Download OckenNotesCh7

7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
CCNY Math Review Chapter 7: Trigonometric identities
7.1:
7.2:
7.3:
7.4:
Pythagorean identities
Sum and difference formulas
Double and half angle formulas
Trigonometric equations
All rights reserved. Copyright 2016 by Stanley Ocken
CCNY Math Review Chapter 7: Trigonometric functions
11/27/16 Frame 1
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Review: Know how to find trig functions of 30◦ , 45◦ , and 60◦ .
π
6
π
4
π
3
Always make sure you know the basic triangles
below, from which you should easily figure out
the information in the table at the right.
θ
= 30◦
= 45◦
sin(θ)
1
2
√1
√2
3
2
◦
= 60
cos(θ)
√
3
2
1
√
2
1
2
tan(θ)
√1
3
1
√
3
30◦
60◦
2
√
1
30◦
√
All rights reserved. Copyright 2016 by Stanley Ocken
3
1
2
45◦
3
√
2
45◦
60◦
1
1
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 2
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Basic trigonometric identities
In the diagram, the terminal side of angle θ goes from the origin to P (x, y). The distance from P to
the origin is r. By the Pythagorean theorem, x2 + y 2 = r2 . The six trig functions are:
• cos(θ) =
• sec(θ) =
x
r
r
x
• sin(θ) =
• csc(θ) =
y
r
r
y
• tan(θ) =
• cot(θ) =
y
x
x
y
It follows from the above that all trig functions can be expressed
in terms of sin(θ) and cos(θ) as follows.
tan(θ) =
sin(θ)
cos(θ)
cot(θ) =
1
tan(θ)
=
y
5
4
P(4,3)
3
r=5
2
cos(θ)
sin(θ)
y=3
1
sec(θ) =
1
cos(θ)
cos(θ) =
1
sec(θ)
csc(θ) =
1
sin(θ)
sin(θ) =
1
csc(θ)
Alternate forms:
0
θ
0
1
x=4
2
3
4
5
x
Dividing x2 + y 2 = r2 by r2 gives ( xr )2 + ( yr )2 = 1 so cos2 (θ) + sin2 (θ) = 1
Dividing by cos2 (θ) gives 1 +
Dividing by sin2 (θ) gives
sin2 (θ)
cos2 (θ)
cos2 (θ)
sin2 (θ)
All rights reserved. Copyright 2016 by Stanley Ocken
=
+1=
1
cos2 (θ)
1
sin2 (θ)
and so 1 + tan2 (θ) = sec2 (θ)
and so 1 + cot2 (θ) = csc2 (θ)
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 3
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Since we derived the last three identities from the Pythagorean Theorem, they are called the
Pythagorean trigonometric identities. The word identity means that the expressions on the two sides of
the equals sign are equal (identical) for every angle θ.
Each of the Pythagorean identities relates two trig functions. You can express each one in terms of the
other.
You could rewrite
cos2 θ + sin2 θ = 1
You could rewrite
1 + tan2 θ = sec2 θ
You could rewrite
1 + cot2 θ = csc2 θ
as cos2 (θ) = 1 − sin2 θ
or as sin2 θ = 1 − cos2 θ
as tan2 θ = sec2 θ − 1
or as sec2 θ − tan2 θ = 1
as cot2 θ = csc2 θ − 1
or as csc2 θ − cot2 θ = 1
You should memorize only the first column of identities, but you need to be prepared to convert any
identity in that column to the corresponding identities in the other columns.
All rights reserved. Copyright 2016 by Stanley Ocken
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 4
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
We sometimes use the following even-odd identities.
cos(−θ) = cos θ and sin(−θ) = − sin θ
All rights reserved. Copyright 2016 by Stanley Ocken
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 5
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
5
4
We sometimes use the following even-odd identities.
cos(−θ) = cos θ and sin(−θ) = − sin θ
To see why these are true, look at the diagram.
The endline of θ goes from the origin to point
(x, y) = (4, 3) in Quadrant 1.
cos θ = xr = 54 and sin θ = yr = 35
P (x, y) = (4, 3)
3
r=5
2
1
θ
X
0
-1
-2
-3
-4
-5
0
All rights reserved. Copyright 2016 by Stanley Ocken
1
2
CCNY Math Review Chapter 7.1 Pythagorean identities
3
4
5
11/27/16 Frame 5
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
5
4
We sometimes use the following even-odd identities.
cos(−θ) = cos θ and sin(−θ) = − sin θ
To see why these are true, look at the diagram.
The endline of θ goes from the origin to point
(x, y) = (4, 3) in Quadrant 1.
cos θ = xr = 54 and sin θ = yr = 35
The endline of angle −θ goes from the origin to point
(x, −y) = (4, −3) in Quadrant 4.
Then cos(−θ) = xr = 54 = cos(θ) and
sin(−θ) = −y
= − yr = − 35 = − sin(θ)
r
It’s easy to check (same argument) that
tan(−θ) = − tan θ
P (x, y) = (4, 3)
3
r=5
2
1
θ
X
0
−θ
-1
r=5
-2
-3
Q(x, −y) = (4, −3)
-4
-5
0
All rights reserved. Copyright 2016 by Stanley Ocken
1
2
CCNY Math Review Chapter 7.1 Pythagorean identities
3
4
5
11/27/16 Frame 5
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Even and odd functions
A function y = f (x) is
• even if for all x in its domain, f (−x) = f (x).
The graph remains unchanged when it is reflected
through the y-axis. Such a graph is also called
y-axis symmetric. If point (x, y) is on the graph,
then so is point (−x, y).
• odd if for all x in its domain, f (−x) = f (x). The
graph remains unchanged when it is reflected
through the origin. Such a graph is also called
origin symmetric. If point (x, y) is on the graph,
then so is point (−x, −y).
To better understand what this means, look at the
graphs below.
The function cos(x) is even: for all x,
cos(−x) = cos(x). Its graph is y-axis symmetric. If
point (x, y) is on the graph, so is point (−x, y)
Y
1
0
−2π
−π
(−x, y)
π
X
2π
(x, y)
-1
We showed on the last slide that cosine is an even
function, while sine and tangent are odd functions.
All rights reserved. Copyright 2016 by Stanley Ocken
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 6
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
Y
1
The function sin(x) is odd: for all x,
sin(−x) = − sin(x). Its graph is
origin symmetric. If point (x, y) is
on the graph, so is point (−x, −y)
7.4 Trigonometric equations
(x, y)
0
−2π
X
π
−π
2π
(−x, −y)
-1
Y
3
2
The function tan(x) is odd: for all
x, tan(−x) = − tan(x). Its graph is
origin symmetric. If point (x, y) is
on the graph, so is point (−x, −y)
(x, y)
1
0
−2π
-1
−π
π
X
2π
(−x, −y)
-2
-3
All rights reserved. Copyright 2016 by Stanley Ocken
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 7
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Simplifying trigonometric expressions
Sometimes it is important to transform one trig expression into another. To do so, you need to build
new identities.
Method 1: Rewrite expressions in terms of sine and cosine
Example 1: Simplify and rewrite cos2 θ(1 + tan2 θ) in terms of sine and cosine.
