Chapter 9
Analytical Trigonometry
1
Overview
• Addition Formulas
• Double angle formulas
• Product to sum and sum to product formulas
• Trig equations
• Inverse trig functions
2
Chapter 9.1
The Addition Formulas
3
Addition
• For numbers, r(s+t) = rs + rt
• However, for functions f(s+t)  f(s) + f(t)
For example cos (/6 + /3) = cos (/2) = 0
But: cos (/6) + cos(/3) = 3/2 + ½  0
• But, we do have addition formulas for sine and cosine
4
For Cosine
Consider two points A and B on the unit circle corresponding to
angles  and . You need not duplicate this proof, just know the
identity
We can write these points as (cos , sin  ) and (cos , sin ). The distance between the
points is sqrt [(cos  – cos  ) 2 + (sin  – sin  ) 2 ] =
sqrt[ cos2  + cos2  - 2 cos  cos  + sin2  + sin2  - 2 sin  sin  ] =
sqrt[ 2 - 2 cos  cos  - 2 sin  sin ] (1)
Now, rotate the point so that A is at (1,0) and B at (cos (-), sin (-)) . The distance
between the two points is: sqrt{ [cos (-) – 1]2 + [sin (-) – 0]2 } = sqrt{ cos2(-) +
sin2(-) – 2 cos(-) + 1} = sqrt {2 - 2 cos(-)} (2)
But the distances are the same, so we have (1) = (2)
2 - 2 cos  cos  - 2 sin  sin  = 2 - 2 cos(-), or
cos(-) = cos  cos  + sin  sin 
5
Sum and Difference Identities for Cosine
cos ( ± ) = cos  cos  + sin  sin 
sin ( ± ) = sin  cos   cos  sin 
Remember: cos (a + b) is not cos a + cos b
6
Example
• cos (2a) cos a + sin (2a) sin a =
cos (2a – a) = cos a
7
Example
Show that cos (/2 – t) = sin t
cos ( - ) = cos  cos  + sin  sin 
cos (/2 – t) = cos (/2) cos t + sin(/2) sin t
= 0 + sin t
8
Example
• Find cos (x - )
9
Solution
• Find cos (x - )
= cos x cos  + sin x sin  = - cos x
10
Example
Find cos 15
= cos (45 - 30) = cos 15
11
Find cos 15
= cos (45 - 30) = cos 45 cos 30 + sin 45 sin 30
= 2/2 (1/2 + 3/2 ) = 2/4 +
6/4
12
Example
• If sin a = 3/5 in Q1 and sin b = - 3 /4 in Q 2, find sin (a-b)
13
Solution
• If sin a = 3/5 in Q1 and sin b = - 3 /4 in Q 2, find sin (a-b)
• Sin (a-b) = sin a cos b – cos b sin a
• Cos a = 4/5, cos b = - 13 /4
• So sin (a-b) = -(3/5) 13 /4 - (4/5) 3 /4
= -[3 13 + 4 3 ] / 20
14
Sum and Difference for Sin and Tan
tan ( ± ) = [tan  ± tan  ] / [1 ∓ tan  tan  ]
15
Verify:
tan ( t - /4) = [tan t – 1] / [tan t + 1]
16
Solution
Show tan ( t - /4) = [tan t – 1] / [tan t + 1]
• Identity is tan ( ± ) = [tan  ± tan  ] / [1 ∓ tan  tan  ]
 = t,  = /4; tan /4 = 1
Tan (t - /4) = [tan t – tan /4 ] / [1 + tan t tan /4 ]
= (tan t – 1)/ (tan t + 1)
17
Verify
sin (a + b) sin (a – b) = sin2a - sin2 b
18
Solution
• sin (a + b) sin (a – b) = sin2a - sin2 b
• Sin(a+b) = sin a cos b + sin b cos a
