Download sin 30˚ (no calculator allowed)

Find the following:
1. sin 30˚ (no calculator allowed)
2. sin 45˚ (no calculator allowed)
3. sin 75˚ (use a calculator)
4. Is sin 30˚ + sin 45˚ = sin 75˚?
5. What two unit circle values combine to be 165˚?
6. What two unit circle values combine to be 105˚?
Find the following:
1. sin 30˚ (no calculator allowed)
=½
2. sin 45˚ (no calculator allowed) =
2
2
3. sin 75˚ (use a calculator) = 0.9659
4. Is sin 30˚ + sin 45˚ = sin 75˚? No, the two2are not
≠ 0.9659
equal ½ +
2
5. What two unit circle angles combine to be 165˚?
120 + 45
6. What two unit circle angles combine to be 105˚?
135 – 30 or 60 + 45
Chapter 5 Trigonometric Equations
5.4
MATHPOWERTM 12, WESTERN EDITION
5.4.1
Sum and Difference Identities
sin(A + B) = sin A cos B + cos A sin B
sin(A - B) = sin A cos B - cos A sin B
cos(A + B) = cos A cos B - sin A sin B
cos(A - B) = cos A cos B + sin A sin B
tan A  tan B
tan( A  B) 
1  tan Atan B
tan A  tan B
tan( A  B) 
1  tan Atan B
5.5.2
Finding Exact Values
1. Find the exact value for sin 750.
Think of the angle measures that produce exact values:
300, 450, and 600.
Use the sum and difference identities.
Which angles, used in combination of addition
or subtraction, would give a result of 750?
sin 750 = sin(300 + 450)
= sin 300 cos 450 + cos 300 sin 450
1
2   3
2 
  
  

2 2   2
2 

2 6
4
5.5.4
Finding Exact Values
2. Find the exact value for cos 150.
cos 150 = cos(450 - 300)
= cos 450 cos 300 + sin 450 sin300
 2
3   2 1 
  
   
 2
2   2
2 
6 2
4

5
3. Find the exact value for sin .
12
5
 
sin
 sin(  )
12
4 6




 sin cos  cos sin
4
6
4
6
 2
3   2 1 
  
   
 2
2   2
2 
6 2

4

3


4
12
3
4 12
 2

6 12

5.5.5
Finding Exact Values
4. Find the exact value for tan 1650.
tan 1650 = tan(1350 + 300)
tan 135  tan 30

1  tan 135 tan 30
3
1
3


3
1   1
 3





3
1
3
3
3



3 
3
3

1 
 3 


Rationalize the denominator
• Multiply top and bottom by the conjugate of
the bottom
3
3
3
3
Finding Exact Values
4. Find the exact value for tan 1650 (continued)
tan 1650
Rationalize the
denominator
3
3

3
3
3

3
96 3 3

93
 12  6

6
 2 
3
3
3
3
Assignment
Simplifying Trigonometric Expressions
1.
Express cos 1000 cos 800 + sin 800 sin 1000 as a
trig function of a single angle.
This function has the same pattern as cos (A - B),
with A = 1000 and B = 800.
cos 100 cos 80 + sin 80 sin 100 = cos(1000 - 800)
= cos 200
2.




Express sin 3 cos 6  cos 3 sin 6 as a single trig function.
This function has the same pattern as sin(A - B), with


A
3
and B 
6
.
  

sin cos  cos sin  sin 
3 6 
3
6
3
6

 sin
6




5.5.3
Using the Sum and Difference Identities



Prove cos    sin .
2


cos
 

2



 cos cos   sin sin
2
2
 (0)(cos  )  (1)(sin )
 sin .
sin 
 sin 
L.S. = R.S.
5.5.6
Using the Sum and Difference Identities
3

Given cos  , where 0    ,
5
2

find the exact value of cos(  ).
x
6
cos 



r
cos(  )  cos cos  sin sin
x=3
6
6
6
r=5
3 3
4 1
 ( )( )  ( )( )
5 2
5 2
5
3 34

10
3
θ
y=±4

3 34
Therefore, cos(  ) 
.
6
10
sinθ = 4/5
5.5.7
Using the Sum and Difference Identities
2
4
and cos B  ,
3
5
where A and B are acute angles,
Given sin A 
find the exact value of sin( A  B ).
sin( A  B )  sin A cos B  cos A sin B
2 4
5 3
 ( )( )  ( )( )
3 5
3 5
x
y
r
A
5
2
3
B
4
3
5
8 3 5

