Math 12 - 4-Trigonometry Math 12 - 4.1-Angle Measurement in Radians, Degrees and Turns 1 full rotation = 360° = 2 π Radians 1 2 rotation = 180° = π Radians 1 4 rotation = 90° = π 2 Radians 1 6 rotation = 60° = π 3 Radians 1 8 rotation = 45° = π 4 Radians 1 12 rotation = 30° = π 6 Radians r r A B a +θ -θ Arc length-a travels along the circumference of a circle of radius -r for a central angle θ in radian a=θ r A co-terminal angle in degree --> θ±n360 ° , n ∈ N A co-terminal angle in radians --> θ±n2 π , n ∈ N
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Math 12 - 4-Trigonometry€¦ · A co-terminal angle in degree --> ... Exp. 4.1.2 Giventhedomainof−1080°≤θ≤1080°, find the co-terminal angles of 160° and express them in
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Math 12 - 4-Trigonometry
Math 12 - 4.1-Angle Measurement in Radians, Degrees and Turns
1 full rotation = 360° = 2π Radians 12
rotation = 180° = π Radians
14
rotation = 90° = π2
Radians 16
rotation = 60° = π3
Radians
18
rotation = 45° = π4
Radians 112
rotation = 30° = π6
Radians
r
r
AB
a
+θ-θ
Arc length-a travels along the circumference of a circle of radius -r for acentral angle θ in radian a=θr
A co-terminal angle in degree --> θ±n360 ° , n∈N
A co-terminal angle in radians --> θ±n2 π , n∈N
Exp. 4.1.1 Convert each angle to a number of rotations and then draw it in standard position of a circle.
(a) 2 π3
(b) 160° (c) 15turns (d) – 320° (e) −7 π
8 (f) −3
5turns
Solution:
(a) 2 π3=180 °
π x 2 π3=120 °=1 turn
2πx2 π
3=+1
3turn→
(b) 160 °= π180 °
x 160 °=8 π9=1 turn
360 °x160 °=+4
9turn→
(c) 15turn= 360 °
1 turnx 1
5turn=72°=( 2π
1 turn) x 1
5turn= 2π
5
(d) −320 °= π180°
x(−320 °)=(−16π)
9=1 turn
360 °x(−320 °)=−8
9turn→
r
A
2 π3
r
A
-320°
r
A
72°
r
A
160°
(e) (−7π)
8=180 °
π x(−7π)
8=−157.5°=1 turn
2πx(−7π)
8=−7
16turn→
(f) −35
turn= 360 °1 turn
x(−35
turn)=−216 °=( 2π1 turn
) x (−35
turn)=−6π5
r
A
−7 π8
r
A
−35
turns
Exp. 4.1.2 Given thedomainof −1080°≤θ≤1080° , find the co-terminal angles of 160° and express
them in general form ?
Solution: Given thedomainof −1080°≤θ≤1080° , the co-terminal angle of θ is θ±n360 °
θ ( ° ) n θ±n360 ° Is the angle in the domain
−1080°≤θ≤1080° ?
160 -4 -1280 No
160 -3 -920 Yes
160 -2 -560 Yes
160 -1 -200 Yes
160 0 160 Yes
160 1 520 Yes
160 2 880 Yes
160 3 1240 No
So the co-terminal angle of 160° is 160 °±n360° where n∈N ,−2≤n≤2
Exp. 4.1.3 Given the domainof −6π≤θ≤6π , find the co-terminal angles of 25π and express
them in general form ?
Solution: Given thedomainof −6π≤θ≤6π , the co-terminal angle of θ is θ±n2π
θ( Radian )
n θ±n2π Is the angle in the domain
−6π≤θ≤6π ?
25π
-4 −385
πNo
25π
-3 −285
πYes
25π
-2 −185
πYes
25π
-1 −85
πYes
25π
0 25π
Yes
25π
1 125
πYes
25π
2 225
πYes
25π
3 325
πNo
So the co-terminal angle of 25π is 2
5π±n2πwhere n∈N ,−4≤n≤3
Exp. 4.1.4 A big circle round about with a radius of 15 m., Jennifer walks along the edge from point A to B as shown in the figure. What is the distance AB ?
Solution: Arc length-a travels along the circumference of a circle
of radius -r for a central angle θ in radian a=θ r
θ in radian is (20 °+40 °) x π180 °
=π3
So
The distance AB=a=π3x15m=5πm=5x3.14168=15.7m
Exp. 4.1.5 John wants to find out the distance PQR along the circumference of a big circle with a radius of 30 m. He draws the diagram as shown in the figure. Can you help John
calculate the distance PQR?