• Sin (a-b) = sin a cos b – sin b cos a
• Sin (a+b)sin(a-b) = sin2a cos 2b – sin a sin b cos a cos b +
sin a sin b cos a cos b – sin2bcos2a =
sin2a cos2b – sin2 b cos2 a =
sin2a (1-sin2b)- sin2 b (1-sin2a) =
sin2a - sin2a sin2b – sin2b + sin2b sin2a =
sin2 a – sin2 b
19
Example
Find sin (-75)
20
Solution
Sin (-75) = - sin(75) = - sin (45 + 30) =
-[sin 45 cos 30 + cos 45 sin 30] = -( 2/2)[ 3 -1]/2
= -[ 6 - 2] /4
21
Example
Find cos (19  / 12 )
22
Solution
Find cos (19  / 12 )
19  / 12 = 2  - 5  /12
5  /12 = 3  /12 + 2  /12 =  / 4 +  / 6
Cos(19 /2) = cos(/4 +/6) = cos(/4)cos(/6) – sin(/4) sin(/6)
= 6/4 - 2/4
23
Verify
Sin (t + /4) = 2/2 [ cos t + sin t ]
24
Solution
• Sin (t + /4) = 2/2 [ cos t + sin t ]
Sin(t + /4) = sin(t) cos(/4) + cos t sin (/4)
= 2/2 [sin t + cos t]
25
Verify
Cos (3t) = 4 cos3 t – 3 cos t
26
Solution
• Cos (3t) = 4 cos3 t – 3 cos t
cos(3t) = cos (t+2t) = cos t cos 2t – sin t sin 2t (i)
cos (2t) = cos t cos t – sin t sin t = cos2t – sin2t
sin (2t) = cos t sin t + sin t cos t = 2 sin t cos t
Substituting back into (i), we have
cos (3t) = cos t [cos2t – sin2t] - sin t [2 sin t cos t] =
cos3t - sin2t [2 cos t + cos t]; but sin2t = 1 - cos2t,
so we have
cos3t – [1 – cos2t] [ 3 cos t] = 4 cos3t – 3 cos t
27
Chapter 9.2
The Double Angle Formulas
28
Double Angles
If we can remember the sum and difference, these are simple
cos (2a) = cos (a+a) = cos2a – sin2 a
sin (2a) = sin a cos a + cos a sin a = 2 cos a sin a
tan (2a) = 2 tan a / [1 - tan2a]
29
Applications
Given sin a = 5/8 find cos (2a)
30
Solution
Given sin a = 5/8 find cos (2a)
cos 2 a = cos2a – sin2 a
cos2a = 1 - sin2 a
So cos 2 a = 1 - sin2 a - sin2 a = 1 – 2 sin2 a
= 1 – 2(5/8) 2 = 1-25/32 = 7/32
31
Example
Verify (sin x + cos x)2 = 1 + sin 2x
32
Solution
Verify (sin x + cos x)2 = 1 + sin 2x
(sin x + cos x)2 = sin x2 + cos x2 + 2 sin x cos x
= 1 + 2 sinx cos x
= 1 + sin 2 x
33
Example
Verify sin (3x) = 3 sin x – 4 sin3 x
34
Solution
Verify sin (3x) = 3 sin x – 4 sin3 x
sin 3x = sin (2x + x) = sin 2x cos x + cos 2x sin x
= 2 (sin x cos x) cos x + sin x ( cos2 x - sin2 x )
= sin x [ 2 cos2 x + cos2 - sin2 x]
= sin x [ 3 cos2 x - sin2 x]
but 3 cos2 x = 3 [1 - sin2 x]
So we have
sin x [ 3 – 4 sin2 x] = 3 sin x – 4 sin3 x
35
Half Angle Formulas
• sin (t/2) = ±
1 −cos 𝑡
2
• cos (t/2) = ±
1+cos 𝑡
2
• tan (t/2) =
sin 𝑡
1+cos 𝑡
36
What can be hard?
• Determining the right quadrant
• If angle A is in Q2, where is angle A/2?
90<A<180 so 45<A/2<90; must be in Q1
• If A is in Q3, where is A/2
180 < A < 270 so 90<A<135; must be in Q2
• If A is in Q4, where is A/2
270<A<360 so 135<A<180; must be in Q2
• Etc.
