Download Trigonometric Identities

Chapter 3:
Trigonometric Identities
Chapter 3 Overview
Two important algebraic aspects of trigonometric functions must be considered:
how functions interact with each other and how they interact with their arguments
⎛π
⎞
π
(the expression inside the function – e.g., ( x −1) in sin ⎜ ( x −1)⎟ ).
4
⎝4
⎠
First, trigonometric functions are inextricably linked because of how they were
defined––via the coordinates of points on the terminal side of an angle. Since the
coordinates are related by the Pythagorean theorem, there are some relationships
that will be true regardless of the angle. These are referred to as identities and will
act as formulas. THESE MUST BE MEMORIZED.
Second, since the trigonometric function is the operation and no multiplication is
present, rules on how the functions work with their arguments will be provided.
Without multiplication, the associative, commutative, and distributive rules do not
hold. In particular, sin ( a + b) ≠ sin a + sin b and sin2x ≠ 2sin x . New rules for the
“algebra” of trigonometric operations will also be developed.
95
3-1: Quotient, Reciprocal, and Pythagorean Identities
REMEMBER:
sinθ =
y
r
cosθ =
x
r
tanθ =
y
x
cscθ =
r
y
secθ =
r
x
cot θ =
x
y
Based on these definitions, here are the basic identities.
The Reciprocal Identities:
cscθ =
1
sinθ
secθ =
1
cosθ
cotθ =
1
tanθ
The Quotient Identities:
tanθ =
sinθ
cosθ
cotθ =
cosθ
sinθ
tanθ =
secθ
cscθ
cot θ =
cscθ
secθ
The Pythagorean Identities*:
tan 2 θ + 1 = sec2 θ
sin 2 θ + cos2 θ = 1
*Recall: sin 2 θ + cos 2 θ = ( sin θ ) + ( cosθ )
2
2
96
1+ cot 2 θ = csc2 θ
The Pythagorean identities have alternative versions as well:
sin 2 θ + cos2 θ = 1
1− cos2 θ = sin 2 θ
1− sin 2 θ = cos2 θ
tan 2 θ +1= sec2 θ
sec2 θ −1= tan 2 θ
sec2 θ − tan 2 θ = 1
1+ cot 2 θ = csc2 θ
csc2 θ −1= cot 2 θ
csc2 θ − cot 2 θ = 1
These alternative forms are very useful because they are “difference of squares”
binomials that can be factored. For example,
1− sin 2 θ = (1− sinθ )(1+ sinθ )
LEARNING OUTCOME
Prove basic trigonometric identities.
The proofs of the identities are algebraic in nature, meaning that the use of
multiplication, addition, and common denominators will cause one side of the
equation to simplify to the other. Similar to proofs in Geometry where one can
work both top down and bottom up, in these proofs the two sides can simplify to
the same thing.
EX 1 Prove csc x tan x cos x = 1
csc x tan x cos x =
1 sin x
i
i cos x =
sin x cos x
1
Notice that the answer is the process, not the final line; the final line was given.
97
EX 2 Prove cot xsec x + csc x = 2csc x
This one can be done quickly if the substitution cot x =
csc x
is used.
sec x
This would give:
csc x
sec x + csc x
sec x
= csc x + csc x
= 2csc x
cot xsec x + csc x =
Another, longer way, would be to turn the whole problem into
sin x and cos x .
cos x 1
1
+
sin x cos x sin x
1
1
=
+
sin x sin x
2
=
sin x
= 2csc x
cot xsec x + csc x =
EX 3 Prove
1
1
+
= 2csc A
csc A+ cot A csc A− cot A
⎛ csc A− cot A ⎞ ⎛ csc A+ cot A ⎞
1
1
1
1
+
=
+⎜
⎜
⎟
⎟
csc A+ cot A csc A− cot A csc A+ cot A ⎝ csc A− cot A ⎠ ⎝ csc A+ cot A ⎠ csc A− cot A
csc A− cot A+ csc A+ cot A
=
(csc A+ cot A)(csc A− cot A)
2csc A
A− cot 2 A
2csc A
=
1
= 2csc A
=
csc2
98
3-1 Free Response Homework
Prove the following identities.
1.
cos2 x + tan 2 xcos2 x = 1
2.
2cosθ =
3.
4 + ( tan σ − cot σ ) = sec2 σ + csc2 σ
4.
(cosθ )(cosθ ) − (−sinθ )(sinθ ) = sec2 θ
5.
(sin x )(−sin x ) − (cos x )(cos x ) = −csc2 x
6.
−
7.
−
8.
(cos a + sin a)(cos a + sin a) − 1 = sin a
9.
tanW (cotW cosW + sinW ) = secW
10.
sec x sin x
−
= cot x
sin x cos x
11.
sin x
1− cos x
+
= 2csc x
1− cos x
sin x
12.
cos2 θ =
13.
tan b + cot b =
14.
cos x
= cot xcsc x + cot 2 x
1− cos x
15.
1+ sin x
= 2sec2 x + 2sec x tan x −1
1− sin x
cosθ tanθ + sinθ
tanθ
2
cos2 θ
sin 2 x
(−sinθ ) = secθ tanθ
cos2 θ
cosϕ
= −csc ϕ cot ϕ
sin 2 ϕ
2cos a
cscb
cosb
99
cotθ cscθ tanθ − sinθ
cscθ
16.
State some strategies used for approaching the previous identities.
3-1 Multiple Choice Homework
1.
(a)
(b)
(c)
(d)
(e)
2.
(a)
(b)
(c)
(d)
(e)
3.
Where defined,
sin x
− sin x
csc x
− csc x
− cos x
If the ratio of sin x to cos x is 1:2, then the ratio of tan x to cot x is
1:4
1:2
1:1
2:1
4:1
If ( sec x ) ( tan x ) < 0 , which of the following must be true?
I. tan x < 0
(a)
(b)
(c)
(d)
(e)
4.
csc x − 1
=
sin x − 1
II. csc x cot x < 0 III. x is in the third or fourth quadrant
I only
II only
III only
II and III
I and II
Simplify
(a) sin∂
1
( tan∂+ cot ∂)
csc∂
(b) sec∂
(c) tan∂
(d) 1
100
(e) cos∂
5.
If 90° < α <180° and 180° < β < 270° , then which of the following must be
FALSE?
(a)
(b)
(c)
(d)
(e)
6.
(a)
(b)
(c)
(d)
(e)
tan α = tan β
cosα = cos β
sin α = tan β
tan α = sin β
sec β = secα
To which of the following is
(sin y − cos y)
cos y
sec y
sec y − 2sin y
sec y − 2 tan y
−2sin y
−2 tan y
101
2
equal?
3-2: Trigonometric Identities and Factoring
Hopefully the answer to #16 in the previous homework looked something like this:
Techniques for Approaching Trigonometric Proofs:
1.
2.
3.
Get down to two or less trig functions. These might be Sin and Cos,
but be aware of how Sec and Tan, or Csc and Cot, work together.
Do Algebra:
a.
Common Denominators
b.
Adding “like terms”
c.
Distribute and/or FOIL
Squares indicate possible Pythagorean identities
Another technique under the “Do Algebra” heading would be Factoring.
LEARNING OUTCOME
Prove trigonometric identities that involve factoring.
It might be best to go back and review factoring at this point. There are three basic
modes of factoring:
1.
2.
3.
Factoring by rules
Factoring by guess and check
Factoring by splitting the middle term
I.
Factoring by Rules
The first technique for factoring is called monomial factoring. It is the most
common and most often forgotten. Monomial factoring is the reverse of
distribution, and involves pulling out factors common to each term.
102
Distribution and Monomial Factoring
a ( b ± c ) = ab ± ac
ab ± ac = a ( b ± c )
EX 1 Factor x 2 y + xy 2
x 2 y + xy 2 =
xy ( x + y )
What would this look like in a trigonometric identity proof?
EX 2 Use factoring to prove cos2 x + tan 2 xcos2 x = 1
cos2 x + tan 2 xcos2 x =
cos2 x 1+ tan 2 x
(
)
cos2 x i sec2 x
1
cos2 x i
cos2 x
1
Factoring Rules for the Sum or Difference of Equal Powers:
1.*
2.
x 2 − y 2 = ( x + y )( x − y )
x 3 + y 3 = ( x + y )( x 2 − xy + y 2 )
3.** x 3 − y 3 = ( x − y )( x 2 + xy + y 2 )
4.** x 4 − y 4 = ( x 2 + y 2 )( x 2 − y 2 )
*NB. x 2 + y 2 does not factor when working with real numbers. The rule for
x 2 + y 2 involves imaginary numbers, which are not of interested at this point.