15
Therefore, sin( A  B ) 
8 3 5
.
15
5.5.8
Assignment
•
Complete #’s 3-12 on worksheet
•
Answers:
9. – sin x
10. cos x
11. sin x
12. – tan x + 2 sin x
Hints:
3. 345 = 300 + 45
4. 195 = 240 – 45
Identities
Prove
2tan x
2
1  tan x

2sin
sin2x.
xcos x
 2sin xcos x
2sin x
 cos2x
sec x
2sin x
x
 cos
1
cos 2 x
2sin x cos 2 x


cos x
1
 2sin xcos x
L.S. = R.S.
5.5.16
Double-Angle Identities
The identities for the sine and cosine of the sum of two
numbers can be used, when the two numbers A and B
are equal, to develop the identities for sin 2A and cos 2A.
sin 2A = sin (A + A)
cos 2A = cos (A + A)
= sin A cos A + cos A sin A
= cos A cos A - sin A sin A
= 2 sin A cos A
= cos2 A - sin2A
Identities for sin 2x and cos 2x:
sin 2x = 2sin x cos x
cos 2x = cos2x - sin2x
cos 2x = 2cos2x - 1
cos 2x = 1 - 2sin2x
5.5.9
Double-Angle Identities
Express each in terms of a single trig function.
a) 2 sin 0.45 cos 0.45
sin 2x = 2sin x cos x
sin 2(0.45) = 2sin 0.45 cos 0.45
sin 0.9 = 2sin 0.45 cos 0.45
b) cos2 5 - sin2 5
cos 2x = cos2 x - sin2 x
cos 2(5) = cos2 5 - sin2 5
cos 10 = cos2 5 - sin2 5
Find the value of cos 2x for x = 0.69.
cos 2x = cos2 x - sin2 x
cos 2(0.69) = cos2 0.69 - sin2 0.69
cos 2x = 0.1896
5.5.10
Double-Angle Identities
Verify the identity tan A 
tan A
1  cos 2 A
.
sin 2A
1  (cos2 A  sin2 A)

2sin Acos A
1  cos2 A  sin2 A)

2 sin Acos A
sin2 A  sin2 A

2sin Acos A
2sin 2 A

2sin Acos A
sin A

cos A
 tan A
L.S = R.S.
5.5.11
Double-Angle Identities
sin 2x
.
Verify the identity tan x 
1  cos 2x
tan x
2sin x cos x

2
1  2 cos x  1
2sin x cos x

2
2cos x
sin x

cos x
 tan x
L.S = R.S.
5.5.12
Double-Angle Equations
2
where 0 A  2.
2
2
For cosA =
where 0 A  2, A   and 7 .
2
4
4

7
 7  9  15
2 A   2 n,
 2  n.
,
,
2 A ,
4
4
4 4
4
4
Find A given cosA2 =
A

8
7
8
9
8

 7 9  15
,
8 8
15
8
,
8
,
8
y
2
2
2
y = cos 2A
5.5.13
Double-Angle Equations
1
Find A given sin A
2 = - where 0 A  2.
2
1
7
11
For sinA = - where 0 A  2, A =
and
.
2
6
6
7  11 19 23 
,
,
2 A
,
6
6
6
6
7
11 
2A
 2 n,
 2 n.
6
6
A
7  11 19 23 
,
,
,
12 12 12 12
y = sin 2A
7
12
11
12
19
12
23 
12
1
y 
2
5.5.14
Using Technology
2 tan x
and predict the period.
Graph the function f (x) 
2
1  tan x
The period is .
Rewrite f(x) as a single trig function: f(x) = sin 2x
5.5.15
Applying Skills to Solve a Problem
The horizontal distance that a soccer ball will travel, when
2
2v
kicked at an angle , is given by d  0 sin  cos  ,
g
where d is the horizontal distance in metres, v0 is the initial
velocity in metres per second, and g is the acceleration due
to gravity, which is 9.81 m/s2.
a) Rewrite the expression as a sine function.
Use the identity sin 2A = 2sin A cos A:
2v02
d
sin  cos 
g
v02
d  sin2
g
5.5.17
Applying Skills to Solve a Problem [cont’d]
b) Find the distance when the initial velocity is 20 m/s.
From the graph, the maximum
2
v0
distance occurs when  = 0.785398.
d  sin2
g
The maximum distance is 40.7747 m.
2
(20)
The graph of sin  reaches its
d
sin2 
9.81
maximum when    .
2
Distance
Sin 2 will reach a maximum when


2  or   .
2
4
Angle 
5.5.18
Suggested Questions:
Pages 272-274
A 1-16, 25-35 odd
B 17-24, 37-40, 43,
47, 52
5.5.19