#### Transcript Chapter 5 Section 3 - Georgia College & State University

```5
Trigonometric
Functions
5.3-1
5 Trigonometric Functions
5.1 Angles
5.2 Trigonometric Functions
5.3 Evaluating Trigonometric Functions
5.4 Solving Right Triangles
5.3-2
5.3 Evaluating Trigonometric
Functions
Right-Triangle-Based Definitions of the Trigonometric Functions ▪
Cofunctions ▪ Trigonometric Function Values of Special Angles ▪
Reference Angles ▪ Special Angles as Reference Angles ▪
Finding Function Values Using a Calculator ▪ Finding Angle
Measures with Special Angles
1.1-3
5.3-3
Right-Triangle-Based Definitions
of Trigonometric Functions
For any acute angle A in standard position,
1.1-4
5.3-4
Right-Triangle-Based Definitions
of Trigonometric Functions
For any acute angle A in standard position,
1.1-5
5.3-5
Right-Triangle-Based Definitions
of Trigonometric Functions
For any acute angle A in standard position,
1.1-6
5.3-6
Example 1
FINDING TRIGONOMETRIC FUNCTION
VALUES OF AN ACUTE ANGLE
Find the sine, cosine,
and tangent values for
angles A and B.
1.1-7
5.3-7
Example 1
FINDING TRIGONOMETRIC FUNCTION
VALUES OF AN ACUTE ANGLE (cont.)
Find the sine, cosine,
and tangent values for
angles A and B.
1.1-8
5.3-8
Cofunction Identities
For any acute angle A in standard position,
sin A = cos(90  A)
csc A = sec(90  A)
tan A = cot(90  A)
cos A = sin(90  A)
sec A = csc(90  A)
cot A = tan(90  A)
1.1-9
5.3-9
Example 2
WRITING FUNCTIONS IN TERMS OF
COFUNCTIONS
Write each function in terms of its cofunction.
(a) cos 52° = sin (90° – 52°) = sin 38°
(b) tan 71° = cot (90° – 71°) = cot 19°
(c) sec 24° = csc (90° – 24°) = csc 66°
1.1-10
5.3-10
30°- 60°- 90° Triangles
Bisect one angle of an equilateral
to create two 30°-60°-90° triangles.
5.3-11
30°- 60°- 90° Triangles
Use the Pythagorean theorem to solve for x.
What would x have been if the shorter side had been 5?
5.3-12
Example
FINDING TRIGONOMETRIC FUNCTION
VALUES FOR 60°
Find the six trigonometric function values for a 60°
angle.
1.1-13
5.3-13
Example
FINDING TRIGONOMETRIC FUNCTION
VALUES FOR 60° (continued)
Find the six trigonometric function values for a 60°
angle.
1.1-14
5.3-14
45°- 45° Right Triangles
Use the Pythagorean theorem to
solve for r.
5.3-15
45°- 45° Right Triangles
Do the trig functions depend on the length of the side?
5.3-16
45°- 45° Right Triangles
5.3-17
Function Values of Special
Angles

sin  cos  tan  cot  sec  csc 
30
45
60
5.3-18
Reference Angles
A reference angle for an angle θ is the positive
acute angle made by the terminal side of angle θ
and the x-axis.
5.3-19
Caution
A common error is to find the
reference angle by using the terminal
side of θ and the y-axis.
The reference angle is always found
with reference to the x-axis.
1.1-20
5.3-20
Example 3(a) FINDING REFERENCE ANGLES
Find the reference angle for an angle of 218°.
by the terminal side of the
angle and the x-axis is
218° – 180° = 38°.
For θ = 218°, the reference angle θ′ = 38°.
1.1-21
5.3-21
Example 3(b) FINDING REFERENCE ANGLES
Find the reference angle for an angle of 1387°.
First find a coterminal angle
between 0° and 360°.
Divide 1387 by 360 to get a
quotient of about 3.9. Begin by
subtracting 360° three times.
1387° – 3(360°) = 307°.
The reference angle for 307° (and thus for 1387°) is
360° – 307° = 53°.
1.1-22
5.3-22
1.1-23
5.3-23
Example 4
FINDING TRIGONOMETRIC FUNCTION
VALUES OF A QUADRANT III ANGLE
Find the values of the six trigonometric functions
for 210°.
The reference angle for a
210° angle is
210° – 180° = 30°.
Choose point P on the
terminal side of the angle so
the distance from the origin
to P is 2.
1.1-24
5.3-24
Example 4
FINDING TRIGONOMETRIC FUNCTION
VALUES OF A QUADRANT III ANGLE
(continued)
1.1-25
5.3-25
Example 5(a) FINDING TRIGONOMETRIC FUNCTION
VALUES USING REFERENCE ANGLES
Find the exact value of cos (–240°).
Since an angle of –240° is coterminal with an
angle of –240° + 360° = 120°, the reference
angle is 180° – 120° = 60°.
1.1-26
5.3-26
Example 5(b) FINDING TRIGONOMETRIC FUNCTION
VALUES USING REFERENCE ANGLES
Find the exact value of tan 675°.
Subtract 360° to find a coterminal angle
between 0° and 360°: 675° – 360° = 315°.
1.1-27
5.3-27
Caution
When evaluating trigonometric
functions of angles given in degrees,
remember that the calculator must
be set in degree mode.
1.1-28
5.3-28
Example 6
FINDING FUNCTION VALUES WITH A
CALCULATOR
Approximate the value of each expression.
(a) sin 53°
(b) sec 97.977°
Calculators do not have a secant
key, so first find cos 97.977° and
then take the reciprocal.
sec 97.977° ≈ –7.20587921
1.1-29
5.3-29
Example 6
FINDING FUNCTION VALUES WITH A
CALCULATOR (continued)
Approximate the value of each expression.
(c)
Use the reciprocal identity
1.1-30
5.3-30
Example 7
USING INVERSE TRIGONOMETRIC
FUNCTIONS TO FIND ANGLES
Use a calculator to find an angle θ in the interval
[0°, 90°] that satisfies each condition.
a) cos θ = .5
We know that cos ____ = .5
To find this on the calculator we use θ =
cos1 0.5  ____
b) Find θ if cos θ = .9211854056
c) Find θ if sec θ = 1.2228
1.1-31
5.3-31
Example 8
FINDING ANGLE MEASURES GIVEN AN
INTERVAL AND A FUNCTION VALUE
Find all values of θ, if θ is in the interval [0°, 360°)
and
Since cos θ is negative, θ must lie in quadrant II or III.
The absolute value of cos θ is
angle is 45°.
so the reference
The angle in quadrant II is 180° – 45° = 135°.
The angle in quadrant III is 180° + 45° = 225°.
1.1-32
5.3-32
Example 9
Problem 141, page 529
When highway curves are designed, the outside of the
curve is often slightly elevated or inclined above the
inside of the curve. This inclination is called
superelevation . For safety reasons, it is important
that both the curve’s radius and superelevation are
correct for a given speed limit. If an automobile is
traveling at velocity V (in feet per second), the safe
radius R for a curve with superelevation θ is modeled
2
by the formula
where
f
and
g
are
V
constants
R
g ( f  tan )
Example 9
2
V
R
g ( f  tan )
a) A roadway is being designed for automobiles traveling at 45mph.
If θ = 3°, g = 32.2, and f = 0.14, calculate R to the nearest foot.
(Hint: You will need to know that 1 mile = 5280 feet)
b) A highway curve has radius R = 1150 ft and a superelevation of
θ = 2.1°. What should the speed limit (in miles per hour) be for
this curve? Use the same values for f and g from part (a).
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