Transcript More Circular Trigonometry

*Special Angles
30°, 45°, and 60° → common reference angles
Memorize their trigonometric functions.
Use the Pythagorean Theorem; triangles below.
60°
45°
2
1
1
2
45°
1
30°
3
*Special Angles
θ
sin θ
cos θ
tan θ
30º
1
3
2
2
1
3

3 3
45º
1
2

2 2
1
2

2 2
1.00
60º
3
1
2
3
2
*Special Angles
θ
sin θ
30º
45º
0.5000
0.7071
0.8660
cos θ
0.8660
0.7071
0.5000
tan θ
0.5774
1.0000
1.7320
60º
Find trig functions of 300° without calculator.
Reference angle is 60°[360° - 300°]; IV quadrant
60°
300°
sin 300° = - sin 60°
cos 300° = cos 60°
tan 300° = - tan 60°
csc 300° = - csc 60°
sec 300° = sec 60°
cot 300° = - cot 60°
Use special angle chart.
sin 300° = - sin 60° = - 0.8660 =
3
2
cos 300° = cos 60° = 0.5000 = 1/ 2
-
3
tan 300° = - tan 60° = -1.732 =
csc 300° = - csc 60° = -1.155 =
2
3
-
3
sec 300° = sec 60° = 2.000 = 2/1
cot 300° = - cot 60° = - 0.5774 =
3
-
3
*Quadrant Angles
•Reference angles cannot be drawn for
quadrant angles 0°, 90°, 180°, and 270°
•Determined from the unit circle; r = 1
•Coordinates of points (x, y)
correspond to (cos θ, sin θ)
*Quadrant Angles
90° (0,1) → (cos θ, sin θ)
r=1
180° (-1,0)
0° (1,0)
270° (0,-1)
*Quadrant Angles
θ
0º
90º
180º
270º
sin θ
0
1
0
-1
cos θ
1
0
-1
0
tan θ
0

0
 sinθ 
 cosθ 






 
1
0
  


-



 -1 
 0 
 
Find trig functions for - 90°.
Reference angle is (360° - 90°) → 270°
sin 270° = -1
cos 270° = 0
tan 270° undefined
csc 270° = -1
sec 270° undefined
cot 270° = 0
Use quadrant angle chart.
-90°
270°
*Coterminal Angle
θ1 = 405º
θ2 = 45º
The angle
between 0º and
360º having the
same terminal
side as a given
angle.
Ex. 405º - 360º =
coterminal angle
45º
*Coterminal Angles
Used with angles greater than 360°,
or angles less than 0°.
•Example
cos 900° =
cos (900° - 720°) =
cos 180° = -1
(See quadrant angles chart)
•Example
tan (-135° ) =
tan (360° -135°) =
tan 225° =
LOOK→ tan (225° - 180°)
tan 45° = 1
(See special angles chart)
Find the value of sec 7π / 4
•Convert from radian to degrees:
sec [(7π/ 4)(180/ π)] = sec 315°
SOLUTION
•Angle in IV quadrant: sec →positive
•sec (360° - 315°) = sec 45°
= 1 /(cos 45°) = √2 = 1.414
•Look how this problem was worked
in previous lesson.
SOLUTION
Express as a function of the reference
angle and find the value.
tan 210°
sec 120 °
SOLUTION
Express as a function of the reference
angle and find the value.
sin (- 330°)
csc 225°
SOLUTION
Express as a function of the reference
angle and find the value.
cot (9π/2)
cos (-5π)
SOLUTION
Inverse Trig Functions
Used to find the angle
•when two sides of right triangle are known...
•or if its trigonometric functions are known
Notation used:
sin-1x or arcsin x
Read: “angle whose sine is …”
Also, cos-1x or arccos x
tan-1x or arctan x
Inverse trig functions have
only one principal value of the angle
defined for each value of x:
-90° < arcsin < 90°
0° < arccos < 180°
-90° < arctan < 90°
Example:
Given tan θ = -1.600, find θ
to the nearest 0.1° for 0° < θ < 360°
•Tan is negative in II & IV quadrants
-1
-1
tan tan  = tan 1.600
-1
 tan 1.600
  
 

reference angle
θ = 180° - 58.0° = 122° II
θ = 360° - 58.0° = 302° IV
Note: On the calculator entering
 tan-1 - 1.600
results in -58.0°
Given sin θ = 33, find θ
to the nearest 0.1° for 0° < θ < 360°
SOLUTION
Given cos θ = - 0.0157, find θ
to the nearest 0.1° for 0° < θ < 360°
SOLUTION
Given sec θ = 1.553 where sin θ < 0,
find θ to the nearest 0.1° for 0° < θ < 360°
SOLUTION
Given the terminal side of θ passes
through point (2, -1), find θ the
nearest 0.1° for 0° < θ < 360°
SOLUTION
The voltage of ordinary house current is
expressed as V = 170 sin 2πft , where
f = frequency = 60 Hz and t = time in seconds.
• Find the angle 2πft in radians when
V = 120 volts and 0 < 2πft < 2π
SOLUTION
• Find t when V = 120 volts
SOLUTION
The angle β of a laser beam is expressed as:
w
β = 2 tan
2d
-1
where w = width of the beam (the diameter)
and d = distance from the source.
Find β if w = 1.00m and d = 1000m.
SOLUTION