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.
Post on 18-Dec-2015
222 Views
Preview:
Transcript
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?
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.
x2 + y2√r = (–4)2 + 32√= = 25√ = 5Using x = –4, y = 3, and r = 5, you can write the following:
sin θ =yr =
35 cos θ =
xr =
45
–
tan θ =yx =
34
– csc θ =ry =
53
sec θ =rx =
54
– cot θ =xy =
43
–
Use the unit circle to evaluate the six trigonometric functions of = 270°.θ
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.
cos θ =xr =
01 = 0 undefined
undefined cot θ =xy
=0
–1tan θ =yx =
–10
sec θ =rx =
10
sin θ =yr
1= 1
–= –1 csc θ =
ry =
11– = –1
= 0
Evaluate the six trigonometric functions of .θ1.
SOLUTION
Use the Pythagorean Theorem to find the value of r.
x2 + y2√r = 32 + (–3)2√= = 18√ = 3√ 2
sin θ =yr cos θ =
xr
tan θ =yx =
33
– csc θ =ry
sec θ =rx cot θ =
xy =
33
–
Using x = 3, y = –3 , and r = 3√ 2, you can write the following:
=3
–3√ 2 = –
3√ 23
= –1 = 3√ 23
– = –√ 2
3√ 23= = √ 2 = –1
= –2
√ 2= 2
√ 2
SOLUTION
Use the Pythagorean theorem to find the value of r.
(–8)2 + (15)2√r = 64 + 225√= = 289√ = 17
Evaluate the six trigonometric functions of .θ
sin θ =yr = 15
17 cos θ =xr = 8
17–
tan θ =yx =
158– csc θ =
ry =
1715
sec θ =rx =
178
– cot θ =xy = 8
15–
Using x = –8, y = 15, and r = 17, you can write the following:
Evaluate the six trigonometric functions of .θ
SOLUTION
Use the Pythagorean theorem to find the value of r.
x2 + y2√r = (–5)2 + (–12)2√= = 25 + 144√ = 13Using x = –5, y = –12, and r = 13, you can write the following:
sin θ =yr cos θ =
xr =
513
–
tan θ =yx = 12
5csc θ =
ry
sec θ =rx cot θ =
xy = 5
12
= 1213
–
=1312
–
=135
–
4. Use the unit circle to evaluate the six trigonometric functions of θ = 180°.
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.
SOLUTION
cos θ =xr = –1
cot θ =xy
tan θ =yx = 0
–1
sec θ =rx
= –11
= –1= –11 = –1
0 undefined
sin θ =yr
0= 1 = 0
csc θ =ry = –1
0 undefined
Find the reference angle θ' for (a) θ = 5π3
and (b) θ = – 130°.
SOLUTION
a. The terminal side of θ lies in Quadrant IV.
So, θ' = 2π – . 5π3
π3
=
b. Note that θ is coterminal with 230°, whose terminal side lies in Quadrant III. So, θ' = 230° – 180° + 50°.
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?
Evaluate (a) tan ( – 240°).
SOLUTION
tan (–240°) = – tan 60° = – √ 3
a.
The angle – 240° is coterminal with 120°. The reference angle is θ' = 180° – 120° = 60°. The tangent function is negative in Quadrant II, so you can write:
30º
60º
3ll2
l
Evaluate (b) csc .17π
6SOLUTION
b. The angle is coterminal
with . The reference
angle is θ' = π – = .
The cosecant function is positive in Quadrant II, so you can write:
17π65π
6 5π6
π6
csc = csc = 217π6
π6
30º
60º
3ll2
l30
6
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.
7π9The angle – is coterminal with . The
terminal side lies in Quadrant III, so θ' = – π =
11π9
11π9
2π9
7.7π9
–
Sketch the angle. Then find its reference angle.
15π4
8.
The terminal side lies in Quadrant IV, so θ' = 2π – = 15π
4π4
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°) = – 2√ 3
30º
150º 30º
60º
3ll2
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:
d = v2
32 sin 2θ
How far can the frogbot jump on Earth?
SOLUTION
d = v2
32 sin 2θ
d = 162
32 sin (2 45°)
= 8
Write model for horizontal distance.
Substitute 16 for v and 45° for θ.
Simplify.
The frogbot can jump a horizontal distance of 8 feet on Earth.
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?
Rock climbing
SOLUTION sin θ =yr
sin 110° =y
5.254.9 y
Use definition of sine.
Solve for y.
The top of the treadmill is about 6 + 4.9 = 10.9 feet above the ground.
Substitute 110° for θ and = 5.25 for r.2
10.5
top related