**The trinomials in rules 4 and 5 are not factorable in the real numbers either.
103
EX 3 Simplify
u3 − v 3
(u − v )
3
u3 − v 3
3 =
u
−
v
( )
(u − v )(u2 + uv + v 2 ) =
(u − v )(u − v )(u − v )
u 2 + uv + v 2
(u − v )(u − v )
EX 4 Prove cot 4 w − csc 4 w = 1− 2 csc 2 w
cot 4 w − csc4 w =
(cot 2 w − csc2 w)(cot 2 w + csc2 w) =
( −1)(cot 2 w + csc2 w) =
(
)
− csc 2 w −1+ csc 2 w =
1− 2 csc 2 w
EX 5 Prove
cos 3 x − sin 3 x
= 1− cos xsin x
cos x − sin x
cos 3 x − sin 3 x
=
cos x − sin x
( cos x − sin x ) cos2 x + cos xsin x + sin2 x
=
cos x − sin x
cos2 x + cos xsin x + sin 2 x =
1− cos xsin x
(
)
104
II.
Factoring by “Guess and Check.”
The second technique is trinomial factoring, and is a guess and check process. It
involves finding numbers that are factors of the first and third terms of a trinomial
and checking the FOILing.
EX 6 Factor y 2 − 4 y − 5
If y 2 − 4 y − 5 is factorable, the first terms must multiply to y 2 . So,
y2 − 4 y − 5 = ( y −
)( y + )
The two numbers must multiply to the last number (–5) and add to the
middle number (–4). In this case, the numbers must be –5 and +1. So,
y 2 − 4 y − 5 = ( y −5)( y +1)
EX 7 Factor 4x 2 −10x − 6
This problem is more complicated than the last because there is more than
one possibility for the factors of the first and the third terms.
4x 2 −10x − 6 = ( 4x
)( x )
or 4x 2 −10x − 6 = ( 2x
)( 2x )
The second numbers might be 2 and –3, –2 and 3, 1 and –6, or –1 and 6.
The guess and check process takes some time, but ultimately yields
4x 2 − 10x − 6 = ( 4x + 2 ) ( x − 3)
= 2 ( 2x + 1) ( x − 3)
105
III.
Factoring by splitting the middle term and/or by grouping.
EX 7 Factor 4x 2 −10x − 6 (EX 4 again)
In order to factor by grouping, –10 needs to be split into two terms. How?
The proper split will be two factors of the product of first and last
coefficients and that add to –10.
4 times –6 is –24. The factors of 24 are
±1 and
±2 and
±3 and
±4 and
 24
12
8
6
The pair that adds to –10 are 2 and −12 .
So, 4x 2 −10x − 6 = 4x 2 + 2x −12x − 6
= 2x ( 2x +1) − 6 ( 2x +1)
= ( 2x − 6)( 2x +1)
= 2 ( x − 3)( 2x +1)
EX 8 Prove
3sin 2 θ − 8sinθ + 4 3sin 2 θ − 5sinθ − 2 3sinθ − 2
÷
=
9sin 2 θ − 4
9sin 2 θ − 3sinθ − 2 3sinθ + 2
3sin 2 θ − 8sinθ + 4 3sin 2 θ − 5sinθ − 2 3sin 2 θ − 8sinθ + 4 9sin 2 θ − 3sinθ − 2
÷
=
⋅
9sin 2 θ − 4
9sin 2 θ − 3sinθ − 2
9sin 2 θ − 4
3sin 2 θ − 5sinθ − 2
=
(3sinθ − 2)(sinθ − 2) ⋅ (3sinθ +1)(3sinθ − 2)
(3sinθ − 2)(3sinθ + 2) (3sinθ +1)(sinθ − 2)
=
3sinθ − 2
3sinθ + 2
106
EX 9 Prove
5sec2 θ −14tanθ + 3 5tanθ − 4
=
5sec2 θ − 8tanθ − 9 5tanθ + 2
5sec2 θ −14tanθ + 3
=
5sec2 θ − 8tanθ − 9
5( tan 2 θ +1) −14tanθ + 3
=
5( tan 2 θ +1) − 8tanθ − 9
5tan 2 θ + 5−14tanθ + 3
=
5tan 2 θ + 5− 8tanθ − 9
5tan 2 θ −14tanθ + 8
=
5tan 2 θ − 8tanθ − 4
(5tanθ − 4)( tanθ − 2) =
(5tanθ + 2)( tanθ − 2)
5tanθ − 4
5tanθ + 2
107
3-2 Free Response Homework
Factor and simplify the following expressions.
1.
x 2 − 6xy + 8y2
2.
3x 3 y +10x 2 y2 − 8xy 3
3.
3x − 5x 2 − 2x 3
4.
x 3 − x 2 y + xy2 − y 3
5.
x 4 − 3x 2 + 2
6.
k4 − r4
2
k2 + r2 (k + r)
7.
x3 + x2 + x + 1
x4 − 1
8.
Prove the following identities.
9.
sin 2 x + cot 2 xsin 2 x = 1
10.
sec 4 β − tan 4 β = 1+ 2 tan 2 β
11.
cos2 w + 2cosw +1 cosw +1
=
cos2 w − 3cosw − 4 cosw − 4
12.
cos3 A− sec3 A
= −sin 2 A+ tan 2 A + 3
cos A− sec A
13.
tan 2 w − secw − 5 secw − 3
=
tan 2 w + 3secw + 3 secw +1
14.
2sin 2 w + 5cosw +1 cosw − 3
=
6sin 2 w + 5cosw − 2 3cosw − 4
15.
3sec2 t − 8 tant +1 3tant − 2
=
tant +1
sec2 t − tant − 3
108
(
)
x 3 + y3
( x + y )3
16.
2csc2 y − 7cot y − 6
cot y − 4
=
2
6csc y − 5cot y −10 3cot y − 4
Simplify the following expressions.
17.
x 2 + 5x
x3
x
÷
i
2
x + 6x + 5 3x + 3 x +1
19.
54 − 2x
6
÷ 2
4
x − 81 x + 9
21.
x 2 + 3wx x 2 − 4wx + 3w 2 x + 3w
i
÷
3w − x
x+w
w2 − x 2
22.
x 2 + 4xy + 3y 2
x+y
−1
i
x
+
2y
÷
(
)
x 2 + 5xy + 6y 2
x 2 + 4xy + 4y 2
3
18.
x 4 −16
1
2 ÷
( x + 2) x2 + 4
20.
(r
(r
(
2
2
)
− rx )
+ rx
2
2
)
−1
x + r r3 + x3
÷
÷
x − r r3 − x3
Prove the following identities.
23.
cot B + 1
1+ tan3 B
=
cot B
− tan B + sec2 B
25.
(8 + cos x ) i (cos
( cos x − 2 ) ( cos
26.
(sec x + 3sec x −10) i (sec x + 2sec x − 3) = sec x + 5
(sec x − 7sec x + 6) (sec x + sec x − 6) sec x − 6
27.
cot 2 x − 4csc x − 11 csc x − 6
=
csc x − 2
cot 2 x − 3
28.
csc6 x − cot 6 x = 1+ 3csc2 xcot 2 x
3
2
2
2
2
24.
) = cos
x − 2cos x + 4 )
x − 4 cos x + 4
2
2
109
2
sin 2 B − sin B + 1 1+ sin3 B
=
1− sin B
1− sin 2 B
x−4
(
29.
− 9 + 3sin x + sin 2 x
54 − 2sin 3 x
6
÷
=
sin 4 x − 81 sin 2 x + 9
3(sin x + 3)
30.
(sin
(sin
2
2
)
x − sin x cos x )
x + sin x cos x
2
2
÷
sin x + cos x cos 3 x + sin 3 x
sin 2 x + sin x cos x + cos2 x
÷
=
−
sin x − cos x cos 3 x − sin 3 x
sin 2 x − sin x cos x + cos2 x
3-2 Multiple Choice Homework
1.
(a)
(b)
(c)
(d)
(e)
2.
(a)
(b)
(c)
(d)
(e)
)
1 ⎞
⎛ 1
cot x ⎜
+
=
⎝ sec x csc x ⎟⎠
cot x
sec x csc x
cot x
sec x + csc x
2cot x
sec x + csc x
cot xsec x + cot x csc x
sec x csc x
1
cot xsec x + cot x csc x
Divide x 3 − 2x 2 + 3x − 18 by x − 3 .
x2 + x + 6
x2 − x − 6
x 2 − 5x + 18
x 2 + 5x − 18
x 2 − 5x − 18
110
3.
(
) (
)
6 tan 2 x −1 2 tan 2 x + 2 tan x +1
÷
Find
tan x + 4
tan 2 x + 4 tan x
2 tan x ( tan x −1)
tan x +1
(b) 3tan x ( tan x −1)
(a)
(c)
(
)(
tan x ( tan x + 2 )
3tan x − 3
(d)
tan x +1
3tan x tan x − 1
(e)
tan x + 1
(
4.
)
12 tan 2 x −1 tan 2 x + 2 tan x +1
2
)
Simplify
cos2 x − 3cos x
2cos x
÷
2
cos x − 5cos x + 6 ( cos x − 2 )2
cos x ( cos x − 3)
2
cos x
(b)
2
cos x − 2
(c)
2
cos x ( cos x − 3)
(d)
cos x − 2
(e) cos x − 2
(a)
5.
