Transcript 9.3 Evaluate Trigonometric Functions of Any Angle

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