. See Figure 2.
B
O 1
θ = 3π/
A
1
C
1
O
ρ = -3π/
A
1
Figure 1. Figure 2.
The point B on the unit circle is associated to the angle θ = 3π/4 and the point C on the unit circle is
associated to the negative angle ρ = − 3 π/4. The radius OB is called the terminal edge of the angle
θ = 3π/4. The terminal edge of the angle ρ = − 3 π/4 is OC.
Reference Angle. The reference angle α of a given angle θ is the acute angle (0 ≤ α ≤ π/2) between
the terminal edge and the horizontal axis in the circular representation of angle θ. The figures below
show the reference angle of an angle whose terminal edge ends in one of the four quadrants.
B
O A
α
B
O A
α
B
α O A
B
O α A
First Quadrant Second Quadrant Third Quadrant Fourth Quadrant
Example 1: The reference angle of θ = 3π/4 and ρ = − 3 π/4 are both π/4. Can you see why?
Trigonometric Functions. The trigonometric functions of an angle are given by the coordinates of
its associated point on the unit circle as follows:
Let the intersection of the terminal edge of angle θ with the unit circle be (x, y). The figure below
illustrate the position of (x, y) for angle θ.
B = (x,y)
O x A
y (x,y) = B
O A
y
x
(x,y) = B
O A
y
x
B = (x,y)
O A
y
x
First Quadrant Second Quadrant Third Quadrant Fourth Quadrant
Then the valuation of the sine, cosine and tangent functions at angle θ associated to point (x, y) on the
unit circle are given by:
sin θ = y; cos θ = x; tan θ =
y
x
Considering the signum (+ or −) of the coordinates (x, y) in the figures above,
(1) If θ has its terminal edge in the first quadrant, then ALL trigonometric functions are positive.
That is sin(θ) > 0, cos(θ) > 0, and tan(θ) > 0.
(2) If θ has its terminal edge in the second quadrant, then only the sine function is positive. That is
sin(θ) > 0, cos(θ) < 0, and tan(θ) < 0.
(3) If θ has its terminal edge in the third quadrant, then only the tangent function is positive. That
is sin(θ) < 0, cos(θ) < 0, and tan(θ) > 0.
(4) If θ has its terminal edge in the fourth quadrant, then only the cosine function is positive. That
is sin(θ) < 0, cos(θ) > 0, and tan(θ) < 0.
You may remember this pattern with the mnemonic illustration below:
CAST Diagram
A
x
y
S
T C
B = (x,y)
O x A
y
α
B = (x,y)
O A
y
x
α
B = (x,y)
O A
y
x α
B = (x,y)
O A
y
x α
First Quadrant Second Quadrant Third Quadrant Fourth Quadrant
Referring to the figures above, if α is the reference angle of θ then the values of sin θ, cos θ, and tan θ
can be expressed in terms of α as follows:
(1) θ is in the first quadrant:
sin θ = sin α; cos θ = cos α; tan θ = tan α
(2) θ is in the second quadrant:
sin θ = sin α; cos θ = − cos α; tan θ = − tan α
(3) θ is in the third quadrant:
sin θ = − sin α; cos θ = − cos α; tan θ = tan α
(4) θ is in the fourth quadrant:
sin θ = − sin α; cos θ = cos α; − tan θ = tan α
Notice that the signs are according to the CAST diagram above.
Example 2: Find the exact values of (1) sin
5 π
3
, (2) cos
5 π
3
and (3) tan
5 π
6
5 π
3
is in the fourth quadrant with reference angle
π
3
. Sine
is negative and cosine is positive in the fourth quadrant. Therefore we
have:
sin
5 π
3
= − sin
π
3
and cos
5 π
3
= cos
π
3
5 π/3 π/
5 π
6
is in the third quadrant with reference angle
π
6
. Tangent is
positive in the third quadrant. Therefore we have:
tan
5 π
6
= tan
π
6
π/
Example 3: Find the values of (a) arctan
, (b) arccos
and (c) arcsin
(a) Let A = arctan
. Then tan(A) = −
3 and −
π
2
< A <
π
2
Since tan(A) is negative, A must be in the fourth quadrant. The value √ 3 says that the reference angle α of A must be
π
3
because tan(α) = √
Therefore A = −
π
3
π/
(b) Let B = arccos
. Then cos(B) = −
and 0 ≤ B ≤ π.
Since cos(B) is negative, B must be in the second quadrant. The value
1
2
says that the reference angle β of B must be
π
3
because cos(β) =
Therefore B = π −
π
3
2 π
3
π/
(c) Let C = arcsin
. Then sin(C) =
and −
π
2
≤ C ≤
π
2
Since sin(C) is positive, C must be in the first quadrant. The only
acute angle C such that sin(C) =
is C =
π
3
π/