Solution:
Original expression is
cos2 θ(1 + tan2 θ)
Use 1 + tan2 θ = sec2 θ
Use cos θ =
1
sec θ
Simplify
= cos2 θ sec2 θ
= cos2 θ · ( cos1 θ )2
= 1
Example 2: Simplify and rewrite
Solution:
Original expression is
Separate into two fractions
Use cos θ =
1
sec θ
Use sin2 θ = 1 − cos2 θ
All rights reserved. Copyright 2016 by Stanley Ocken
sec2 θ−1
sec2 θ
in terms of sine and cosine.
sec2 θ−1
sec2 θ
=
sec2 θ
sec2 θ
−
1
sec2 θ
= 1 − cos2 θ
=
sin2 θ
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 8
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Sometimes the instruction is more general:
u
Example 3: Combine fractions and simplify 1+sin
cos u +
Solution:
1+sin u
cos u
Original expression is
cos u + 1+sin u
Add fractions
=
Expand numerator
=
2
Use sin u +
cos2 u
=1
=
Rewrite numerator
=
Cancel
=
cos u
1+sin u
(1+sin u)(1+sin u)+cos u cos u
cos u(1+sin u)
1+2 sin u+sin2 u+cos2 u
cos u(1+sin u)
1+2 sin u+1
cos u(1+sin u)
2(1+sin u)
cos u(1+sin u)
2
cos u = 2 sec u
The last step was a required simplification, since it removed a fraction from the answer.
All rights reserved. Copyright 2016 by Stanley Ocken
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 9
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
In the next example, we use the important principle that a fraction can be rewritten as its numerator
3
1
u
1
times the reciprocal of its denominator:
= 3 · and = u ·
4
4
v
v
u
sin u
Example 4: Combine fractions and simplify cos
+ csc
sec u
u
Solution:
Since sec u = cos1 u , then sec u cos u = 1 and so cos u =
Since csc u =
1
,
sin u
then csc u sin u = 1 and so sin u =
Original expression is
cos u
sec u
Rewrite fractions
cos u ·
sin u
csc u
1
+
sec u
+
sin u ·
1
csc u
Use remarks above
= cos u cos u + sin u sin u
Pythagorean identity
= cos2 u + sin2 u = 1
All rights reserved. Copyright 2016 by Stanley Ocken
1
.
sec u
1
.
csc u
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 10
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
In the next example, we first create a complex fraction. We then simplify the fraction by multiplying its
numerator and denominator by the LCD of the nested (little) fractions that it contains.
sin u+1
Example 5: Simplify csc
u+1
Solution:
Original expression is
sin u+1
csc u+1
Rewrite csc u
= sin1 u+1
+1
Multiply top and bottom by sin u
=
sin u(sin u+1)
1 +1)
sin u( sin
u
Multiply out on bottom
=
sin u(sin u+1)
1+sin u
Cancel: (since 1 + sin u = sin u + 1)
= sin u
sin u
All rights reserved. Copyright 2016 by Stanley Ocken
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 11
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 6 : Simplify
7.3 Double and half-angle formulas
7.4 Trigonometric equations
2+cot2 u
csc2 u
Solution: (the hard way)
2+cot2 u
csc2 u
Original expression is
u )2 u
2+( cos
sin u
Rewrite cot u and csc u
=
Square the fractions
=
Multiply top and bottom by sin2 u
=
2 sin2 u+cos2 u
1
Use cos2 u = 1 − sin2 u
=
2 sin2 u+1−sin2 u
1
( sin1 u 2 )
2
2+ cos2 u
sin u
1
sin2 u
= sin2 u + 1
All rights reserved. Copyright 2016 by Stanley Ocken
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 12
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 7 : Simplify
7.3 Double and half-angle formulas
7.4 Trigonometric equations
2+cot2 u
csc2 u
Solution: Hint: think about the identity csc2 u = cot2 u + 1
2+cot2 u
csc2 u
Original expression is
Use the hint, but rewrite the identity
as csc2 u − 1 = cot2 u
Rewrite the numerator
=
2+csc2 u−1
csc2 u
Rewrite the numerator
=
1+csc2 u
csc2 u
Split the fraction in two
=
1
csc2 u
Use
1
csc u
= sin u
All rights reserved. Copyright 2016 by Stanley Ocken
+
csc2 u
csc2 u
= sin2 u + 1
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 13
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Quiz Review for 7.1
Example 1: Simplify and rewrite cos2 θ(1 + tan2 θ) in terms of sine and cosine.
Example 2: Simplify and rewrite
sec2 θ−1
sec2 θ
in terms of sine and cosine.
Example 3: Combine fractions and simplify
1+sin u
cos u
Example 4: Combine fractions and simplify
cos u
sec u
Example 5: Simplify
+
+
cos u
1+sin u
sin u
csc u
sin u+1
csc u+1
Example 6 : Simplify
2+cot2 u
csc2 u
(use complex fraction simplification)
Example 7 : Simplify
2+cot2 u
csc2 u
(use a trig identity involving cot u and csc u.)
All rights reserved. Copyright 2016 by Stanley Ocken
CCNY Math Review Chapter 7.1 Pythagorean identities
11/27/16 Frame 14
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Finding trig functions of general angles.
Procedure: finding sin(θ), cos(θ), tan(θ)
1. Draw the endline of angle θ.
2. The reference angle θ is the acute angle between the endline of the angle and the x-axis.
• cos(θ) = cos(θ) if θ’s endline is in Q1 or Q4,where cosine is + , or
cos(θ) = − cos(θ) if θ’s endline is in Q2 or Q3, where cosine is − .
• sin(θ) = sin(θ) if θ’s endline is in Q1 or Q2, where sine is + , or
sin(θ) = − sin(θ) if θ’s endline is in Q3 or Q4, where sine is − .
• tan(θ) = tan(θ) if θ’s endline is in Q1 or Q3, where tangent is + , or
tan(θ) = − tan(θ) if θ’s endline is in Q2 or Q4, where tangent is − .
Review example: Find cos 120◦ .
Solution: θ = 120◦ has endline in Q2.
Its (yellow) reference angle is θ = 180◦ − 120◦ = 60◦ .
The triangle at the far right shows that cos 60◦ = 21 .
Since cosine is negative in Q2,
cos 120◦ = cos θ = − cos θ = − cos 60◦ = − 21 .
All rights reserved. Copyright 2016 by Stanley Ocken
.
Y
Q2
θ = 60◦
Q1
30◦
θ = 120◦
2
√
X
Q3
Q4
M19500 Precalculus Chapter 7.2 Sum and difference formulas
3
60◦
1
11/27/16 Frame 15
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Sum and difference trig formulas
Trig functions of the sum of angles.
Let u and v be any angles, both expressed in radians, or both in degrees. Then:
• cos(u + v) = cos(u) cos(v) − sin(u) sin(v)
• sin(u + v) = sin(u) cos(v) + cos(u) sin(v)
tan u+tan v
• tan(u + v) = 1−tan
u tan v
Exercise: Use the first two formulas above to derive the third.
sin(u+v)
sin(u) cos(v)+cos(u) sin(v)
Hint: start with tan(u + v) = cos(u+v)
= cos(u)
cos(v)−sin(u) sin(v) .
