9.3 Evaluate Trigonometric Functions of Any Angle

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9.3 Evaluate Trigonometric
Functions of Any Angle
How can you evaluate trigonometric
functions of any angle?
What must always be true about the
value of r?
Can a reference angle ever have a
negative measure?
General Definitions of Trigonometric Functions
y
x
Sometimes called circular functions
Let (–4, 3) be a point on the
terminal side of an angle θ in
standard position. Evaluate the
six trigonometric functions of θ.
SOLUTION
Use the Pythagorean theorem to find the value of r.
r = √ x2 + y2 = √ (–4)2 + 32 = √ 25 = 5
Using x = –4, y = 3, and r = 5, you can write the following:
3
4
y
x
sin θ =
cos θ =
= 5
=– 5
r
r
3
5
y
r
tan θ =
csc θ =
=– 4
= 3
x
y
5
r
x
4
–
sec θ =
cot θ =
=
=–
4
x
y
3
The Unit Circle
y
r=1
x
Quadrantal Angle
Use the unit circle to evaluate the six
trigonometric functions of θ = 270°.
SOLUTION
Draw the unit circle, then draw the angle θ =
270° in standard position. The terminal side of θ
intersects the unit circle at (0, –1), so use x = 0
and y = –1 to evaluate the trigonometric
functions.
sin θ =
cos θ =
tan θ =
1
–1
y
r
csc θ =
= 1 = –1
= – 1 = –1
r
y
1
0
r
x
sec θ =
= 0 undefined
= 1 =0
x
r
–1
0
y
x
= 0 undefined cot θ =
= –1 = 0
x
y
Evaluate the six trigonometric functions of θ.
1.
SOLUTION
Use the Pythagorean Theorem
to find the value of r.
r = √ x2 + y2 = √ 32 + (–3)2 = √ 18 = 3√ 2
Using x = 3, y = –3 , and r = 3√ 2, you can write the
following:
3
y
x
√2
–
–
– 3 = √2
=
sin θ =
cos θ =
=
=
2
r
3√ 2
r
3√ 2
2
3
y
r
–
– 3√ 2 = –√ 2
tan θ =
csc θ =
=
=
=
–1
3
3
x
y
r
x
3
3√
2
sec θ =
cot θ =
=–
=
= –1
=
√
2
x
3
y
3
Evaluate the six trigonometric functions of θ.
SOLUTION
Use the Pythagorean theorem to find
the value of r.
r = √ (–8)2 + (15)2 = √ 64 + 225 = √ 289 = 17
Using x = –8, y = 15, and r = 17, you can write the following:
y
x
15
8
sin θ =
cos θ =
= 17
= – 17
r
r
y
r
15
17
–
tan θ =
csc θ =
=
= 15
8
x
y
r
x
17
8
–
sec θ =
–
cot θ =
=
=
8
x
y
15
Evaluate the six trigonometric functions of θ.
SOLUTION
Use the Pythagorean theorem to
find the value of r.
r = √ x2 + y2 = √ (–5)2 + (–12)2 = √ 25 + 144 = 13
Using x = –5, y = –12, and r = 13, you can write the following:
y
12
sin θ =
= – 13
r
y
tan θ =
= 12
x
5
r
13
–
sec θ =
=
5
x
x
5
–
cos θ =
=
13
r
r
13
–
csc θ =
=
12
y
x
5
cot θ =
=
y
12
4. Use the unit circle to evaluate the six trigonometric
functions of θ = 180°.
SOLUTION
Draw the unit circle, then draw the angle θ = 180° in
standard position. The terminal side of θ intersects the
unit circle at (–1, 0), so use x = –1 and y = 0 to evaluate the
trigonometric functions.
y
r
y
x
sin θ =
tan θ =
sec θ =
r
x
0
= 1 =0
0
= –1
–1
= 1 = –1
cos θ =
x
–1
=
r
1
csc θ =
r
y
–1
= 0
undefined
cot θ =
x
y
–1
= 0
undefined
= –1
Reference Angle Relationships
5π
Find the reference angle θ' for (a) θ =
3
and (b) θ = – 130°.
SOLUTION
a. The terminal side of θ lies in Quadrant IV.
π
5π
=
So, θ' = 2π –
.
3
3
b. Note that θ is coterminal with 230°, whose terminal
side lies in Quadrant III. So, θ' = 230° – 180° + 50°.
9.3 Assignment
Page 574, 4-15 all
9.3 Evaluate Trigonometric
Functions of Any Angle
• How can you evaluate trigonometric
functions of any angle?
• What must always be true about the
value of r?
• Can a reference angle ever have a
negative measure?
Evaluating Trigonometric Functions
Reference Angle Relationships
Evaluate (a) tan ( – 240°).
SOLUTION
The angle – 240° is coterminal
a. with 120°. The reference angle is
θ' = 180° – 120° = 60°. The tangent
function is negative in Quadrant
II, so you can write:
tan (–240°) = – tan 60° = – √ 3
30º
2l
l 3
60º
l
17π
Evaluate (b) csc
.
6
SOLUTION
b. The angle 17π is coterminal
6
5π
with
. The reference
6
5π
π
angle is θ' = π –
=
.
6
6
The cosecant function is
positive in Quadrant II, so you
can write:
30º
π
=2
6
csc 17π = csc
6

6
2l
l 3
 30 
60º
l
Sketch the angle. Then find its reference angle.
5. 210°
The terminal side of θ lies in Quadrant III,
so θ' = 210° – 180° = 30°
Sketch the angle. Then find its reference angle.
6. – 260°
– 260° is coterminal with 100°, whose terminal side
of θ lies in Quadrant II, so θ' = 180° – 100° = 80°
Sketch the angle. Then find its reference angle.
7π
–
7.
9
11π
7π
The angle – 9 is coterminal with 9 . The
terminal side lies in Quadrant III,
2π
11π
so θ' =
–π=
9
9
Sketch the angle. Then find its reference angle.
8.
15π
4
The terminal side lies in Quadrant IV,
π
15π
so θ' = 2π –
= 4
4
9. Evaluate cos ( – 210°) without using a calculator.
– 210° is coterminal with 150°. The terminal side lies in
Quadrant II, which means it will have a negative value.
So, cos (– 210°) = – √ 3
2
30º
150º
30º
2l
l 3
60º
l
Robotics
The “frogbot” is a robot designed for exploring rough
terrain on other planets. It can jump at a 45° angle and
with an initial speed of 16 feet per second. On Earth,
the horizontal distance d (in feet) traveled by a
projectile launched at an angle θ and with an initial
speed v (in feet per second) is given by:
2
v
d = 32 sin 2θ
How far can the frogbot jump on Earth?
SOLUTION
2
v
sin 2θ
d=
Write model for horizontal distance
32
2
16
d = 32 sin (2 45°) Substitute 16 for v and 45° for θ.
= 8
Simplify.
The frogbot can jump a horizontal distance of 8 feet
on Earth.
Rock climbing
A rock climber is using a rock
climbing treadmill that is 10.5 feet
long. The climber begins by lying
horizontally on the treadmill,
which is then rotated about its
midpoint by 110° so that the rock
climber is climbing towards the
top. If the midpoint of the
treadmill is 6 feet above the
ground, how high above the
ground is the top of the
treadmill?
y
sin θ =
SOLUTION
Use definition of sine.
r
y
10.5
sin 110° =
Substitute 110° for θ and
= 5.25 for r.
5.25
2
4.9 y
Solve for y.
The top of the treadmill is about 6 + 4.9 = 10.9 feet above the ground.
9.3 Assignment day 2
P. 574, 16-30 all

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