(a)
(b)
5x 3 y + 20x 2 y 2 + 20xy 3
Simplify
5xy
( x + 2)
( x + 2 y)
2
2
(c) x 2 + y 2
(d) x 2 + 4 y 2
(e) x 2 − 4xy + 4 y 2
111
3-3: Composite Argument and Even/Odd Rules
As stated in the chapter overview, sin ( a + b) ≠ sin a + sin b . This is easily
demonstrated with a pair of the special angles. If a = 30° and b = 60° , then
1
3 1+ 3
=
. Clearly these
sin (30° + 60°) = sin90° = 1 . But sin30° + sin60° = +
2 2
2
are not equal. So how does a trigonometric operation apply to its argument when
the argument is a composite (i.e. ( a + b) )?
The Composite Argument Identities:
cos ( A− B ) = cos Acos B + sin Asin B
cos ( A+ B ) = cos Acos B − sin Asin B
sin ( A− B ) = sin Acos B − cos Asin B
sin ( A+ B ) = sin Acos B + cos Asin B
tan A− tan B
1+ tan Atan B
tan A+ tan B
tan ( A+ B ) =
1− tan Atan B
tan ( A− B ) =
112
Some functions are called “odd” functions because they act like odd powers––they
preserve negative signs: ( −x ) = − ( x ) . Others are called “even” functions because
3
3
they act like even powers and cancel negative signs: ( −x ) = ( x ) .
2
2
Odd Functions:
Even Functions:
sin ( − A) = − sin A
cos ( − A) = cos A
csc ( − A) = − csc A
sec ( − A) = sec A
tan ( − A) = − tan A
cot ( − A) = − cot A
LEARNING OUTCOMES
Find exact trigonometric values for composite arguments.
Solve equations involving composite argument rules.
Prove identities involving composite rules.
There are basically three tasks that can be completed using these rules:
1.
2.
3.
Find exact values,
Solve equations, and
Prove other identities.
113
5
in Quad II, and (6, − 8) is on the terminal side of α , find
13
the exact value of sin (α + β ) .
EX 1 Given sin β =
5
in Quad II, then by the Pythagorean identity sin 2 θ + cos2 θ = 1 ,
13
12
cos β = − .
13
If sin β =
If (6, − 8) is on the terminal side of α , then sin α = −
cosα =
6 3
= .
10 5
sin (α + β ) = sin α cos β + cosα sin β
⎛ 4 ⎞ ⎛ −12 ⎞ ⎛ 3⎞ ⎛ 5 ⎞
= ⎜− ⎟⎜
⎟+
⎜ ⎟
⎝ 5 ⎠ ⎝ 13 ⎠ ⎜⎝ 5 ⎟⎠ ⎝ 13⎠
63
=
65
114
8
4
= − and
10
5
π
π
3
EX 2 Solve for x ∈{All Reals} if cos xcos − sin xsin =
.
3
3 2
π
π
cos xcos − sin xsin =
3
3
⎛
π⎞
cos ⎜ x + ⎟ =
3⎠
⎝
3
2
3
2
π
3
= cos –1
3
2
⎧ π
⎪
± 2π n
π ⎪ 6
x+ = ⎨
3 ⎪ π
− ± 2π n
⎪⎩ 6
x+
⎧ π
⎪ − ± 2π n
⎪
x=⎨ 6
⎪ − π ± 2π n
⎪ 2
⎩
The final answer here is known as the general solution. It is a condensed way to
express and infinite number of coterminal answers.
⎧ π
⎪ − ± 2π n
⎪
x=⎨ 6
⎪ − π ± 2π n
⎪⎩ 2
⎫ ⎧
25π 13π π
⎪ ⎪ ... −
,−
,− ,
⎪ ⎪
6
6
6
⎬= ⎨
9π 5π π
⎪ ⎪
... − , − ,− ,
⎪⎭ ⎪⎩
2
2 2
11π 23π
,
...
6
6
3π 7π
,
...
2 2
⎫
⎪
⎪
⎬
⎪
⎪⎭
Vocabulary:
1.
General Solution – all (the infinite number of) coterminal solutions
2.
Particular Solution – the specific solutions in a given domain
115
Sometimes the request is for the particular solution.
π
π 1
EX 3 Solve for x ∈⎡⎣ 0, 2π ⎤⎦ if sin xcos − cos xsin = .
3
3 2
π
π 1
sin xcos − cos xsin =
3
3 2
⎛
π⎞ 1
sin ⎜ x − ⎟ =
3⎠ 2
⎝
x−
⎛ 1⎞
π
= sin –1 ⎜ ⎟
3
⎝ 2⎠
⎧
⎪
π ⎪
x− = ⎨
3 ⎪
⎪⎩
⎧
⎪
⎪
x=⎨
⎪
⎪⎩
π
± 2π n
6
5π
± 2π n
6
π
± 2π n
2
7π
± 2π n
6
Of all the answers summarized by the general solution above, only two fall
within the domain x ∈⎡⎣ 0, 2π ⎤⎦ :
⎧ π 7π ⎫
x=⎨ ,
⎬
⎩2 6 ⎭
116
EX 4 Prove cos ( x + 30° ) − sin ( x + 60° ) = − sin x
cos ( x + 30°) − sin ( x + 60°) = cos xcos30° − sin xsin30° − (sin xcos60° + cos xsin60°)
3
1
1
3
cos x − sin x − sin x − cos x
2
2
2
2
= −sin x
=
117
3-3 Free Response Homework
1.
Given tan α =
15
in Quad III, and (3, − 5) is on the terminal side of µ , find
8
the exact values of
a)
b)
c)
sin ( µ − α )
d)
cos ( µ + α )
e)
tan (α − µ )
f)
sec (α − µ )
csc (α + µ )
cot (α + µ )
5
in Quad III, and (12, − 5) is on the terminal side of ω ,
4
find the exact values of
2.
Given sec λ = −
a)
b)
c)
sin ( λ + ω )
d)
cos ( λ − ω )
e)
tan (ω + λ )
f)
sec (ω + λ )
csc (ω − λ )
cot ( λ − ω )
Find the general solutions to the following equations.
1
for x ∈{All Reals}
2
3.
sin3θ cos12° − cos3θ sin12° =
4.
sin xsin60° − cos xcos60° =
5.
sin φ cos15° = cos φ sin15° for x ∈{All Reals}
6.
tan3x + tan x = 1− tan3x tan x for x ∈⎡⎣0, 2π ⎤⎦
7.
⎛
⎛
π⎞
π⎞
sec ⎜ x − ⎟ = 2 + 2sec ⎜ x − ⎟ for x ∈⎡⎣ −π , π ⎤⎦
4⎠
4⎠
⎝
⎝
8.
(cos Acos B − sin Asin B) + (sin Acos B + cos Asin B)
1
for x ∈⎡⎣0°, 360° ⎤⎦
2
2
118
2
= 1 for x ∈{All Reals}
Prove the following identities.
9.
sin ( A+ B ) + sin ( A− B )
= tan Acot B
sin ( A+ B ) − sin ( A− B )
10.
cos ( a + b) cosb + sin ( a + b) sinb = cosa
11.
sin Acos A+ cos Asin A
2tan A
=
cos Acos A− sin Asin A 1− tan 2 A
12.
sin Acos B + cos Asin B tan A+ tan B
=
cos Acos B − sin Asin B 1− tan Atan B
13.
cot Acot B −1
= cot ( A+ B )
cot A+ cot B
3-3 Multiple Choice Homework
1.
Which of the following is equivalent to sin ( A + 30°) + cos ( A + 60°) for what
values of A?
(a) sin A
(b) cos A
(c)
3sin A + cos A
(d)
3sin A
(e)
3 cos A
2.
Which of the following is equivalent to sin (α + β ) + sin (α − β ) ?
(a) sin2α
(b) sin α 2 − β 2
(
(c) 2sin α sin β
(d) 2sin α cos β
(e) 2cosα sin β
)
119
3.
(a)
(b)
(c)
(d)
(e)
4.
(a)
(b)
(c)
(d)
(e)
5.
3
1
If sin A = , 90° ≤ A ≤180°, cos B = , and 270° ≤ B ≤ 360°, sin ( A + B ) =
5
3
–0.333
0.733
–0.832
0.954
–0.554
tan ( x1 − x2 ) =
tan x1 − tan x2
1− tan x1 tan x2
tan x1 − tan x2
−
1+ tan x1 tan x2
tan x1 + tan x2
1− tan x1 tan x2
tan x1 + tan x2
1+ tan x1 tan x2
None of these
cos
7π
π
7π
π
cos − sin sin =
8
8
8
8
(a) –1
(b) 1
(c) 0
1
(d)
2
1
(e) −
2
120
3-4: Double Angle Rules
As stated in the chapter overview sin2x ≠ 2sin x , and, as with the composite rules,
this is easily demonstrated with a pair of the special angles. If a = 30° , then
⎛ 1⎞
3
sin ( 2 i 30°) = sin60° =
. But 2sin30° = 2 ⎜ ⎟ = 1 . Clearly these are not equal.
