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Section 6
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Here we extend
the trigonometric ratios to all angles by defining the trigonometric functions of angles
...
Place θ in
standard position as shown in Figure (b) below
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In triangle P OQ, the opposite side has
length y and the adjacent side has length x
...
So
y
r
r
csc θ =
y
sin θ =

x
r
r
sec θ =
x

y
x
x
cot θ =
y

cos θ =

tan θ =

These observations allow us to extend the trigonometric ratios to any angle
...
But cos 45◦ = x/r, and since
√
√
2

...
From the second Figure below it’s clear that
√
√
3

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That acute angle will be called the reference angle
...

(b) The angles 870◦ and 150◦ are coterminal [because 870 − 2(360) = 150]
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So the reference angle is
θ = 180◦ − 150◦ = 30◦

EXAMPLE: Find the reference angle for
13π
17π
(b) θ =
(a) θ =
6
7
Solution:
18π − π
18π π
π
17π
=
=
− = 3π − , therefore θ =
(a) We have
6
6
6
6
6
13π
14π − π
14π π
π
(b) We have
=
=
− = 2π − , therefore θ =
7
7
7
7
7

EXAMPLE: Find
(a) sin 240◦

(b) cot 495◦
3

π
(Quadrant II)
...

7

EXAMPLE: Find
(a) sin 240◦

(b) cot 495◦

Solution:
(a) This angle has its terminal side in Quadrant III, as shown in the first Figure below
...
Thus

(b) The angle 495◦ is coterminal with the angle 135◦ (since 495◦ − 360◦ = 135◦ ), and the
terminal side of this angle is in Quadrant II, as shown in the second Figure below
...
We have

EXAMPLE: Find
16π
(a) sin
3

(b) sec −

π
4

4

EXAMPLE: Find
16π
(a) sin
3
Solution:
(a) Since

(b) sec −

π
4

16π
15π + π
15π π
π
=
=
+ = 5π +
3
3
3
3
3
the reference number for 16π/3 is π/3 (see the first Figure below) and the terminal point of
16π/3 is in Quadrant III
...
Since secant is positive in this quadrant, we get

5

Trigonometric Identities

EXAMPLE:
(a) Express sin θ in terms of cos θ
...

Solution:
(a) From the first Pythagorean identity we get
√
sin θ = ± 1 − cos2 θ
where the sign depends on the quadrant
...
By part (a)
1 − sin2 θ

cos θ = ±

and since cos θ is negative in Quadrant II, the negative sign applies here
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Solution: The triangle has sides of length 10 cm and 3 cm, with included angle 120◦

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