Solution: Arc length-a travels along the circumference of a circle
of radius -r for a central angle θ in radian a=θ r
θ in radian is (360°−120 °) x π180 °
=4π3
So
The distance PQR=a=4π3
x 30m=40x3.14168=125.66 m
r = 15 m.
A
r = 15 m.
B
20°
40°
r = 30 m.
Pr = 30 m.
R
Q
120°
Math 12 - 4.2-The Unit Circle – Trigonometric Ratios
A unit circle is a circle that has the radius of 1, from the general circle equation x2+ y2=r2 , replace a
radius with 1, so x2+ y2=1
Trigonometric Ratios Definition Reciprocal
sine sin θ= yr
1cscθ
cosine cosθ= xr
1secθ
tangent tanθ= yx
1cotθ
cosecant cscθ= ry
1sin θ
secant sec θ= rx
1cosθ
cotangent cotθ= xy
1tanθ
R=1 unitr
S(x, y)
θ
sin, csc --> +
cos, sec --> +tan, cot --> +
ALL --> +
R=1 unit
θ= π6
S( , )12
√32
R=1 uni
t
S( , )1√2
1√2
θ= π4
R=1
uni
t
θ= π3
S( , )12
√32
Exp. 4.2.1 On a unit circle as shown in the figure, point S(0.91, 0.42) is on the circumference. Determine
sin θ ,cosθ , tanθ , cscθ , secθ ,cot θ ?
Solution:
Find the Trigonometric Ratios
sin θ= yr=0.42
1=0.42
cosθ= xr=0.91
1=0.91
tanθ= yx=0.42
0.91=0.4615
cscθ= ry= 1
0.42=2.3809
sec θ= rx= 1
0.91=1.0989
cotθ= xy=0.91
0.42=2.1666
R=1 unit
S(0.91, 0.42)
θ
Exp. 4.2.2 Find the exact value from the unit circle of
(a) csc ( 43
π)
(b) sec(−32
π)
(c) cot(−0.8turn)
(d) sin (109
π)
(e) cos(−23
π)
(f) tan (−2.25 turns )
Solution:
(a) csc( 43π) , 4
3π → 180°
π x 43
π=240 °
From the Unit Circle --> cscθ= ry= 1
sin θ --> csc( 4
3π)= 1
sin( 43π)
= 1
( yr)
Draw the angle in the Unit Circle and then find y.
csc( 43π)= 1
sin( 43π)
= 1
(−0.8661
)=1.1547
r=1
43π →240 °
(b) sec (−32
π) , −32
π→ 180 °π x(−3
2π)=−270 °
From the Unit Circle --> sec θ= rx= 1
cosθ --> sec (−3
2π)= 1
cos(−32
π)= 1
( xr)
Draw the angle in the Unit Circle and then find y.
sec (−32
π)= 1
( xr)= 1
(01)→ It can ' t be defined.
(c) cot(−0.8turns ) , (−0.8 turn)→ 360 °1 turn
x (−0.8 turn)=−288 °
From the Unit Circle --> cot θ= xy= 1
tanθ --> cot(−0.8turns )= 1
tan(−0.8turns )= 1
( yx)
Draw the angle in the Unit Circle and then find x, y.
cot(−0.8turns )= 1
(0.950.3
)=0.3154
r=1
−32
π→−270°
r=1
−0.8 turns→−288°
(d) sin (109
π) , 109
π=π+19
π→180°+(180 °π x
19
π)=200 °
From the Unit Circle --> sin θ= yr
Draw the angle in the Unit Circle and then find y.
sin (109
π)= yr=−0.3
1=−0.3
(e) cos(−23
π) , −23
π→ 180°π x (−2
3π)=−120 °
From the Unit Circle --> cosθ= xr
Draw the angle in the Unit Circle and then find y.
Exp. 4.6.2 An AIR Defence station aims the STA missile to attack the intruder DRONE as shown in the figure. The intruder DRONE flies at the constant attitude at 2,000 m. High. Describe the distance d in term of ɵ ?
Solution: As the DRONE approaches the station horrizontally, the relation between d and ɵ is described
tan θ= d2000
then sketch the graph as shown.