2
⎝ 2⎠
So if sin2x ≠ 2sin x , what does sin2x equal?
sin2x = sin ( x + x )
= sin xcos x + cos xsin x
= 2sin xcos x
The Double Angle Argument Identities:
sin ( 2 A) = 2 sin A cos A
cos ( 2 A) = cos2 A− sin 2 A
= 1− 2sin 2 A
= 2cos2 A−1
tan ( 2 A) =
2tan A
1− tan 2 A
or
tan ( 2 A) =
sin ( 2 A)
cos ( 2 A)
LEARNING OUTCOMES
Find exact trigonometric values for double angle trigonometric functions.
Solve equations involving double angle rules.
Prove identities involving double angle rules.
121
The same three types of problems will be done with the double angle rules as were
done with the composite rules.
EX 1 Given sin β =
tan ( 2β ) .
If sin β =
5
in Quad II, find the exact value of sin ( 2β ) , cos ( 2β ) , and
13
5
12
in Quad II, then by the Pythagorean identity, cos β = − .
13
13
sin ( 2β ) = 2sin β cos β
⎛ 5 ⎞ ⎛ 12 ⎞
−
⎝ 13⎟⎠ ⎜⎝ 13 ⎟⎠
= 2⎜
=−
120
169
cos ( 2β ) = cos2 β − sin 2 β
2
⎛ 12 ⎞
⎛ 5⎞
= ⎜− ⎟ −⎜ ⎟
⎝ 13 ⎠
⎝ 13⎠
2
144 25
−
169 169
119
=
169
=
tan ( 2β ) =
sin ( 2β )
cos ( 2β )
120
= 169
119
169
120
=−
119
−
122
EX 2 Solve for x if 4sin xcos x = 2 .
4sin xcos x = 2
2
2sin xcos x =
2
2
sin ( 2x ) =
2
2x = sin –1
⎧
⎪
⎪
2x = ⎨
⎪
⎪⎩
⎧
⎪
⎪
x=⎨
⎪
⎪⎩
EX 3 Prove tan x =
2
2
π
± 2π n
4
3π
± 2π n
4
π
±πn
8
3π
±πn
8
sin 2x
1+ cos2x
It would be appropriate in this problem to work both sides of the equation
toward the same goal.
sin2x
1+ cos2x
sin x
2sin xcos x
=
cos x 1+ 2cos2 x −1
tan x =
(
2sin xcos x
2cos2 x
sin x
=
cos x
=
123
)
3-4 Free Response Homework
1.
Given sin A =
1
in Quad II, and ( −7, 24) is on the terminal side of B, find
3
the exact values of
a)
b)
c)
sin ( 2 A)
d)
cos ( 2B )
e)
tan ( 2 A)
f)
sec ( 2B )
csc ( 2 A)
cot ( 2B )
Find the general solutions to the following equations. Use exact values wherever
possible.
2.
4cosθ sinθ = 2 for θ ∈{All Reals}
3.
cos2 x − sin 2 x − 1 = 2cos ( 2x ) for x ∈ ⎡⎣0, 2π ⎤⎦
4.
2sin ( 2x − 30°) cos ( 2x − 30°) = −
5.
csc2x +
6.
2tan x
= 1 for x ∈⎡⎣ −2π , 2π ⎤⎦
1− tan 2 x
7.
1− tan 2 x
= 3 for x ∈{All Reals}
2tan x
3
for x ∈{All Reals}
2
1
= 6 for x ∈{All Reals}
sin xcos x
Prove the following identities.
1− tan 2 B
1+ tan 2 B
8.
cos2B =
9.
cos2β = cos4 β − sin 4 β
124
10.
cot A + tan A = 2csc ( 2 A)
11.
cot 2 φ −1= cos2φ cot 2 φ + cos2φ
12.
⎛ 1− tan 2 x cos2 x − sin 2 x ⎞
2
⎜ 2tan x + 2sin xcos x ⎟ = 4csc 2x − 4
⎝
⎠
13.
δ
(1+ tanδ ) tan2δ = 1−2tan
tan δ
2
3-4 Multiple Choice Homework
1.
Which of the following is NOT equal to
(a) cos x − sin x tan x
2sin x
(b) sec x −
cot x
(c) 2cos x − sec x
(d) sec x − 2sin 2 x tan x
cos2x
(e) All are equal to
cos x
2.
(a)
(b)
(c)
(d)
(e)
Which expression equals cot 2x ?
2cot x
1+ cot 2 x
2 cot x
1− cot 2 x
2sin x
1− tan 2 x
2 tan x
1− tan 2 x
1
1
cot x − tan x
2
2
125
cos2x
?
cos x
3.
If cos x =
4
3π
≤ x ≤ 2π , then tan 2x =
and
5
2
7
24
24
−
25
3
−
4
24
−
7
7
25
(a) −
(b)
(c)
(d)
(e)
4.
(a)
(b)
(c)
(d)
(e)
5.
(a)
(b)
(c)
(d)
(e)
For what value between 0° and 360° does cos2x = 2 cos x ?
68.5° or 291.5°
68.5° only
103.9° or 256.1°
90° or 270°
111.5° or 248.5°
If
3sin 2θ
1
= and 0° ≤ θ ≤180° , then θ =
1− cos2θ 2
0°
0° or 180°
80.5°
0° or 80.5°
99.5°
126
3
1
If sin A = , 90° < A <180°, cos B = , and 270° < B < 360° , the value of
5
3
tan ( A + B ) is
6.
(a)
(b)
(c)
(d)
(e)
0.30
3.19
3.34
1.05
0.31
127
3-5: Half Angle Rules
Just as there are Double Angle Identities, there are half angle formulas.
The Half Angle Argument Identities:
⎛1 ⎞
1
sin ⎜ A⎟ = ± (1− cos A)
2
⎝2 ⎠
⎛1 ⎞
1
cos ⎜ A⎟ = ± (1+ cos A)
2
⎝2 ⎠
⎛1 ⎞
1− cos A
tan ⎜ A⎟ = ±
1+ cos A
⎝2 ⎠
⎛1 ⎞
sin A
tan ⎜ A⎟ =
⎝ 2 ⎠ 1+ cos A
⎛ 1 ⎞ 1− cos A
tan ⎜ A⎟ =
sin A
⎝2 ⎠
Note the ± on three of the formulas. This does really mean “plus OR minus,”
not “plus AND minus” as usually meant in mathematics. That is, in these
formulas, one OR the other is correct, but not both. Determine the sign from the
1
size and quadrant of A .
2
The same three types of problems will be done with the half angle rules as were
done with the double angle and composite rules.
LEARNING OUTCOMES
Find exact trigonometric values for half angles.
Solve equations involving half angle rules.
Prove identities involving half angle rules.
128
EX 1 Given sin β = −
⎛1 ⎞
5
and 540° < β < 630° , find the exact value of sin ⎜ β ⎟ ,
13
⎝2 ⎠
⎛1 ⎞
⎛1 ⎞
cos ⎜ β ⎟ , and tan ⎜ β ⎟ .
⎝2 ⎠
⎝2 ⎠
If sin β = −
5
12
and 540° < β < 630° , then cos β = − .
13
13
1
1
Also, since 540° < β < 630° , then 270° < β < 315° . This means that β is
2
2
in QIV, and the cosine will be positive and the sine negative.
⎛1 ⎞
1
sin ⎜ β ⎟ = ± (1− cos β )
2
⎝2 ⎠
⎛1 ⎞
1
cos ⎜ β ⎟ = ± (1+ cos β )
2
⎝2 ⎠
=−
1 ⎛ ⎛ 12 ⎞ ⎞
1− −
2 ⎜⎝ ⎜⎝ 13 ⎟⎠ ⎟⎠
=+
=−
1 ⎛ 25 ⎞
2 ⎜⎝ 13 ⎟⎠
=
1⎛ 1 ⎞
2 ⎜⎝ 13⎟⎠
=
1
26
=
−5
26
⎛1 ⎞
sin β
tan ⎜ β ⎟ =
⎝ 2 ⎠ 1+ cos β
5
−
13
=
⎛ 12 ⎞
1+ ⎜ − ⎟
⎝ 13 ⎠
5
−
= 13
1
13
= −5
129
1 ⎛ ⎛ 12 ⎞ ⎞
1+ −
2 ⎜⎝ ⎜⎝ 13 ⎟⎠ ⎟⎠
EX 2 Solve for x if
sin x
= 3.
1− cos x
The left side of the equation is not one of the formulas, but it is the
reciprocal of one of the formulas.
sin x
= 3
1− cos x
1− cos x 1
=
sin x
3
⎛1 ⎞
1
tan ⎜ x ⎟ =
⎝2 ⎠
3
1
1
x = tan –1
2
3
1
π
x = ±πn
2
6
π
x = ± 2π n
3
⎛1 ⎞⎛
⎛ 1 ⎞⎞
EX 3 Prove tan ⎜ x ⎟ ⎜ 2cot x + tan ⎜ x ⎟ ⎟ = 1
⎝2 ⎠⎝
⎝ 2 ⎠⎠
This is an interesting problem because there are two places to use a half
angle rule, and the same formula is not needed both times.