Then divide top and bottom of the fraction by cos(u) cos(v) and simplify.
It is best to memorize the above three identities. Every other identity in 7.2 and 7.3 can be
easily derived from them. For example, recall that cos(−u) = cos(u) and sin(−v) = − sin(v).
Then we can figure out cos(u − v) by writing u − v as the sum u+ -v. The result is:
cos(u − v) = cos(u+ -v) = cos(u) cos(−v) − sin(u) sin(−v)
= cos(u) cos(v) + sin(u) sin(v) since sin(−v) = − sin(v).
This method yields three slightly new identities:
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.2 Sum and difference formulas
11/27/16 Frame 16
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Trig functions of the difference of angles.
Let u and v be any angles, both expressed in radians, or both in degrees. Then:
• cos(u − v) = cos(u) cos(v) + sin(u) sin(v)
• sin(u − v) = sin(u) cos(v) − cos(u) sin(v)
tan u−tan v
• tan(u − v) = 1+tan
u tan v
In previous sections, we found trig functions of any angle θ with reference angle
30◦ , 45◦ , or 60◦ . By using the sum and difference formulas, we can also handle angles
with reference angle 15◦ and 75◦ .
Example 1: Rewrite each of the four angles 15◦ , 75◦ , 105◦ , 165◦ as a sum or difference
of angles with reference angle 30◦ , 45◦ , or 60◦ .
Solution:
15◦ = 45◦ − 30◦ ; 75◦ = 45◦ + 30◦ ; 105◦ = 45◦ + 60◦ ; 165◦ = 120◦ + 45◦ .
Example 2: Use these answers to find cos(15◦ ), sin(75◦ ), cos(105◦ ), and sin(165◦ ).
Solution: Just apply the appropriate sum or difference formula. See the next slide.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.2 Sum and difference formulas
11/27/16 Frame 17
7.1 Pythagorean identities
•
cos 15◦
=
7.2 Sum and difference formulas
cos(45◦
−
30◦ )
=
√1
2
• sin 75◦ = sin(45◦ + 30◦ )
7.3 Double and half-angle formulas
cos 45◦ cos 30◦
√
3
2
·
+
√1
2
·
1
2
+
7.4 Trigonometric equations
sin 45◦ sin 30◦
√
1+√ 3
2 2
=
= sin 45◦ cos 30◦ + cos 45◦ sin(30◦ )
=
• cos 105◦ = cos(45◦ + 60◦ )
√1
2
√
·
3
2
+
√1
2
·
√
1
2
3+1
√
2 2
=
= cos 45◦ cos 60◦ − sin 45◦ sin 60◦
=
√1
2
·
1
2
−
√1
2
√
·
3
2
=
√
1−√ 3
2 2
In the following example we use the fact that 120◦ is a Quadrant 2 angle with reference angle
180◦ − 120◦ = 60◦ . See the diagram on the first slide in this lesson.
• sin 165◦ = sin(45◦ + 120◦ )
= sin 45◦ cos 120◦ + cos 45◦ sin 120◦
= √12 · (− cos 60◦ ) + √12 · sin 60◦ =
=
√1
2
· (− 21 ) +
√1
2
√
·
3
2
=
√
−1+
√ 3
2 2
You could multiply numerator and denominator of each answer above by
sin 75◦ =
√
3+1
√
2 2
=
√
√
( 3+1) 2
√ √
2 2( 2)
All rights reserved. Copyright 2016 by Stanley Ocken
=
√ √ √
3 2+ 2
2(2)
√
=
√
6+ 2
4
√
2. For example
.
M19500 Precalculus Chapter 7.2 Sum and difference formulas
11/27/16 Frame 18
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
In all of the above examples, we have used degree measures for angles. Of course, you
should also be able to find trig functions of angles expressed in radians.
7π
11π
π
, sin 5π
Example 3. Find cos 12
12 , cos 12 , and sin 12
2π
3π
π
π
Solution: If you know, for instance, that 5π
12 = 12 + 12 = 6 + 4 , you can apply the
identity for sin( π6 + π4 ). But it’s easier to convert this sort of problem to degrees by
◦
π
◦
using the easy fact: 12
radians = 180
12 = 15 .
π
• cos 12
= cos(15◦ ) = cos(45◦ − 30◦ )
◦
◦
◦
◦
• sin 5π
12 = sin(5 · 15 ) = sin 75 = sin(45 + 30 )
◦
◦
◦
◦
• cos 7π
12 = cos(7 · 15 ) = cos 105 = cos(45 + 60 )
◦
◦
◦
◦
• sin 11π
12 sin(11 · 15 ) = sin 165 = sin(45 + 120 )
All of the rewritten problems were done in Example 2.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.2 Sum and difference formulas
11/27/16 Frame 19
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
We can now find trig functions of any angle that is a multiple of 15◦ =
π
12 .
Example 4: Find sin( 67π
12 )
Solution: First find a coterminal angle between 0 and 2π by subtracting (twice)
(67−24−24)π
67π
24π
24π
2π = 24π
= 19π
12 . The result is 12 − 12 − 12 =
12
12 .
19π
◦
◦
This is a bit better: sin 67π
12 = sin 12 = 19 · 15 = 285 . It’s still not obvious how to
break this up as a sum or difference. One way is to find the reference angle, as follows.
Y
285◦ is between 270◦ and 360◦ . Thus
285◦ is in Quadrant 4, and so its reference angle
is 360◦ − 285◦ = 75◦ .
Since 285◦ is in Quadrant 4, its sine is negative.
Thus sin(285◦ ) = − sin 75◦ .
Q2
Q1
θ = 285◦
X
θ = 75◦
Q3
Q4
Now all you need to do is use √
a sum formula to find
√ 3 , one of the answers found in Example 2.
sin 75◦ = sin(45◦ + 30◦ ) = 1+
2 2
Now put it all together:
√
1+√ 3
19π
◦
◦
Answer: sin 67π
12 = sin 12 = sin 285 = − sin 75 = − 2 2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.2 Sum and difference formulas
11/27/16 Frame 20
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Sometimes you need to work backwards, from a numerical expression to a formula:
Example 5: Find the exact value of sin 16◦ cos 44◦ + cos 16◦ sin 44◦ .
Solution: This is a situation where variables are easier to understand than numbers.
Let u = 16◦ and v = 44◦ . You are being asked to find the exact value of
sin 16◦ cos 44◦ + cos 16◦ sin 44◦ = sin u cos v + cos u sin v, which is exactly sin(u + v).
Therefore you are being asked to find sin(u + v) = sin(16◦ + 44◦ ) = sin(60◦ ) =
√
3
2
.
Finally, you can use addition/subtraction formulas to prove new identities, as follows.
Example 6: Prove the identity sin(x − π2 ) = − cos x.
Solution: Since this is a formula involving the sine of a difference of angles, write
sin(u − v) = sin u cos v − cos u sin v. Substitute u = x and v = π2 to obtain
sin(x − π2 ) = sin x cos π2 − cos x sin π2 = − cos(x) because cos π2 = 0 and sin π2 = 1.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.2 Sum and difference formulas
11/27/16 Frame 21
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Quiz Review for 7.2
Review example: Find cos 120◦ .