37
Power Reduction Identities
From cos 2 a = cos2 a – sin2 a
We have cos 2 a = (1 – sin2 a ) - sin2 a = 1 – 2 sin2 a
so sin2 a = (1 - cos 2a) / 2
We also have cos2a = cos2 a – (1 - cos2 a) = 2 cos2 a - 1
so cos2 a = (1 + cos 2a) /2
From their ratio, we have
tan2 a = (1 – cos2a) / (1 + cos2a)
38
Verify
tan (b/2) = sin b / (1 + cos b)
39
Solution
tan (b/2) = sin b / (1 + cos b)
tan2 (b/2) = (1-cosb)/(1+cosb)
=[(1+cosb)(1-cosb)]/[(1+cosb) 2
= (1 – cos2 b)/(1+cosb) = sin2 b / (1 + cosb) 2
40
Uses
Each of the half and double angle identities can be used to find
values of additional trig functions
If cos a = -7/25, find sin (a/2) and cos (a/2) if a is in Q3
41
Solution
• If cos a = -7/25, find sin (a/2) and cos (a/2)
• Cos a/2 = ±
• Sin a/2 = ±
1+cos 𝑎
2
1−cos 𝑎
2
=±
=±
25−7
50
25+7
50
=±
=±
9
25
16
25
• What are the correct signs? We know a is in Q3; where is
a/2? Q3 goes from 180 to 270. half of 180 is 90, half of 270 is
135, The angles has to be in Q2; sin >0, cos and tan <0
42
Examples
Find cos (t/2), sin (t/2), tan (t/2) if cos t = 15/17 in Q 1
43
Solutions
• t is in Q1 so t/2 must also be; cos t = 15/17
• Cos(t/2) =
• Sin(t/2) =
• Tan(t/2) =
17+15
34
17−15
34
1
16
=
32
34
=
16
17
=
2
34
=
1
17
= 1/4
44
Which do you need to remember?
• Sum and difference identities
• See page 616 for a summary of identities
45
Chapter 9.3
Product to Sum and Sum to Product Formulas
46
Sum to Product Identities
• Sin a + sin b = 2 sin
• Sin a – sin b = 2 cos
𝑎+𝑏
2
cos
𝑎−𝑏
2
𝑎+𝑏
𝑎−𝑏
sin
2
2
• Cos a + cos b = 2 cos
• Cos a – cos b =-2 sin
𝑎+𝑏
𝑎−𝑏
cos
2
2
𝑎+𝑏
2
sin
𝑎−𝑏
2
47
Product to Sum
• sin A sin B = ½ [ cos(A - B) – cos (A + B)]
• sin A cos B = ½ [ sin (A - B) + sin (A + B) ]
• cos A cos B = ½ [cos ( A – B) + cos (A + B) ]
48
What do you need to know?
• How to use these identities
49
Chapter 9.4
Trigonometric Equations
50
Solving and Checking Equations
• Like algebraic equation but…
Now we have trig functions too!
• And, because of periodicity, we may have multiple solutions
51
Checking Solutions
• Consider sin x + cos x = 1. Is x =  / 4 a solution?
Is  /2 ?
52
Substitute and Verify
• Consider sin x + cos x = 1. Is x =  /4 a solution?
sin ( /4) =
2
2
, cos ( /4) =
2
2
, no
•  /2?
Sin ( /2) = 1, cos ( /2) = 0, yes
53
Some Solution Techniques
• Solve directly
• Change functions to sine and cosine
• Factoring
• Identities
• Square both sides
• Substitution
54
Solving an Equation Directly
• Sin x = 1/2
• Find all solutions in the interval (0, 2 )
• Solutions in any interval?
55
Solution
• We know sin ( /6 ) = ½
• Can there be a solution in other quadrants?
Q2, 5  /6
Q3 and Q4 are <0, so no sols
• If we go beyond 2 ?
 /6 + 2n  and 5 /6 + 2n 
56
Factors of 2
• In general, if t is a solution, so is t + 2n
57
Solving by Factoring
• cos2 x + cos x – 2 = 0
58
Solution
• cos2 x + cos x – 2 = 0
• (cos x – 1)(cos x + 2) = 0
• Cos x  2
• Cos x = 1 for x = 2n 
59
Example
• 2 sin2x -3 sin x + 1 = 0
60
Solution
• 2 sin2x – 3 sinx + 1 = 0
• Factor: (2 sin x – 1)(sin x – 1) = 0
• Sin x = ½, sin x = 1
• X = /6 + 2n  or 5 /6 + 2n ,
• X = /2 + 2n 
61
Solving Using Identities
• 3 tan2x – sec2x – 5 = 0
62
Solution
• 3 tan2x – sec2x – 5 = 0
• It turns out that tan2x + 1 = sec2x
• Suppose you didn’t remember that?