⎛1 ⎞⎛
⎛ 1 ⎞⎞
sin x ⎛ 2cos x 1− cos x ⎞
tan ⎜ x ⎟ ⎜ 2cot x + tan ⎜ x ⎟ ⎟ =
+
sin x ⎟⎠
⎝2 ⎠⎝
⎝ 2 ⎠ ⎠ 1+ cos x ⎜⎝ sin x
sin x ⎛ 1+ cos x ⎞
1+ cos x ⎜⎝ sin x ⎟⎠
=1
=
130
3-5 Free Response Homework
Given the following values and quadrant ranges, find the exact values of
⎛1 ⎞
a) sin ⎜ x ⎟
⎝2 ⎠
⎛1 ⎞
b) cos ⎜ x ⎟
⎝2 ⎠
⎛1 ⎞
c) tan ⎜ x ⎟
⎝2 ⎠
4
and 0° < x < 90°
5
1.
cos x =
3.
cos x = −
2.
cos x = −
4
and 90° < x < 180°
5
24
and 180° < x < 270°
25
Given the following values and quadrant ranges, find the exact values of
a)
⎛1 ⎞
csc ⎜ x ⎟
⎝2 ⎠
4.
sin x = −
5.
cos x =
b)
⎛1 ⎞
sec ⎜ x ⎟
⎝2 ⎠
c)
⎛1 ⎞
cot ⎜ x ⎟
⎝2 ⎠
24
and − 90° < x < 0°
25
4
and 630° < x < 720°
5
Find the general solutions to the following equations. Use exact values wherever
possible.
1
1
1− cos x ) = for x ∈{All Reals}
(
2
2
6.
7.
−
1
1
1+ cos x ) = for x ∈ ⎡⎣0, 2π ⎤⎦
(
2
2
8.
9.
1
3
for x ∈ ⎡⎣0, 4π ⎤⎦
1− cos x ) = −
(
2
2
−
1
(1+ cos x ) = 0 for x ∈{All Reals}
2
131
10.
sin x
= −1 for x ∈{All Reals}
1+ cos x
11.
1− cos x
= − 3 for x ∈ ⎡⎣ −2π , 2π ⎤⎦
sin x
Prove the following identities.
12.
⎛1 ⎞
⎛1 ⎞
tan ⎜ x ⎟ + cot ⎜ x ⎟ = 2csc x
⎝2 ⎠
⎝2 ⎠
13.
⎛1 ⎞
tan x tan ⎜ x ⎟ = sec x −1
⎝2 ⎠
14.
⎛1 ⎞
2 tan ⎜ x ⎟
⎝2 ⎠
= sin x
2⎛1 ⎞
1+ tan ⎜ x ⎟
⎝2 ⎠
15.
⎛1 ⎞
tan ⎜ x ⎟ = csc x − cot x
⎝2 ⎠
16.
⎛
1
1 ⎞
sin
x
+
cos
x = 1+ sin x
⎜
2
2 ⎟⎠
⎝
17.
⎛1 ⎞
⎛1 ⎞
cos ⎜ x ⎟ + sin ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
= sec x + tan x
⎛1 ⎞
⎛1 ⎞
cos ⎜ x ⎟ − sin ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
18.
⎛1 ⎞
⎛1 ⎞
cos ⎜ x ⎟ − sin ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
cos x
=
⎛1 ⎞
⎛ 1 ⎞ 1+ sin x
cos ⎜ x ⎟ + sin ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
2
132
3-5 Multiple Choice Homework
1.
(a)
(b)
(c)
(d)
(e)
2.
(a)
(b)
(c)
(d)
(e)
3.
If
1− cosθ
3
=
, then θ =
sin θ
3
15°
30°
45°
60°
75°
θ
θ
tan + cot =
2
2
1
cot θ
2
−2cscθ
−2cot θ
2cot θ
2cscθ
Which of the following is not equivalent to sin 40° ?
(a)
1− cos 2 40°
(b) 2sin 20°cos20°
(c)
1− cos80°
2
1+ cos80°
2
(e) All are equivalent to sin 40°
(d)
133
4.
If sin θ = −
(a) −
20
θ
, θ lies in QIV, find cos .
29
2
3 10
10
2
5
3 10
(c)
10
2
(d) −
5
(e) None of the above
(b)
5.
(a)
(b)
(c)
(d)
(e)
x
2 =
x
1+ tan 2
2
2 tan
sin x
cos x
tan x
sin x + 1
2cos x − 1
134
3-6: Solving Equations with Identities and Algebra
One of the main reasons for studying the identities in the last several sections is to
use them in solving equations. These trigonometric identities will sometimes
occur in the midst of a calculus question, and they will often be mixed in with
factoring or other algebraic operations.
Much of the carryover from Trigonometry to Calculus is in the process of solving
equations. Therefore, this process will be addressed specifically. Since so many
trigonometric identities are used to solve equations involving trigonometric
functions, THEY MUST BE MEMORIZED.
The Reciprocal Identities: cscθ =
The Quotient Identities:
1
sinθ
secθ =
sinθ
cosθ
secθ
tanθ =
cscθ
1
cosθ
cot θ =
1
tanθ
cosθ
sinθ
cscθ
cot θ =
secθ
tanθ =
cot θ =
The Pythagorean Identities:
sin 2 θ + cos2 θ = 1
1− cos2 θ = sin 2 θ
1− sin 2 θ = cos2 θ
tan 2 θ +1= sec2 θ
sec2 θ −1= tan 2 θ
sec2 θ − tan 2 θ = 1
1+ cot 2 θ = csc2 θ
csc2 θ −1= cot 2 θ
csc2 θ − cot 2 θ = 1
The Composite Argument Identities: cos ( A − B ) = cos A cos B + sin Asin B
cos ( A+ B ) = cos Acos B − sin Asin B
sin ( A− B ) = sin Acos B − cos Asin B
sin ( A+ B ) = sin Acos B + cos Asin B
tan A− tan B
tan ( A− B ) =
1+ tan Atan B
tan A+ tan B
tan ( A+ B ) =
1− tan Atan B
135
Even Functions:
cos ( −A ) = cos A
Odd Functions:
sin ( −A ) = − sin A
sec ( − A) = sec A
csc ( − A) = −csc A
tan ( − A) = − tan A
cot ( − A) = −cot A
sin ( 2A ) = 2sin A cos A
The Double Angle Argument Identities:
cos ( 2 A) = cos2 A− sin 2 A
= 1− 2sin 2 A
= 2cos2 A−1
2tan A
tan ( 2 A) =
1− tan 2 A
1
⎛1 ⎞
sin ⎜ A⎟ = ± (1− cos A )
⎝2 ⎠
2
⎛1 ⎞
1
cos ⎜ A⎟ = ± (1+ cos A)
2
⎝2 ⎠
The Half Angle Argument Identities:
⎛1 ⎞
1− cos A
tan ⎜ A⎟ = ±
1+ cos A
⎝2 ⎠
sin A
=
1+ cos A
1− cos A
=
sin A
LEARNING OUTCOME
Solve equations involving the trigonometric identities.
136
EX 1 Solve 4sin 2 x = 3 exactly for x ∈⎡⎣0, 2π ⎤⎦ .
This is a fairly straightforward “isolate the variable” problem. Be careful
when square rooting both sides of the equation.
4sin 2 x = 3
3
sin 2 x =
4
± 3
sin x =
2
3
2
π
± 2π n
3
2π
± 2π n
3
3
2
sin x =
sin x = −
⎧
⎪
⎪
x=⎨
⎪
⎪⎩
⎧ 4π
⎪
± 2π n
⎪ 3
x=⎨
⎪ 5π ± 2π n
⎪⎩ 3
Now, choose the particular answers that occur in the interval stated:
x=
π 2π 4π
5π
,
,
, or
3 3 3
3
137
)
EX 2 Solve sec2 x − sec x = 2 exactly for x ∈⎡⎣0, 2π .
sec2 x − sec x = 2
sec2 x − sec x − 2 = 0
(sec x − 2)(sec x + 1) = 0
sec x − 2 = 0
sec x = 2
1
cos x =
2
⎧ π
⎪
± 2π n
⎪
x=⎨ 3
⎪ − π ± 2π n
⎪ 3
⎩
So, x =
sec x +1= 0
sec x = −1
cos x = −1
x = π ± 2π n
π 5π
,
, or π
3 3
)
EX 3 Solve sec2 x − 2sec x = 3 exactly for x ∈⎡⎣0, 2π .
sec2 x − 2sec x = 3
sec2 x − 2sec x − 3 = 0
(sec x − 3)(sec x +1) = 0
sec x +1= 0
sec x = −1
cos x = −1
x = π ± 2π n
sec x − 3 = 0
sec x = 3
1
cos x =
3
x = ±1.231± 2π n
So, x = 1.231, 5.052, or π
138
As with proofs in Geometry or with trigonometric identities, there is no
algorithmic approach to solving trigonometric equations. Each problem is
different and only experience can really guide the approach to a particular problem.