Example 1: Rewrite each of the four angles 15◦ , 75◦ , 105◦ , 165◦ as a sum or difference
of angles with reference angle 30◦ , 45◦ , or 60◦ .
Example 2: Use the answers to Example 1 to find cos(15◦ ), sin(75◦ ), cos(105◦ ), and
sin(165◦ ).
7π
11π
π
, sin 5π
Example 3. Find cos 12
12 , cos 12 , and sin 12
Example 4: Find sin( 67π
12 ).
Example 5: Find the exact value of sin 16◦ cos 44◦ + cos 16◦ sin 44◦ .
Example 6: Prove the identity sin(x − π2 ) = − cos x.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.2 Sum and difference formulas
11/27/16 Frame 22
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Review: Expressing one trig function in terms of a given trig function
Review example: Assume θ is an acute angle with cos θ = x.
Find the other five trig functions of θ.
Solution: We are given cos θ = x. To find out
other trig functions of θ, draw the acute triangle
A
at the right, with cos θ = H
= x1 = x as required.
2
2
2
Solve√A + O = H√ for
O = H 2 − A2 = 1 − x2 .
Now we can find the other trig functions of θ.
√
√
2
O
= 1−x
= 1 − x2 .
• sin θ = H
1
• tan θ =
• cot θ =
• sec θ =
• csc θ =
O
A
A
O
H
A
H
O
Hypotenuse = 1
p
1 − x2 = Opposite
θ
Adjacent = x
√
=
=
=
=
1−x2
.
x
1
√ x
tan θ = 1−x2 .
1
1
cos θ = x .
√ 1
.
1−x2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
11/27/16 Frame 23
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Double-angle formulas
Recall the sum formulas:
• cos(u + v) = cos(u) cos(v) − sin(u) sin(v)
• sin(u + v) = sin(u) cos(v) + cos(u) sin(v)
tan u+tan v
• tan(u + v) = 1−tan
u tan v
In each formula, substitute u for v.
Double-angle formulas.
Let u be any angle , expressed in radians or degrees. Then:
• cos(2u) = cos2 u − sin2 u • sin(2u) = 2 sin u cos u • tan(2u) =
2 tan u
1−tan2 u
Solving cos2 u + sin2 u = 1 for cos2 θ = 1 − sin2 θ yields an alternate double angle
formula for cosine: cos(2u) = cos2 u − sin2 u = 1 − sin2 θ − sin2 θ = 1 − 2 sin2 θ.
Solving cos2 u + sin2 u = 1 for sin2 θ = 1 − cos2 θ yields cos(2u) = 2 cos2 u − 1.
More double-angle formulas for cosine
Let u be any angle , expressed in radians or degrees. Then:
• cos(2u) = 2 cos2 u − 1
• cos(2u) = 1 − 2 sin2 u
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
11/27/16 Frame 24
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Using the double-angle formulas
Example 1: Suppose cos θ = − 23 and θ is in Quadrant 3. Find cos(2θ) and sin(2θ).
Solution:
• To find cos(2θ), use the double-angle formula and the identity sin2 θ = 1 − cos2 θ.
Then cos(2θ) = cos2 θ − sin2 θ = cos2 θ − (1 − cos2 θ) = 2 cos2 θ − 1 = 2(− 23 )2 − 1 =
2( 49 ) − 1 =
8
9
−
9
9
= − 19 .
• To find sin(2θ), use the double-angle formula sin(2θ) = 2 sin θ cos θ. We are given
cos θ = − 23 but need to figure out sin θ.
√
√
For this you need to remember that if A2 = 7, then A = ± 7, NOT 7.
Since sin2 θ = 1 − cos2 θ, we
q have
q
q
√
√
2 2
4
2
sin θ = ± 1 − cos θ = ± 1 − (− 3 ) = ± 1 − 9 = ± 59 ± 35 .
To decide between plus and minus, use
√ the fact that θ is a Quadrant 3 angle, and so its
sine is negative. Therefore sin θ = − 35 . Now we can apply the double-angle formula:
√
sin(2θ) = 2 sin θ cos θ = 2(−
5
2
3 )(− 3 )
2
=
√
4 5
9
.
2
If these answers are correct, then cos 2θ + sin 2θ should be 1. Please check that this is so.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
11/27/16 Frame 25
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Half-angle formulas
We now combine the identities cos 2θ = cos2 θ − sin2 θ and cos2 θ + sin2 θ = 1
Adding them yields 2 cos2 θ = 1 + cos(2θ). Dividing by 2 gives cos2 θ = 1+cos(2θ)
.
2
1−cos(2θ)
2
2
2
2
2
Subtracting cos 2θ = cos θ − sin θ from cos θ + sin θ = 1 yields sin θ =
2
Formulas for lowering powers
Let θ be any angle , expressed in radians or degrees. Then:
2θ
2θ
• sin2 θ = 1−cos
• tan2 θ =
• cos2 θ = 1+cos
2
2
Substituting
u
2
for θ and solving for the three trig functions of
u
2
1−cos 2θ
1+cos 2θ
yields
Half-angle formulas
Let u be anyqangle , expressed in radians
q or degrees. Then:
q
u
1+cos u
u
1−cos u
u
1−cos u
• cos 2 = ±
•
sin
=
±
•
tan
=
±
2
2
2
2
1+cos u
The choice of sign depends on which quadrant angle u2 is in.
You need to know and apply these six formulas. Rather than memorizing the formulas
themselves, memorize how to derive them, as shown above.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
11/27/16 Frame 26
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Using the half-angle formulas
Example 2: Find the exact value of sin π8 .
45
◦
Solution: Since π radians = 180◦ , the angle is 180
8 = 2 = 22.5 . The .5 should alert
you to the fact that you need to use a half-angle formula for sin( u2 ). In other words,
set u2 = 22.5◦ and solve for u to obtain u = 45◦ .
r
q
q
1− √1
◦
u
1−cos
u
1−cos
45
2
Then sin 22.5◦ = sin 2 = ±
=±
=±
2
2
2 . To figure out the
sign, note that 22.5◦ris in Quadrant 1, where sine is positive. Therefore
1− √1
2
You can make the answer look nicer by multiplying top
sin π8 = sin 22.5◦ =
2
and bottom
the radical by 2.
r under
r of the fraction
q √
1
1
p
√
2 1− √
1− √
2
2− 2
π
1
2
Answer:
=
=
=
sin
=
2− 2 .
2
2·2
4
8
2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
11/27/16 Frame 27
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Example 3: Find the exact value of cos 112.5◦ .
Solution: The .5 should alert you to the fact that you need to use a half-angle formula
for cos( u2 ). In other words, set u2 = 112.5◦ and solve for u to obtain u = 225◦ .
q
q
u
1+cos 225◦
Then cos 112.5◦ = cos u2 = ± 1+cos
=
±
2
2
To proceed further, we need to pause and figure out cos 225◦ . First, u = 225◦ is a Quadrant 3
angle. Draw a third quadrant picture to see that the reference angle of u = 225◦ is
225◦ − 180◦ = 45◦ = u. Now recall the Reference Angle Principle: cos u = ± cos u. The sign is
negative because cosine is negative in Quadrant 3. Therefore, setting u = 225◦ , we obtain
cos 225◦ = − cos(45◦ ) = − √12 .
r
q
1− √1
◦
2
Now we go back and calculate cos 112.5◦ = ± 1+cos2 225 = ±
2 .
