Section 6.3 Trigonometric Functions of Angles

In the preceding section we defined the trigonometric ratios for acute angles. Here we extendthe trigonometric ratios to all angles by defining the trigonometric functions of angles.

Trigonometric Functions of Angles

Let POQ be a right triangle with acute angle θ as shown in Figure (a) below. Place θ instandard position as shown in Figure (b) below.

Then P = P (x, y) is a point on the terminal side of θ. In triangle POQ, the opposite side haslength y and the adjacent side has length x. Using the Pythagorean Theorem, we see that thehypotenuse has length r =

√

x2 + y2. So

sin θ =y

rcos θ =

x

rtan θ =

y

x

csc θ =r

ysec θ =

r

xcot θ =

x

y

These observations allow us to extend the trigonometric ratios to any angle. We define thetrigonometric functions of angles as follows

1

Evaluating Trigonometric Functions at Any Angle

EXAMPLE: Find

(a) cos 135◦ (b) tan 390◦

Solution:

(a) From the first Figure below we see that cos 135◦ = −x/r. But cos 45◦ = x/r, and since

cos 45◦ =√2/2, we have cos 135◦ = −

√2

2.

(b) The angles 390◦ and 30◦ are coterminal. From the second Figure below it’s clear that

tan 390◦ = tan 30◦, and since tan 30◦ =√3/3, we have tan 390◦ =

√3

3.

From the Example above we see that the trigonometric functions for angles that aren’t acutehave the same value, except possibly for sign, as the corresponding trigonometric functions ofan acute angle. That acute angle will be called the reference angle.

EXAMPLE: Find the reference angle for

(a) θ =5π

3(b) θ = 870◦

2

EXAMPLE: Find the reference angle for

(a) θ =5π

3(b) θ = 870◦

Solution:

(a) Since5π

3=

6π − π

3=

6π

3−

π

3= 2π −

π

3

the reference number for 5π/3 is π/3 and the terminal point of 5π/3 is in Quadrant IV.

(b) The angles 870◦ and 150◦ are coterminal [because 870− 2(360) = 150]. Thus, the terminalside of this angle is in Quadrant II (see the second Figure below). So the reference angle is

θ = 180◦ − 150◦ = 30◦

EXAMPLE: Find the reference angle for

(a) θ =17π

6(b) θ =

13π

7

Solution:

(a) We have17π

6=

18π − π

6=

18π

6−

π

6= 3π −

π

6, therefore θ =

π

6(Quadrant II).

(b) We have13π

7=

14π − π

7=

14π

7−

π

7= 2π −

π

7, therefore θ =

π

7(Quadrant IV).

EXAMPLE: Find

(a) sin 240◦ (b) cot 495◦

3

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. Thereference angle is therefore 240◦ − 180◦ = 60◦, and the value of sin 240◦ is negative. Thus

(b) The angle 495◦ is coterminal with the angle 135◦ (since 495◦ − 360◦ = 135◦), and theterminal side of this angle is in Quadrant II, as shown in the second Figure below. So thereference angle is 180◦ − 135◦ = 45◦, and the value of cot 495◦ is negative. We have

EXAMPLE: Find

(a) sin16π

3(b) sec

(

−π

4

)

4

EXAMPLE: Find

(a) sin16π

3(b) sec

(

−π

4

)

Solution:

(a) Since16π

3=

15π + π

3=

15π

3+

π

3= 5π +

π

3

the reference number for 16π/3 is π/3 (see the first Figure below) and the terminal point of16π/3 is in Quadrant III. Thus sin(16π/3) is negative and

(b) The angle −π/4 is in Quadrant IV, and its reference angle is π/4 (see the second Figurebelow). Since secant is positive in this quadrant, we get

5

Trigonometric Identities

EXAMPLE:

(a) Express sin θ in terms of cos θ.

(b) Express tan θ in terms of sin θ where θ is in Quadrant II.

Solution:

(a) From the first Pythagorean identity we get

sin θ = ±√1− cos2 θ

where the sign depends on the quadrant. If θ is in Quadrant I or II, then sin θ is positive, andhence

sin θ =√1− cos2 θ

whereas if θ is in Quadrant III or IV, sin θ is negative, and so

sin θ = −√1− cos2 θ

(b) Since tan θ = sin θ/ cos θ, we need to write cos θ in terms of sin θ. By part (a)

cos θ = ±√

1− sin2 θ

and since cos θ is negative in Quadrant II, the negative sign applies here. Thus

tan θ =sin θ

cos θ=

sin θ

−√

1− sin2 θ

6

Areas of Triangles

EXAMPLE: Find the area of triangle ABC shown in the Figure below.

Solution: The triangle has sides of length 10 cm and 3 cm, with included angle 120◦. Therefore

A =1

2ab sin θ

=1

2(10)(3) sin 120◦

= 15 sin 60◦

= 15

√3

2≈ 13 cm2

7