However, there is a helpful strategy sequence. Ask these questions:
1.
2.
3.
Are the arguments different?
Are the trig functions different?
Is there any algebra to perform?
Strategies for Solving Trigonometric Equations:
I.
If the arguments are different, make them the same. Usually use:
Composite Rules
Double or Half Angle Rules
II.
If the arguments are the same but the trigonometric functions are different,
make them the same, by using:
Reciprocal or Quotient Rules
Pythagorean Identities
1
tan x formulas
2
Double Angle Argument Formulas
III.
Do the algebra:
Set equation equal to 0 and factor
(Remember: Do NOT divide by a variable!!!)
Find common denominators (then recheck for Pythagorean identities)
IV. Get to a place where there is one trigonometric function to inverse away,
thus isolating the variable.
As with proofs in Geometry or with trigonometric identities, there is no shortcut to
understanding. Most people need a great deal of experience with these before
enlightenment occurs.
139
)
EX 4 Solve cos3θ cos12° = sin3θ sin12° exactly for θ ∈⎡⎣0°, 360° .
The arguments in this equation are different from one another, so they must
be made the same. Since there are no triple angle rules, there must be
something else. Four trigonometric functions in these pairs look like a
composite rule. Do some minor rearranging to see it more clearly.
cos3θ cos12° = sin3θ sin12°
cos3θ cos12° − sin3θ sin12° = 0
cos (3θ + 12°) = 0
⎪⎧
90° ± 360n°
⎩⎪ −90° ± 360n°
3θ + 12° = ⎨
⎧⎪
78° ± 360n°
⎪⎩ −102° ± 360n°
3θ = ⎨
⎧⎪
26° ±120n°
⎩⎪ −34° ±120n°
θ =⎨
θ = 26°, 86°, 146°, 206°, 266° or 326°
)
EX 5 Solve sin 2 x + cos x − 1 = 0 exactly for x ∈⎡⎣0°, 360° .
This cannot be factored like EX 2 because the trigonometric functions are
different. But the Pythagorean identities can be used to make them the
same.
sin 2 x + cos x − 1 = 0
1− cos2 x + cos x − 1 = 0
−cos2 x + cos x = 0
cos x ( −cos x + 1) = 0
cos x = 0
x = ±90 ± 360n°
−cos x + 1 = 0
cos x = 1
x = 0° ± 360n°
x = 0°, 90° or 270°
140
)
The interval will not always be a standard x ∈⎡⎣0, 2π .
EX 6 Solve cos2 x − sin 2 x =
)
1
exactly for x ∈⎡⎣ −π , π .
2
cos2 x − sin 2 x =
cos2x =
1
2
1
2
π
± 2π n
4
π
x = ± ±πn
8
2x = ±
x=
π
7π
π
7π
,−
, − , or
8
8
8
8
141
3-6 Free Response Homework
Solve the following equations in the given interval. Use exact values when
possible.
3
in θ ∈⎡⎣0°, 360° ⎤⎦
2
1.
cos2 2θ − sin 2 2θ =
2.
3− 3sin x − 2cos2 x = 0 in x ∈⎡⎣ −2π , 2π ⎤⎦
3.
sin3θ cos12° − cos3θ sin12° =
4.
⎛
⎛
π⎞
π⎞
sec ⎜ x − ⎟ = 2 + 2sec ⎜ x − ⎟ in x ∈(0, 2π ⎤⎦
4⎠
4⎠
⎝
⎝
5.
⎛1 ⎞
2cos2 ⎜ x ⎟ − 2 = 3cos x in x ∈( −π , π )
⎝2 ⎠
6.
cos4θ − cos2θ = 0 in θ ∈⎡⎣0°, 360° ⎤⎦
7.
4sin xcos x = 3 in x ∈⎡⎣0, 2π ⎤⎦
8.
sin 2 x − cos2 x =
9.
cosθ cos20° − sinθ sin 20° =
10.
2cos2 x = 3 + 2sin 2 x in x ∈⎡⎣ −180°, 0° ⎤⎦
11.
tan 4 x − 4 tan 2 x + 3 = 0 in x ∈⎢
12.
3
3
sin 2 3t − 2sin t cos t − 2 = 0 in t ∈⎡⎣0°, 360°
2
2
1
in θ ∈⎡⎣0°, 360° ⎤⎦
2
3
in x ∈⎡⎣0, 2π ⎤⎦
2
1
in θ ∈⎡⎣0°, 360° ⎤⎦
2
⎡ −π
⎣ 4
,
π⎞
4 ⎟⎠
142
)
13.
3− 3sin x − 2cos2 x = 0 in x ∈{All Reals}
14.
2cot 2 x − 5csc x + 5 = 0 on x ∈⎡⎣ −2π , 2π
15.
tan 4 x − sec4 x = −3 in x ∈⎡⎣ −45°, 0° ⎤⎦
16.
tan 2 x − sec2 x − sec x − 2 = 0 in x ∈⎡⎣ −2π , 2π ⎤⎦
17.
sin 2 A − 2sin (3A + π ) = −cos2 A in A ∈ π , 3π
18.
2sin ( 2x − 30°) cos ( 2x − 30°) = −
19.
⎡ π π⎞
sec2 ( x − 4) + sin (3x ) = 1+ tan 2 ( x − 4) in x ∈⎢ − , ⎟
⎣ 2 2⎠
20.
cos2xsin x + sin2xcos x − 2sin3x =
21.
csc2x +
22.
1− cos x
= 3 in x ∈ 0, 2π
sin x
23.
1− tan 2 x cos2 x − sin 2 x
+
= 1 in x ∈(0, 2π )
2tan x
2sin xcos x
)
(
)
3
in x ∈⎡⎣0°, 720° ⎤⎦
2
1
in x ∈⎡⎣0, 2π
2
)
1
= 6 in x ∈⎡⎣0, π ⎤⎦
sin xcos x
(
)
24.
1
2
in x ∈⎡⎣0, 2π ⎤⎦
1− cos x ) =
(
2
2
25.
4tan x + 3 = tan x in x ∈(0, 2π ⎤⎦
26.
⎛π
⎞
π
π
sin3acos + cos3asin + tan 2 a = sec2 a in a ∈⎜ , π ⎟
2
2
⎝2
⎠
143
27.
(cos
28.
1
sin x tan x
2 − sin 2 1 x = 0 on x ∈⎡ −2π , 2π )
⎣
2
2
29.
2csc2 θ + 5cotθ = 0 in θ ∈⎡⎣ –180°, 180° ⎤⎦
30.
sin 2 θ − cos2 θ =
31.
tan3x + tan x = tan3x tan x − 1 in x ∈⎡⎣0, 2π ⎤⎦
32.
1− cos x
= −1 in x ∈{All Reals}
sin x
33.
sinθ = sin2θ in θ ∈( –90°, 270°)
34.
2tan 2 θ − 3secθ = 0 in θ ∈⎡⎣ –180°, 180° ⎤⎦
35.
⎛
⎛
π⎞
π⎞
sec ⎜ 3x − ⎟ = 2 + 2sec ⎜ 3x − ⎟ in x ∈(0, 2π ⎤⎦
4⎠
4⎠
⎝
⎝
36.
cos2x − sin4x = 0 in x ∈( −π , π )
37.
⎛1 ⎞
2cos2 ⎜ x ⎟ − 2 = 2cos x in x ∈( −π , π )
⎝2 ⎠
2
)
x − sin 2 x − ( 2sin xcos x ) = 0 in x ∈⎡⎣ −2π , 2π
2
2
1
in θ ∈( –90°, 270°)
2
144
)
3-6 Multiple Choice Homework
1.
(a)
(b)
(c)
(d)
(e)
2.
(a)
(b)
(c)
(d)
(e)
If 2 sin 2x = 3cos 2x and 0 ≤ 2x ≤
π
, then x =
2
0.25
0.52
0.49
0.39
0.63
If 4 sin x + 3 = 0 on 0 ≤ x ≤ 2π , then x =
5.435
0.848
3.990 or 5.435
0.848 or 5.435
–0.848
3.
Solve the equation sin15x + cos15x = 0 . What is the sum of the three
smallest positive solutions?
(a)
(b)
(c)
(d)
(e)
π
20
π
3
7π
20
21π
10
21π
4
145
4.
(a)
(b)
(c)
(d)
(e)
For what positive value(s) x ≤ 180° of does tan 2x = 2 cot 2x ?
54.7°
25° and 155°
27.4° and 117.4°
27.4°, 62.6°, 117.4°, and 152.6°
None of the above
5.