◦
The sign is negative since
r 112.5 is a Quadrant 2 angle, where cosine is negative.
Thus cos 112.5◦ = −
1− √1
2
. Multiplying top and bottom of the fraction under the
q √
p
√
radical by 2 gives the neater form − 2−4 2 = cos 112.5◦ = − 12 2 − 2 .
2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
11/27/16 Frame 28
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Example 4: Find cos u2 if sin u = − 35 and u is in Quadrant 3.
Solution:
q
u
.
Begin with the half-angle formula cos u2 = ± 1+cos
2
First figure out cos u. Sinceq
cos2 u + sin2 u =q
1, solve for q
p
3 2
9
4
2
cos u = ± 1 − sin u = ± 1 − − 5 = ± 1 − 25
= ± 16
25 = ± 5 .
Since u is in Quadrant 3, where cosine is negative, cos u = − 45 .
q 4
q
q
1− 5
u
1
=
±
=
±
The half-angle formula tells us cos u2 = ± 1+cos
2
2
10 . To figure out
the sign, we need to be careful and find out what quadrant u2 is in. Since u is in
Quadrant 3, 180◦ < u < 270◦ . Divide by 2 to get 90◦ < u2 < 135◦ .
q
1
Therefore u2 is in Quadrant 2, where cosine is negative, and so cos u2 = − 10
.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
11/27/16 Frame 29
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Example 5: Find cos(2 arccos(x)).
Solution: Let θ = arccos(x). Then cos θ = x. Then
cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 2x2 − 1 .
Example 6. Find tan(2 arcsin x).
Solution: Let θ = arcsin x. Then sin θ = x. We want to calculate tan 2θ =
We only know sin θ = x. To find out tan θ or other
trig functions of θ, draw the acute triangle at the
O
= x1 = x as required.
right, with sin θ = H
√
√
Solve A2 + O2 = H 2 for A = H 2 − O2 = 1 − x2 .
√ x
Then tan θ = O
.
A =
1−x2
Hypotenuse = 1
2 tan θ
.
1−tan2 θ
x = Opposite
θ
Adjacent =
p
1 − x2
2 tan θ
Now we are ready to find tan 2θ = 1−tan
2 θ.
x
x2
1−x2
x2
1−2x2
)2 = 1 − 1−x
• Find the denominator 1 − tan2 θ = 1 − ( √1−x
2 = 1−x2 − 1−x2 = 1−x2 .
2
2x
• Find the numerator 2 tan θ = √1−x
.
2
√
√ 2x
√
2 tan θ
2x·( 1−x2 )2
1−x2
2x 1−x2
1−x2
√ 2x
√
Then tan 2θ =
=
=
·
=
=
.
2
2
1−2x2
1−2x
1−x2 1−2x
( 1−x2 )·(1−2x2 )
1 − tan2 θ
1−x2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
11/27/16 Frame 30
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Quiz Review for 7.3
Review example: Assume θ is an acute angle. If cos θ = x, find the other five trig
functions of θ.
Example 1: Suppose cos θ = − 32 and θ is in Quadrant 3. Find cos(2θ) and sin(2θ).
Example 2: Find the exact value of sin π8 .
Example 3: Find the exact value of cos 112.5◦ .
Example 4: Find cos u2 if sin u = − 35 and u is in Quadrant 3.
Example 5: Find cos(2 arccos(x)).
Example 6. Find tan(2 arcsin x).
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
11/27/16 Frame 31
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Rewriting trigonometric equations
In Chapter 5.5, we showed how to solve some simple equations such as sin θ = 12 ,
which ask for an angle whose trig function value is specified. We call equations of this
form basic trig equations. If you know how to solve basic trig equations, you can solve
more complicated equations involving trig functions, such as the following:
Example 1: Rewrite the equation sin2 θ = 21 as two basic trig equations.
√
Reminder: the solution of x2 = 5 is x = ± 5.
q
Solution: The equation says (sin θ)2 = 12 . Then sin θ = ± 12 = ± √12 .
Answer: sin2 θ =
1
2
can be rewritten as sin θ =
√1
2
; sin θ = − √12 .
Example 2: Rewrite the equation 2 sin2 θ + sin θ = 0 as basic trig equations.
Solution: Rewrite the equation as (sin θ)(2 sin θ + 1) = 0. Solve this equation by
setting each factor to zero: sin θ = 0 or 2 sin θ + 1 = 0.
Answer: 2 sin2 θ + sin θ = 0 can be rewritten as sin θ = 0 ; sin θ = − 21 .
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 32
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Example 3: Rewrite the equation sin2 θ − 5 sin θ = 6 as basic trig equations.
Solution: It might make things simpler to use the abbreviation u = sin θ. Then
sin2 θ − 5 sin θ = 6 becomes
u2 − 5u = 6 ⇒ u2 − 5u − 6 = 0 ⇒ (u − 6)(u + 1) = 0 ⇒ u = 6, u = −1.
Now plug in sin θ for u to obtain:
Answer: sin2 θ − 5 sin θ = 6 can be rewritten as sin θ = −1 ; sin θ = 6.
Example 4: Rewrite the equation 2 cos2 θ − 2 sin2 θ = 1 as basic trig equations.
Solution: This is trickier since it involves two different trig functions. However, we
know that cos2 θ = 1 − sin2 θ and so 2 cos2 θ − 2 sin2 θ = 1 becomes
2(1 − sin2 θ) − 2 sin2 θ = 1
1 − sin2 θ − sin2 θ = 12
1 − 2 sin2 θ = 12
1 − 21 = 2 sin2 θ
1
2
2 = 2 sin θ
1
2
= sin θ
4q
± 14 = sin θ = ± 12 .
Answer: 2 cos2 θ − 2 sin2 θ = 1 can be rewritten as sin θ = − 21 ; sin θ = 12 .
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 33
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Solving for an angle whose trig function is given
Two reminders:
• Reference angle: For any angle θ, we write θ for the reference angle of θ.
θ is the acute angle between the endline of θ and the x-axis.
• Absolute value: |x| = 4 and x = ±4 say the same thing.
Reference Angle Principle (RAP) : Let α be an acute angle: 0 < α < 90◦ .
Suppose sin θ = ± sin α. Then α is the reference angle of θ.
In other words: if | sin θ| = sin α then θ has reference angle θ = α.
This principle is true for any trig function, not just for sines. Applying this principle
together with ASTC sign information yields statements such as:
• If sin θ = 12 = sin 30◦ , then θ = 30◦ and θ is in Q1 or Q2.
• If sin θ = − 12 = − sin 30◦ , then θ = 30◦ and θ is in Q3 or Q4.
• If cos θ =
• If cos θ =
√1 = cos 45◦ , then θ = 45◦
2
− √12 = − cos 45◦ , then θ =
and θ is in Q1 or Q4.