3 (sin/cos) 2 – 1/cos2 – 5 = 0
Multiply by cos2
3 sin2 – 1 – 5 cos2 = 0
3 sin2 – 5(1 – sin2) = 1
8 sin2 = 6, sin2 = ¾, sin =  3 /2, x= /3 + n , 2/3 + n
63
Solving by Squaring Both Sides
• sin x + cos x = 1;
• May introduce extraneous solutions
64
Solution
• Square both sides
• Sin2 x + 2sin cos + cos2 = 1
• 2sin cos = 0, sin = 0 or cos = 0, x = /2 + n /2
• However, we introduced an additional root when we squared
– The actual solution is /2 + 2 n and 2 n
65
Be Careful with Substitution
• sin 3x = 1
66
Solution
• sin 3 x = 1
we know that sin y = 1 if y = /2
y = 3x, x = /6
Now, about the multiples! The variable we used was y;
y + 2  k are the solutions, not x + 2  k
y = /2 + 2  k = ( + 4 k) / 2
3x =( + 4 k) / 2
x = ( + 4 k) / 6
67
Example
• sin (x/2) = 1/2
68
Solution
• Sin (x/2) = ½
• Consider y = x/2
• Sin y = ½ for y = /6+ 2n, 5 /6 + 2n
• So, x/2 = /6+ 2n, 5 /6 + 2n
• x = /3+ 4n, 5 /3 + 4n
69
Example
• Cos 3a = 1
70
Solution
• Cos 3a = 1
• Let 3a = b
• Cos b = 1, b = 0 + 2n
• What is a?
• a = b/3 = 2n/3
71
Example
• 2sin2 a – cos (2a) = 0
72
Solution
• 2sin2 a – cos (2a) = 0
• Will need an identity
• Cos (2a) = cos2a – sin2a so
• 2sin2 a – cos (2a) = 2sin2 a - cos2a + sin2a
• And cos2a = 1 - sin2a
• Thus we have 3sin2 a -1 + sin2 a =0
• 4sin2 a = 1 or sin2 a = ¼
• Sin a =  ½ ; a = /6 + n , 5 /6 + n 
73
Example
• Solve sin3x cos x – sin x cos3x = -1/4
74
Solution
• Solve sin3x cos x – sin x cos3x = -1/4
• Factor: sinx cos x (sin2x – cos2x) = -1/4
• Use double angle formula: -sin x cos x (cos2x) = -1/4
• And again, for sin 2x: (1/2 ) sin 2x cos 2x = ¼
• Sin y cos y = 1/2; if y = /4, sin and cos are 1/ 2
• So y = /4 and 5 /4 + 2n 
• X = /8 + n  and 5 /8 + n 
75
Solve
• sinx cosx = 1
76
Solution
• sin x cos x = 1
• We have two ways to get 1: 1 x 1 or something bigger than 1
times 1. Neither are possible. There is no solution
77
What do you need to know?
How to solve equations with trig functions
• Solve directly
• Change functions to sine and cosine
• Factoring
• Identities
• Square both sides
• Substitution
78
Chapter 9.5
Inverse Trig Functions
79
Functions and Inverse Functions
• We require a function, f(x) to produce a unique answer for
each value, x. We cannot have, for example, f(2) = 3 or 4
• For a function to have an inverse that is a function, it must be
one-to-one. That is, if f(x1) = y, then f(x2) ≠ y unless x1 = x2
• Inverse f-1(f(x))=x
• Sometimes we use the “vertical line test” for functions and the
“horizontal line test” for inverses
• Clearly the inverse trig functions are not one-to-one, and to
invert them we must restrict their domain
80
Defining Inverse Trig Functions
• If cos x = a then arccos a = x;
x is the angle whose cosine is a*
• If sin t = 1/2, arcsine 1/2 = t, t is the angle whose sine is ½
• For the Inverse Trig functions to be functions, we have to
pass the vertical line test
– For example, arcsine ½ = /6, 5/6, 13/6, etc. if not
restricted;
*Also use the notation sin-1 a and cos-1 a
81
Range and Domain of Inverse Functions
Function
Range
Domain
Sin -1 or arcsin
[-/2, /2]
[-1, 1]
Cos -1 or arccos
[0, ]
[-1, 1]
Tan -1 or arctan
[-/2, /2]

Note: Sin -1 ≠ 1/Sin !!!!