From the graph, the bigger the angle ɵ, the shorter the distance dWhen the angle ɵ approaches to 90°, the distance changes rapidly.
d
2000
0 m
ɵ
π2
−π2
π4−π
4
Math 12 - 4.7-Trigonometric Functions – Equations and Graphs
To find the the graph that fit characteristics of the specific phenomenon,the following sinusoidal equations are very useful.
y=a cosb(θ−c)+d
ory=asin b(θ−c)+d
The parameters a, b, c, d can be adjusted to fit the equations for each phenomenon such as- Wave shapes of ocean waves, sound waves, heart beats
-Circular motion of ferris wheel-Swing of pendulum-Pistons of machine
Exp. 4.7.1 Solve these equations over the intervals by graphing ?
(a) sin θ=√32
,−3π≤θ≤3π
(b) cos3θ= 12,−360 °≤θ≤360°
(c) 1=5sin [3(θ−1)]+3,−2π≤θ≤2π
Solution:
(a) sin θ=√32
,−3π≤θ≤3π --> sketch the graphs of y=sin θ and y=√32
From the graphs --> θ=−5π3
,− 4π3
, π3,2π3
,5π3
,8π3
,
y=sin θ
y=√32
(b) cos3θ= 12,−360 °≤θ≤360° --> sketch the graphs of y=cosβ ,β=3θ and y=1
2
From the graphs --> β=3θ=−300 °→θ=−100 °
β=3θ=−60° →θ=−20 °
β=3θ=60° →θ=20°
β=3θ=300 °→θ=100°
y=cosβ ,β=3θ
y=12
(c) 1=5sin [3(θ−1)]+3,−2π≤θ≤2π --> −25
=sin [3(θ−1)] ,−2π≤θ≤2π -->
sin β=−25
,β=3(θ−1) ,−2π≤θ≤2π
--> sketch the graphs of y=sinβ ,β=3(θ−1) and y=−25
From the graphs --> β=3(θ−1)=−78
π →θ= 1724
π
β=3(θ−1)=−18
π →θ= 2324
π
β=3(θ−1)=98
π→θ=3324
π
β=3(θ−1)= 158
π→θ= 3924
π
y=sin β ,β=3(θ−1)
y=−25
Exp. 4.7.2 Given the average temperatures of 12 months in Chamonix, France. Find the sinusoidal function equation of these temperatures ?
Month (°C) Month (°C) Month (°C)
January -20.5 May 5 September 6.6
February -15 June 9.7 October 1.9
March -9.5 July 12.4 November -11.5
April -2 August 10.6 December -14.9
Solution: Sketch the graph of Month(x axis) – Temperature(°C – y axis)
The graph looks like a cos function, assume y=a cosb( x−c)+d ,a<0
∣a∣=max y value−min y value2
=12.4 °C−(−20.5° C )
2=16.45° C≈ 33
2
The estimated Period: 12
3.1416π≈3.8π =
2π∣b∣
--> ∣b∣= 2π3.8π
≈ 12
The phase shift c , notice at the minimum Jan- -20.5C – it shifts to the right almost 13π≈1unit
c = 1, d=max yvalue+min y value2
=12.4° C+(−20.5° C )
2=−4° C
S0 y=−332
cos(12( x−1))−4 --> sketch this grpah and compare with the first graph which is from the
values from the table as hsown in the next figure.
y=−332
cos(12(x−1))−4
Math 12 - 4.8-Trigonometric Functions
Equations from Pythagorean Theorem
Reciprocal and Quotient Equations
Equations from Pythagorean Theorem
x2+ y2=r2 where x=r cosθ , y=r sinθ
Replace a radius with 1 in the unit circle. --> x2+ y2=1, x=cosθ , y=sinθ
(sin θ)2+(cosθ)2=1
(sinθ)2
(cosθ)2 +(cosθ)2
(cosθ)2 =1
(cosθ)2 →( tanθ)2+1=(secθ)2
(sinθ)2
(sinθ)2 +(cosθ)2
(sinθ)2 =1
(sin θ)2 →(cotθ)2+1=(csc θ)2
Reciprocal Equations
sin θ= 1cscθ
, cosθ= 1sec θ
, tan θ= 1cotθ
R=1 unit
S(x, y)
R
θ
Quotient Equations
tan θ= sin θcosθ
, cotθ= cosθsin θ
Sum Equationssin (θ+β)=sin θ cosβ+cosθsinβ , cos(θ+β)=cosθcosβ−sinθ sinβ ,