45◦ and θ is in Q2 or Q3.
√
3 = tan 60◦ , then θ = 60◦ and θ is in Q1 or Q3.
• If tan θ = √
• If tan θ = − 3 = − tan 60◦ , then θ = 60◦ and θ is in Q2 or Q4.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 34
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 5: Find all solutions of sin θ = − 12
with 0 ≤ θ ≤ 2π.
Solution: We are given sin θ = − 12 . Then
| sin θ| = 12 = sin 30◦ . By RAP, every solution
of sin θ = − 12 has reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦ .
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
Quadrant 1
√
P1 ( 3, 1)
θ = 30◦
0
0
-2
2
X
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 35
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 5: Find all solutions of sin θ = − 12
with 0 ≤ θ ≤ 2π.
Solution: We are given sin θ = − 12 . Then
| sin θ| = 12 = sin 30◦ . By RAP, every solution
of sin θ = − 12 has reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦ .
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
Quadrant 1
√
P1 ( 3, 1)
-2
√
P3 (− 3, −1)
30◦
θ = 30◦
0
0
2
X
Quadrant 3
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 35
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 5: Find all solutions of sin θ = − 12
with 0 ≤ θ ≤ 2π.
Solution: We are given sin θ = − 12 . Then
| sin θ| = 12 = sin 30◦ . By RAP, every solution
of sin θ = − 12 has reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦ .
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
Quadrant 2 Quadrant 1
√
P2 (− 3, 1)
-2
√
P3 (− 3, −1)
√
P1 ( 3, 1)
30◦
30◦
θ = 30◦
0
0
2
X
Quadrant 3
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 35
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 5: Find all solutions of sin θ = − 12
with 0 ≤ θ ≤ 2π.
Solution: We are given sin θ = − 12 . Then
| sin θ| = 12 = sin 30◦ . By RAP, every solution
of sin θ = − 12 has reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦ .
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
Quadrant 2 Quadrant 1
√
P2 (− 3, 1)
-2
√
P3 (− 3, −1)
√
P1 ( 3, 1)
30◦
30◦
0
0
θ = 30◦
30◦
Quadrant 3 Quadrant 4
2
X
√
P4 ( 3, −1)
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 35
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 5: Find all solutions of sin θ = − 12
with 0 ≤ θ ≤ 2π.
Solution: We are given sin θ = − 12 . Then
| sin θ| = 12 = sin 30◦ . By RAP, every solution
of sin θ = − 12 has reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦ .
Since sin θ is negative, θ will end in Q3 or Q4.
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
Quadrant 2 Quadrant 1
√
P2 (− 3, 1)
-2
√
P3 (− 3, −1)
√
P1 ( 3, 1)
30◦
30◦
0
0
θ = 30◦
30◦
Quadrant 3 Quadrant 4
2
X
√
P4 ( 3, −1)
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 35
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 5: Find all solutions of sin θ = − 12
with 0 ≤ θ ≤ 2π.
Solution: We are given sin θ = − 12 . Then
| sin θ| = 12 = sin 30◦ . By RAP, every solution
of sin θ = − 12 has reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦ .
Since sin θ is negative, θ will end in Q3 or Q4.
In Q3, θ = 180◦ + 30◦ = 210◦ .
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
Quadrant 2 Quadrant 1
√
P2 (− 3, 1)
-2
√
P3 (− 3, −1)
θ = 210◦ =
30◦
30◦
0
0
7π
6
θ = 30◦
30◦
Quadrant 3 Quadrant 4
√
P1 ( 3, 1)
2
X
√
P4 ( 3, −1)
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 35
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 5: Find all solutions of sin θ = − 12
with 0 ≤ θ ≤ 2π.
Solution: We are given sin θ = − 12 . Then
| sin θ| = 12 = sin 30◦ . By RAP, every solution
of sin θ = − 12 has reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦ .
Since sin θ is negative, θ will end in Q3 or Q4.
In Q3, θ = 180◦ + 30◦ = 210◦ .
In Q4, θ = 360◦ − 30◦ = 330◦ .
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
Quadrant 2 Quadrant 1
√
P2 (− 3, 1)
-2
√
P3 (− 3, −1)
θ = 210◦ =
30◦
30◦
0
0
7π
6
θ = 30◦
30◦
Quadrant 3 Quadrant 4
θ = 330◦ =
√
P1 ( 3, 1)
2
X
√
P4 ( 3, −1)
11π
6
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 35
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 5: Find all solutions of sin θ = − 12
with 0 ≤ θ ≤ 2π.
Solution: We are given sin θ = − 12 . Then
| sin θ| = 12 = sin 30◦ . By RAP, every solution
of sin θ = − 12 has reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦ .
Since sin θ is negative, θ will end in Q3 or Q4.
In Q3, θ = 180◦ + 30◦ = 210◦ .
In Q4, θ = 360◦ − 30◦ = 330◦ .
11π
Answer: θ = 7π
6 , 6 . The answers must be
expressed in radians, since the question
specified 0 ≤ θ ≤ 2π.
All rights reserved. Copyright 2016 by Stanley Ocken
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
Quadrant 2 Quadrant 1
√
P2 (− 3, 1)
-2
√
P3 (− 3, −1)
θ = 210◦ =
30◦
30◦
0
0
7π
6
θ = 30◦
30◦
Quadrant 3 Quadrant 4
θ = 330◦ =
√
P1 ( 3, 1)
2
X
√
P4 ( 3, −1)
11π
6
-2
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 35
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 5: Find all solutions of sin θ = − 12
with 0 ≤ θ ≤ 2π.
Solution: We are given sin θ = − 12 . Then
| sin θ| = 12 = sin 30◦ . By RAP, every solution
of sin θ = − 12 has reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦ .
Since sin θ is negative, θ will end in Q3 or Q4.
In Q3, θ = 180◦ + 30◦ = 210◦ .
In Q4, θ = 360◦ − 30◦ = 330◦ .
11π
Answer: θ = 7π
6 , 6 . The answers must be
expressed in radians, since the question
specified 0 ≤ θ ≤ 2π.
Example 6: Find all solutions of sin θ = − 12 .
All rights reserved. Copyright 2016 by Stanley Ocken
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
Quadrant 2 Quadrant 1
√
P2 (− 3, 1)
-2
√
P3 (− 3, −1)
θ = 210◦ =
30◦
30◦
0
0
7π
6
θ = 30◦
30◦
Quadrant 3 Quadrant 4
θ = 330◦ =
√
P1 ( 3, 1)
2
X
√
P4 ( 3, −1)
11π
6
-2
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 35
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 5: Find all solutions of sin θ = − 12
with 0 ≤ θ ≤ 2π.
Solution: We are given sin θ = − 12 . Then
| sin θ| = 12 = sin 30◦ . By RAP, every solution
of sin θ = − 12 has reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦ .
Since sin θ is negative, θ will end in Q3 or Q4.
In Q3, θ = 180◦ + 30◦ = 210◦ .
In Q4, θ = 360◦ − 30◦ = 330◦ .
11π
Answer: θ = 7π
6 , 6 . The answers must be
expressed in radians, since the question
specified 0 ≤ θ ≤ 2π.
Example 6: Find all solutions of sin θ = − 12 .