82
Arcsin
arcsin
2
1
0
-1.5
-1
-0.5
0
0.5
1
1.5
-1
-2
83
Arccos
arcos
3.5
3
2.5
2
1.5
1
0.5
0
-1.5
-1
-0.5
0
0.5
1
1.5
84
Arctan
arctan
2
1.5
1
0.5
0
-15
-10
-5
0
5
10
15
-0.5
-1
-1.5
-2
85
Examples
• Arcsin( sin (/4)) = /4
• Cos(arcos(2/3)) = 2/3
• Tan(arctan 5 ) = 5
• Arcsin(1/2) = /6
• Arccos (0) = /2
• Arctan(-1) = -/4
86
Examples
Evaluate:
• sec(arcos(
2
2
)) + arcsin(-1)
• cos(arctan( 3 )
• cos(arcsin(2/3))
87
Solution
Evaluate:
• sec(arcos(
2
2
)) + arcsin(-1) ; 2 + 3/2
• cos(arctan( 3 ) ; arctan 3 = /3, cos(/3) = 1/2
• cos(arcsin(2/3)); If the sine is 2/3, two sides of the triangle are
2 and 3, the third is 32 − 22 =  5;
The cosine is  5/3;
88
Example
• Show sin (arcos x) = 12 − 𝑥 2
89
Solution
• Show sin (arcos x) = 12 − 𝑥 2
• If the cos is x, then the 3rd side is 12 − 𝑥 2 , and
the sin is 12 − 𝑥 2
90
Example
• arcsin x + arcos x = /2
91
Solution
• arcsin x + arcos x = /2
• Let arcsin x = a and arcos x = b, each being an angle.
• The three angles of the triangle are a, b, and /2
• Since they must sum to , a + b = /2
• If a + b = /2, then arcsin x + arcos x = /2
92
Example, arctan
• Find arctan(-1)
93
Solution
• Find arctan(-1)
• We know we need an angle that is a multiple of /4;
which angle is in the range of tan?
• The range of tan is (- /2, /2)
•
So, we have arctan(-1) = - /4
94
Example
• Tan(arctan 5)
95
Solution
• Tan(arctan 5)
• Tan(arctanx) = x, so our answer is 5
96
Example
• Tan a = x/3, 0<a<. Find a – tan 2a as a function of x
97
Solution
• Tan a = x/3, 0<a<. Find a – tan 2a as a function of x
• We know that a = arctan(x/3)
• What is tan 2a? Tan 2a =
2 tan 𝑎
1−𝑡𝑎𝑛2 a
• Tan(arctan(x/3)) = x/3, tan 2a = (2x/3)/(1-(x/3)2) = 6x/(9-x2)
• a – tan 2a = arctan (x/3) - 6x/(9-x2)
98
Example
• Simplify cos[ /6 + arctan 1/2 )]
99
Solution
• Simplify cos[ /6 + arctan ½ )]
• Cos(a+b) = cos a cos b – sin a sin b
• What are cos (arctan ½) and sin (arctan ½)?
• If the tan is ½, then the sides are 1 and 2 and 1 + 4
• Cos (arctan ½) = 2/ 5 , sin(arctan ½) = 1/ 5
• cos[ /6 + arctan ½ )] = ( 3/2) (2/ 5) –(1/2) (1/ 5) =
1/(2 5)[2 3 − 1]
100
Solve Using Identities
• If sin x = y/3, find x – sin (2x)
101
Solution
• Sin 2x =2 sinx cos x = 2 sinx 1 − 𝑠𝑖𝑛2 x
• 2
𝑦
3
1 −
𝑦2
9
=
2𝑦 9 − 𝑦 2
9
102
What do you need to know
• How to work with arcsin and arcos
• The range and domain of the inverse functions
103