For all positive angles less than 360°, if csc ( 2x + 30° ) = cos ( 3y − 15° ) , the
sum of x and y is
(a)
(b)
(c)
(d)
(e)
185°
5°
30°
215°
95°
146
Trigonometric Identities Practice Group Test––Page 1
Find EXACT values (no decimals). Show all work.
For problems #1-6, (3, − 4) is on the terminal side of A and 270° ≤ A ≤ 360°, csc B =
90° ≤ B ≤ 180°, and tanC =
7
and −180° ≤ C ≤ −90° . Find the EXACT value of:
24
1.
sin ( A − B )
2.
cos2 A
3.
⎛1 ⎞
tan ⎜ C ⎟
⎝2 ⎠
4.
csc ( A + B )
5.
tan2B
6.
⎛1 ⎞
cos ⎜ A⎟
⎝2 ⎠
147
13
and
5
7.
1
1
1
1
Prove: tan x + cot x = csc xsec x
2
2
2
2
9.
Solve EXACTLY: sin 2 2x − 2sin xcos x = 2 for x ∈⎡⎣ −π , π
10.
Solve EXACTLY: −cos2 x = sin 2 x − 2sin (3x + π ) for x ∈⎡⎣0, 2π
8.
148
1+ tan 2 x
= sec2x
Prove:
1− tan 2 x
)
)
Trigonometric Identities Practice Group Test––Page 2
Find EXACT values (no decimals). Show all work.
For problems #1-6, (3, − 4) is on the terminal side of A and 270° ≤ A ≤ 360°, csc B =
90° ≤ B ≤ 180°, and tanC =
7
and −180° ≤ C ≤ −90° . Find the EXACT value of:
24
1.
cos ( A − B )
2.
sin2 A
3.
⎛1 ⎞
cot ⎜ C ⎟
⎝2 ⎠
4.
sec ( A + B )
5.
cot 2B
6.
⎛1 ⎞
sin ⎜ A⎟
⎝2 ⎠
149
13
and
5
7.
⎛1 ⎞
2 tan ⎜ x ⎟
⎝2 ⎠
Prove:
= sin x
2⎛1 ⎞
1+ tan ⎜ x ⎟
⎝2 ⎠
9.
Solve EXACTLY: sin 2 2x − 2sin xcos x = 2 for x ∈⎡⎣ −3π , − π
10.
Solve EXACTLY: −cos2 x = sin 2 x − 2sin (3x + π ) for x ∈⎡⎣ −2π , 0
8.
150
Prove: cos2β =
)
)
1− tan 2 β
1+ tan 2 β
Trigonometric Identities Practice Group Test––Page 3
Find EXACT values (no decimals). Show all work.
For problems #1-6, (3, − 4) is on the terminal side of A and 270° ≤ A ≤ 360°, csc B =
90° ≤ B ≤ 180°, and tanC =
7
and −180° ≤ C ≤ −90° . Find the EXACT value of:
24
1.
sin ( A + B )
2.
sec2 A
3.
⎛1 ⎞
tan ⎜ B⎟
⎝2 ⎠
4.
csc ( A − B )
5.
tan2 A
6.
⎛1 ⎞
cos ⎜ C ⎟
⎝2 ⎠
151
13
and
5
7.
1
1
Prove: tan x + cot x = 2csc x
2
2
9.
Solve EXACTLY: sin 2 2x − 2sin xcos x = 2 for x ∈⎡⎣ −2π , 0
10.
Solve EXACTLY: −cos2 x = sin 2 x − 2sin (3x + π ) for x ∈⎡⎣ −3π , − π
8.
152
Prove: (1+ tan ∂ ) ( tan 2∂ ) =
)
)
2 tan ∂
1− tan ∂
Trigonometric Identities Practice Group Test––Page 4
Find EXACT values (no decimals). Show all work.
For problems #1-6, (3, − 4) is on the terminal side of A and 270° ≤ A ≤ 360°, csc B =
90° ≤ B ≤ 180°, and tanC =
7
and −180° ≤ C ≤ −90° . Find the EXACT value of:
24
1.
cos ( A + B )
2.
csc2 A
3.
⎛1 ⎞
cot ⎜ B⎟
⎝2 ⎠
4.
sec ( A − B )
5.
cot 2 A
6.
⎛1 ⎞
sin ⎜ C ⎟
⎝2 ⎠
153
13
and
5
7.
Prove:
sin A + B + sin A − B
8.
Prove:
cos ( a + b) cosb + sin(a + b)sin b = cos a
( ) ( ) = tan Acot B
sin ( A + B ) − sin ( A − B )
)
9.
Solve EXACTLY: sin 2 2x − 2sin xcos x = 2 for x ∈⎡⎣0, 2π
10.
Solve EXACTLY: −cos2 x = sin 2 x − 2sin (3x + π ) for x ∈⎡⎣ −π , π
154
)
Trigonometric Identities Homework Answer Key
3-1 Free Response Homework
1.
2.
3.
4.
cos2 x + tan 2 xcos2 x
sin 2 x
= cos2 x + cos2 x
cos2 x
= cos2 x + sin 2 x
=1
cosθ tanθ + sinθ
tanθ
cosθ tanθ sinθ
=
+
tanθ
tanθ
sinθ
= cosθ +
tanθ
sinθ
= cosθ +
sinθ
cosθ
cosθ
= cosθ + sinθ
sinθ
= cosθ + cosθ
= 2cosθ
4 + ( tan σ − cot σ )
5.
(cosθ )(cosθ ) − (−sinθ )(sinθ )
cos2 θ
cos2 θ + sin 2 θ
=
cos2 θ
1
=
cos2 θ
= sec2 θ
(sin x )(−sin x ) − (cos x )(cos x )
sin 2 x
−sin 2 x − cos2 x
=
sin 2 x
− sin 2 x + cos2 x
=
sin 2 x
−1
= 2
sin x
= −csc2 x
(
6.
2
= 4 + tan 2 σ − 2 tan σ cot σ + cot 2 σ
= 4 + tan 2 σ − 2 (1) + cot 2 σ
= 2 + tan 2 σ + cot 2 σ
= 2 + sec2 σ − 1+ csc2 σ − 1
= sec2 σ + csc2 σ
7.
155
−
(−sinθ )
cos2 θ
sinθ 1
=
cosθ cosθ
= tanθ secθ
cosϕ
sin 2 ϕ
1 cosϕ
=−
sin ϕ sin ϕ
= −csc ϕ cot ϕ
−
)
8.
(cos a + sin a)(cos a + sin a) − 1
11.
2cos a
cos a + 2sin acos a + sin 2 a −1
=
2cos a
2
2
cos a + sin a + 2sin acos a −1
=
2cos a
1+ 2sin acos a −1
=
2cos a
2sin acos a
=
2cos a
= sin a
2
9.
10.
sin x
1− cos x
+
1− cos x
sin x
=
(1− cos x ) (1− cos x )
sin x sin x
+
sin x (1− cos x ) (1− cos x ) sin x
sin 2 x + 1− 2cos x + cos2 x
=
sin x (1− cos x )
=
=
tanW (cotW cosW + sinW )
= cosW + tanW sinW
sinW
= cosW +
sinW
cosW
cos2 W sin 2 W
=
+
cosW cosW
1
=
cosW
= secW
=
1+ 1− 2cos x
sin x (1− cos x )
2 − 2cos x
sin x (1− cos x )
2 (1− cos x )
sin x (1− cos x )
2
sin x
= 2csc x
=
12.
sec x sin x
−
sin x cos x
cos x sec x sin x sin x
=
−
cos x sin x cos x sin x
1− sin 2 x
=
cos xsin x
cos2 x
=
cos xsin x
cos x
=
sin x
= cot x
13.
156
cot θ cscθ tanθ − sinθ
csc θ
cscθ − sinθ
=
cscθ
sinθ
= 1−
cscθ
= 1− sin 2 θ
= cos2 θ
tanb + cot b
sin b cosb
=
+
cosb sin b
sin b sin b cosb cosb
=
+
sin b cosb sin b cosb
1
=
sin bcosb
cscb
=
cosb
14.
cot xcsc x + cot 2 x
cos x 1
cos2 x
=
+
sin x sin x sin 2 x
cos x + cos2 x
=
1− cos2 x
cos x 1+ cos x
=
1− cos x 1+ cos x
(
(
=
15.
)
)(
3-2 Free Response Homework
)
cos x
1− cos x
2sec2 x + 2sec x tan x − 1
2
1 sin x
=
−1
2 +2
cos x cos x
cos x
2 + 2sin x − cos2 x
=
cos2 x
2 + 2sin x − 1− sin 2 x
=
1− sin 2 x
2 + 2sin x − 1− sin 2 x
=
1− sin x 1+ sin x
(
(
=
16.
(
)(
1+ 2sin x + sin 2 x
(1− sin x )(1+ sin x )
=
(1+ sin x )(1+ sin x )
(1− sin x )(1+ sin x )
=
1+ sin x
1− sin x
( x − 4 y )( x − 2 y )
2.
xy ( 3x − 2 y ) ( x + 4 y )
3.