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
Quadrant 2 Quadrant 1
√
P2 (− 3, 1)
-2
√
P3 (− 3, −1)
θ = 210◦ =
30◦
30◦
0
0
7π
6
θ = 30◦
30◦
Quadrant 3 Quadrant 4
θ = 330◦ =
√
P1 ( 3, 1)
2
X
√
P4 ( 3, −1)
11π
6
-2
Solution: To find all angles θ with the endlines shown in Q3 and Q4, add/subtract any whole number
k of complete circles to/from each Example 5 answer.
Answer: θ =
7π
6
+ 2kπ and θ =
All rights reserved. Copyright 2016 by Stanley Ocken
11π
6
+ 2kπ, where k is any integer.
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 35
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
Y
Example 7: Find all solutions of tan θ = √13 .
Solution: Since tan θ = √13 = tan 30◦ , RAP
says: every solution of tan θ = √13 has
reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦
7.4 Trigonometric equations
2
√
P1 ( 3, 1)
Quadrant 1
θ = 30◦
0
-2
0
2
X
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 36
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
Y
Example 7: Find all solutions of tan θ = √13 .
Solution: Since tan θ = √13 = tan 30◦ , RAP
says: every solution of tan θ = √13 has
reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦
7.4 Trigonometric equations
2
√
P1 ( 3, 1)
Quadrant 1
-2
√
P3 (− 3, −1)
30◦
θ = 30◦
0
0
2
X
Quadrant 3
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 36
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 7: Find all solutions of tan θ = √13 .
Solution: Since tan θ = √13 = tan 30◦ , RAP
says: every solution of tan θ = √13 has
reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
√
P2 (− 3, 1)
√
P1 ( 3, 1)
Quadrant 2 Quadrant 1
-2
√
P3 (− 3, −1)
30◦
30◦
θ = 30◦
0
0
2
X
Quadrant 3
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 36
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 7: Find all solutions of tan θ = √13 .
Solution: Since tan θ = √13 = tan 30◦ , RAP
says: every solution of tan θ = √13 has
reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
√
P2 (− 3, 1)
√
P1 ( 3, 1)
Quadrant 2 Quadrant 1
-2
√
P3 (− 3, −1)
30◦
30◦
0
0
θ = 30◦
30◦
Quadrant 3 Quadrant 4
2
X
√
P4 ( 3, −1)
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 36
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 7: Find all solutions of tan θ = √13 .
Solution: Since tan θ = √13 = tan 30◦ , RAP
says: every solution of tan θ = √13 has
reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦
Since tan θ is positive, θ will end in Q1 or Q3.
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
√
P2 (− 3, 1)
√
P1 ( 3, 1)
Quadrant 2 Quadrant 1
-2
√
P3 (− 3, −1)
30◦
30◦
0
0
θ = 30◦
30◦
Quadrant 3 Quadrant 4
2
X
√
P4 ( 3, −1)
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 36
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 7: Find all solutions of tan θ = √13 .
Solution: Since tan θ = √13 = tan 30◦ , RAP
says: every solution of tan θ = √13 has
reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦
Since tan θ is positive, θ will end in Q1 or Q3.
In Q1, θ = θ = 30◦
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
√
P2 (− 3, 1)
√
P1 ( 3, 1)
Quadrant 2 Quadrant 1
θ = 30◦ =
◦
-2
√
P3 (− 3, −1)
30
30◦
◦
0
0
θ = 30
30◦
Quadrant 3 Quadrant 4
2
π
6
X
√
P4 ( 3, −1)
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 36
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 7: Find all solutions of tan θ = √13 .
Solution: Since tan θ = √13 = tan 30◦ , RAP
says: every solution of tan θ = √13 has
reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦
Since tan θ is positive, θ will end in Q1 or Q3.
In Q1, θ = θ = 30◦
In Q3, θ = 180◦ + 30◦ = 210◦ .
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
θ = 210◦ =
7π
6
√
P2 (− 3, 1)
√
P1 ( 3, 1)
Quadrant 2 Quadrant 1
θ = 30◦ =
◦
-2
√
P3 (− 3, −1)
30
30◦
◦
0
0
θ = 30
30◦
Quadrant 3 Quadrant 4
2
π
6
X
√
P4 ( 3, −1)
-2
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 36
7.1 Pythagorean identities
7.2 Sum and difference formulas
Example 7: Find all solutions of tan θ = √13 .
Solution: Since tan θ = √13 = tan 30◦ , RAP
says: every solution of tan θ = √13 has
reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦
Since tan θ is positive, θ will end in Q1 or Q3.
In Q1, θ = θ = 30◦
In Q3, θ = 180◦ + 30◦ = 210◦ .
Answer: θ = π6 + 2kπ and θ = 7π
6 + 2kπ,
where k is any integer.
All rights reserved. Copyright 2016 by Stanley Ocken
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Y
2
θ = 210◦ =
7π
6
√
P2 (− 3, 1)
√
P1 ( 3, 1)
Quadrant 2 Quadrant 1
θ = 30◦ =
◦
-2
√
P3 (− 3, −1)
30
30◦
◦
0
0
θ = 30
30◦
Quadrant 3 Quadrant 4
2
π
6
X
√
P4 ( 3, −1)
-2
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 36
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
Y
Example 7: Find all solutions of tan θ = √13 .
Solution: Since tan θ = √13 = tan 30◦ , RAP
says: every solution of tan θ = √13 has
reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦
Since tan θ is positive, θ will end in Q1 or Q3.
In Q1, θ = θ = 30◦
In Q3, θ = 180◦ + 30◦ = 210◦ .
Answer: θ = π6 + 2kπ and θ = 7π
6 + 2kπ,
where k is any integer.
Example 8: Find all solutions of tan θ =
All rights reserved. Copyright 2016 by Stanley Ocken
− √13
7.4 Trigonometric equations
2
θ = 210◦ =
7π
6
√
P2 (− 3, 1)
√
P1 ( 3, 1)
Quadrant 2 Quadrant 1
θ = 30◦ =
◦
30
30◦
-2
√
P3 (− 3, −1)
◦
0
0
θ = 30
30◦
Quadrant 3 Quadrant 4
2
π
6
X
√
P4 ( 3, −1)
-2
with
− π2
<θ<
π
2.
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 36
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
Y
Example 7: Find all solutions of tan θ = √13 .
Solution: Since tan θ = √13 = tan 30◦ , RAP
says: every solution of tan θ = √13 has
reference angle θ = 30◦ .
First draw the Quadrant 1 angle θ = 30◦ ,
which is its own reference angle.
Then draw the red ”X” to find all endlines of
angles with reference angle 30◦
Since tan θ is positive, θ will end in Q1 or Q3.
In Q1, θ = θ = 30◦
In Q3, θ = 180◦ + 30◦ = 210◦ .
Answer: θ = π6 + 2kπ and θ = 7π
6 + 2kπ,
where k is any integer.
Example 8: Find all solutions of tan θ =
− √13
7.4 Trigonometric equations
2
√
P2 (− 3, 1)
√
P1 ( 3, 1)
Quadrant 2 Quadrant 1
30◦
30◦
-2
0
0
θ = 30◦
30◦
2
X
θ = −30◦ = − π6
√
P3 (− 3, −1)
Quadrant 3 Quadrant 4
√
P4 ( 3, −1)
-2
with
− π2
<θ<
π
2.
Solution: To have a negative tangent, we need θ in Q2 or Q4. To have − π2 < θ < π2 we must have θ
in Q1 or Q4. Therefore we want a negative Q4 angle with reference angle 30◦ . The idea here is:
Any acute angle θ is the reference angle of the Q4 angle −θ. Look at Q4 in the diagram.
Answer: θ = − π6 .
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 36
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Example 9: Solve sin θ = + 12 for 0 ≤ θ ≤ 2π.
Solution: Since sin θ is positive, θ is in Q1 or Q2 and has reference angle
Answer: In Q1, θ = π6 . In Q2, θ = π − π6 = 5π
6 .
π
6.
Example 10: Solve sin θ = − 21 for 0 ≤ θ ≤ 2π.
Solution: Since sin θ is negative , θ is in Q3 or Q4 and has reference angle
π
11π
Answer: In Q3, θ = π + π6 = 7π
6 . In Q4,θ = 2π − 6 = 6 .
π
6.
Example 11: Find arcsin(− 12 ).
Solution: The definition of arcsin tells us that we require
• sin θ = − 12 , and so θ is in Q3 or Q4, where sine is negative, and
• − π2 ≤ θ ≤ π2 and so θ is in Q4 or Q1.
Therefore we need θ in Q4, and so it seems from Example 10 that we should pick
π
π
θ = 11π
6 . However, this choice does not satisfy − 2 ≤ θ ≤ 2 . No problem:
π
11π
6 − 2π = − 6 has the same sine and the same endline. Alternatively, as in Example
8, θ = − π6 is in Q4 and has reference angle π6 .
Answer: arcsin(− 12 ) = − π6 .
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 37
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Solving general trigonometric equations
Example 12 : Solve cos2 θ =
Solution: First solve cos2 θ =
|±
√1 |
2
=
√1
2
1
2
1
2
for 0 ≤ θ ≤ 2π.
q
for cos θ = ± 12 = ± √12 . Since
= cos 45◦ , RAP tells us that the reference angle of θ is 45◦ . Therefore
• In Q1, θ = π4
• In Q3, , θ = π + π4 = 5π
4
5π 7π
Answer: θ = π4 , 3π
,
,
4
4
4 .
• In Q2, θ = π − π4 = 3π
4
• In Q4, θ = 2π − π4 = 7π
4
Example 13: Solve sin2 θ = 34 .
q
√
Solution: Solve the stated equation for sin θ = ± 34 = ± 23 . Since
√
√
| ± 23 | = 23 = sin 60◦ , RAP tells us that the reference angle of θ is 60◦ = π3 .
• In Q2, θ = π − π3 = 2π
• In Q1, θ = π3
3
π
5π
• In Q3, , θ = π + π3 = 4π
•
In
Q4,
θ
=
2π
−
=
3
3
3
There is no restriction on θ, so we must add/subtract any whole number of circles
to/from these answers.
4π
5π
Answer : θ = π3 + 2kπ, 2π
3 + 2kπ,
3 + 2kπ,
3 + 2kπ, where k is any integer.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 38
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Example 14: Solve sin θ = 12 .
q
Solution: sin θ = ± 12 = ± √12 , which is just two basic trig equations: sin θ = √12 ;
sin θ = − √12 . This is similar to Example 12 above.
5π
7π
Answer : θ = π4 + 2kπ, 3π
4 + 2kπ,
4 + 2kπ,
4 + 2kπ, where k is any integer.
2
2
Example 15: Solve 2 sin θ + sin θ = 0.
Solution: Rewrite 2 sin2 θ + sin θ = 0 as (sin θ)(2 sin θ + 1) = 0. Solved by setting
each factor to zero: sin θ = 0 or 2 sin θ + 1 = 0.
Therefore 2 sin2 θ + sin θ = 0 can be rewritten as sin θ = 0 ; sin θ = − 12
x
1
To solve sin θ = 0, just
draw the sine graph:
0
−2π
−π
0
π
2πθ
-1
x = sin(θ)
Clearly sin θ = 0 precisely when θ = 0, ±π, ±2π, ±3π, ..., ±kπ.
We solved sin θ = − 21 back in Example 10, and include its solutions in our
11π
Answer: θ = kπ; 7π
6 + 2kπ;
6 + 2kπ, where k is any integer.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 39
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Example 16: Solve sin2 θ − 5 sin θ = 6.
Solution: You may like using the abbreviation u = sin θ. Then sin2 θ − 5 sin θ = 6
becomes u2 − 5u = 6 ⇒ u2 − 5u − 6 = 0 ⇒ (u − 6)(u + 1) = 0 ⇔ u = 6, u = −1.
Replace u by sin θ to obtain:
sin2 θ − 5 sin θ = 6 can be rewritten as sin θ = −1 ; sin θ = 6.
Of course, sin θ = 6 has no solutions, since −1 ≤ sin θ ≤ 1 for all angles θ.
x
1
To solve sin θ = −1, just
draw the sine graph:
0
−4π
−2π
−π − π
2
0
π
2
π
3π
2
2π
4πθ
-1
x = sin(θ)
− π2 , 3π
2
Therefore sin θ = −1 at θ =
and every 2π radians before or after these
solutions. In other words sin θ = −1 at θ = − π2 + 2kπ, where k is any integer.
Answer : θ = − π2 + 2kπ, where k is any integer.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 40
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Quiz Review for 7.4
Example 1: Rewrite the equation sin2 θ =
1
2
as two basic trig equations.
2
Example 2: Rewrite the equation 2 sin θ + sin θ = 0 as basic trig equations.
Example 3: Rewrite the equation sin2 θ − 5 sin θ = 6 as basic trig equations.
Example 4: Rewrite the equation 2 cos2 θ − 2 sin2 θ = 1 as basic trig equations.
Example 5: Find all solutions of sin θ = − 12 with 0 ≤ θ ≤ 2π.
Example 6: Find all solutions of sin θ = − 12 .
Example 7: Find all solutions of tan θ =
√1 .
3
Example 8: Find all solutions of tan θ = − √13 with − π2 < θ < π2 .
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 41
7.1 Pythagorean identities
7.2 Sum and difference formulas
7.3 Double and half-angle formulas
7.4 Trigonometric equations
Example 9: Solve sin θ = + 12 for 0 ≤ θ ≤ 2π.
Example 10: Solve sin θ = − 21 for 0 ≤ θ ≤ 2π.
Example 11: Find arcsin(− 21 ).
Example 12: Solve cos2 θ =
Example 13: Solve sin2 θ =
Example 14: Solve sin2 θ =
1
2 for
3
4.
1
2.
0 ≤ θ ≤ 2π.
Example 15: Solve 2 sin2 θ + sin θ = 0.
Example 16: Solve sin2 θ − 5 sin θ = 6.
All rights reserved. Copyright 2016 by Stanley Ocken
M19500 Precalculus Chapter 7.4 Trigonometric equations
11/27/16 Frame 42