−x ( 2x − 1) ( x + 3)
4.
(x
5.
( x − 1)( x + 1)( x
6.
k−r
k+r
7.
1
x −1
)
)
)
8.
9.
2
)
+ y2 ( x − y)
2
−2
)
x 2 − xy + y 2
( x + y)
2
sin 2 x + cot 2 xsin 2 x
= sin 2 x (1+ cot 2 x )
= sin 2 x (csc2 x )
=1
Answers vary
10.
3-1 Multiple Choice Homework
1.
2.
3.
4.
5.
6.
1.
sec4 β − tan 4 β
= sec2 β − tan 2 β sec2 β + tan 2 β
(
)(
= sec2 β + tan 2 β
= 1+ tan 2 β + tan 2 β
= 1+ 2tan 2 β
D
A
C
B
A
B
156
)
11.
12.
cos2 w + 2cosw +1
cos2 w − 3cosw − 4
( cosw +1)( cosw +1)
=
( cosw +1)( cosw − 4 )
cosw +1
=
cosw − 4
=
(
)
6 (1− cos w ) + 5cosw − 2
2
2 − 2cos2 w + 5cosw +1
=
6 − 6cos2 w + 5cosw − 2
−2cos2 w + 5cosw + 3
cos3 A− sec3 A
=
−6cos2 w + 5cosw + 4
cos A− sec A
2
cos A− sec A) cos2 A+ cos Asec A+ sec2 A = 2cos w − 5cosw − 3
(
=
6cos2 w − 5cosw − 4
cos A− sec A
( 2cosw +1)( cosw − 3)
= cos2 A+1+ sec2 A
=
( 2cosw +1)( 3cosw − 4 )
= 1− sin 2 A +1+ 1+ tan 2 A
cosw − 3
=
= −sin 2 A+ tan 2 A+ 3
3cosw − 4
(
(
13.
2sin 2 w + 5cosw +1
6sin 2 w + 5cosw − 2
2 1− cos2 w + 5cosw +1
14.
)
)
(
)
tan 2 w − secw − 5
tan 2 w + 3secw + 3
sec 2 w − 1) − sec w − 5
(
=
(sec2 w − 1) + 3sec w + 3
15.
(
(
sec 2 w − sec w − 6
=
sec 2 w + 3sec w + 2
(sec w + 2 )(sec w − 3)
=
(sec w + 2 )(sec w + 1)
=
3sec2 t − 8 tant +1
sec2 t − tant − 3
3 tan 2 t +1 − 8 tant +1
=
tan 2 t +1 − tant − 3
)
)
3tan 2 t + 3− 8 tant +1
tan 2 t +1− tant − 3
3tan 2 t − 8 tant + 4
=
tan 2 t − tant − 2
( 3tant − 2 )( tant − 2 )
=
( tant +1)( tant − 2 )
3tant − 2
=
tant +1
=
sec w − 3
sec w + 1
156
16.
2csc2 y − 7cot y − 6
6csc2 y − 5cot y −10
=
(
6 ( cot
1+ tan3 B
− tan B + sec2 B
1+ tan B ) 1− tan B + tan 2 B
(
=
− tan B + 1+ tan 2 B
23.
(
)
y +1) − 5cot y −10
2 cot 2 y +1 − 7cot y − 6
2
2cot 2 y + 2 − 7cot y − 6
6cot 2 y + 6 − 5cot y −10
2cot 2 y − 7cot y − 4
=
6cot 2 y − 5cot y − 4
( 2cot y +1)( cot y − 4 )
=
( 2cot y +1)( 3cot y − 4 )
cot y − 4
=
3cot y − 4
18.
= (1+ tan B )
=
cot B
cot B
cot B + 1
cot B
1+ sin3 B
1− sin 2 B
1+ sin B ) 1− sin B + sin 2 B
(
=
(1+ sin B)(1− sin B)
24.
(
3
x ( x +1)
=
x−2
x+2
(8 + cos x ) i (cos
( cos x − 2 ) ( cos
19.
− ( 9 + 3x + x
3( x + 3)
20.
−
21.
22.
)
25.
)
sin 2 B − sin B + 1
1− sin B
3
2
)
= 1+ tan B
=
17.
(
)
2
2
)
x − 2cos x + 4 )
x − 4 cos x + 4
( 2 + cos x )( cos2 x − 2cos x + 4 ) ( cos x − 2 )( cos x − 2 )
=
i
( cos x − 2 )
(cos2 x − 2cos x + 4 )
= ( 2 + cos x )( cos x − 2 )
r 2 + rx + x 2
r 2 − rx + x 2
= cos2 x − 4
x
26.
sec2 x + 3sec x −10
(
(sec
1
=
=
156
2
) i (sec x + 2sec x − 3)
x − 7sec x + 6 ) (sec x + sec x − 6 )
2
2
(sec x + 5 )(sec x − 2 ) i (sec x −1)(sec x + 3)
(sec x − 6 )(sec x −1) (sec x + 3)(sec x − 2 )
sec x + 5
sec x − 6
27.
cot 2 x − 4csc x − 11
cot 2 x − 3
csc 2 x − 1 − 4csc x − 11
=
csc 2 x − 1− 3
csc 2 x − 4csc x − 12
=
csc 2 x − 4
(csc x + 2)(csc x − 6)
=
(csc x + 2)(csc x − 2)
(
=
28.
)
csc x − 6
csc x − 2
csc6 x − cot 6 x
( ) ( )
= ( csc x − cot x ) ( csc
3
= csc 2 x − cot 2 x
2
2
3
4
x + csc 2 x cot 2 x + cot 4 x
= csc 4 x + csc 2 x cot 2 x + cot 4 x
(
)
(
)
)
= csc 2 x 1+ cot 2 x + csc 2 x cot 2 x + cot 2 x csc 2 x − 1
= csc 2 x − cot 2 x + 3csc 2 x cot 2 x
= 1+ 3csc 2 x cot 2 x
29.
54 − 2sin 3 x
6
÷ 2
4
sin x − 81 sin x + 9
2 27 − sin 3 x
sin 2 x + 9
=
i
6
sin 2 x − 9 sin 2 x + 9
(
(
)
(
)
)(
)
2 ( 3− sin x )( 9 + 3sin x + sin x ) (sin x + 9 )
=
i
6
(sin x − 3)(sin x + 3)(sin x + 9 )
− ( 9 + 3sin x + sin x )
=
2
2
2
2
3(sin x + 3)
156
30.
(sin
(sin
2
2
)
x − sin x cos x )
x + sin x cos x
2
2
÷
sin x + cos x cos 3 x + sin 3 x
÷
sin x − cos x cos 3 x − sin 3 x
sin x − cos x ) ( cos x − sin x )( cos2 x + sin x cos x + sin 2 x )
(
=
i
2 i
2
sin
x
+
cos
x
(
)
sin x (sin x − cos x )
( cos x + sin x )( cos2 x − sin x cos x + sin2 x )
sin 2 x (sin x + cos x )
=−
2
cos2 x + sin x cos x + sin 2 x
cos2 x − sin x cos x + sin 2 x
3-2 Multiple Choice
1.
2.
3.
4.
5.
D
A
E
C
B
3-3 Free Response Homework
1a.
b.
c.
d.
e.
f.
2a.
5
34
−99
17 34
−5
3
34
3
−17 34
5
99
5
b.
c.
d.
e.
f.
156
−16
65
−33
65
16
63
−65
63
65
56
33
56
3.
⎪⎧14° ± 120n
θ=⎨
⎪⎩54° ± 120n
4.
x = {60°, 180°}
5.
φ = 15° ± 180n
6.
⎧ π 5π 9π 13π
3π 7π 11π 15π ⎫
x=⎨ ,
,
,
,−
, −
, −
, −
⎬
16 16
16
16 ⎭
⎩16 16 16 16
7.
⎧11π
x=⎨
⎩ 12
, −
12.
5π ⎫
⎬
12 ⎭
= tan ( A + B )
8.
A and B = {All Reals}
9.
sin ( A+ B ) + sin ( A− B )
=
13.
sin ( A+ B ) − sin ( A− B )
2sin Acos B
2cos Asin B
= tan Acot B
=
cos ( a + b) cosb + sin ( a + b) sinb
= cot ( A + B )
= (cos acosb − sin asinb) cosb + (sin acosb + cos asinb) sinb
= cos acos2 b − sin asinbcosb + sin acosbsinb + cos asin 2 b
= cos acos2 b + cos asin 2 b
= cos a cos2 b + sin 2 b
= cos a
11.
tan Atan B (cot Acot B − 1)
tan Atan B (cot A + cot B )
1− tan Atan B
tan B + tan A
1
=
tan ( A + B )
=
(
tan A + tan B
1− tan Atan B
cot Acot B − 1
cot A + cot B
=
sin Acos B + cos Asin B + sin Acos B − cos Asin B
=
sin Acos B + cos Asin B − (sin Acos B − cos Asin B )
10.
sin Acos B + cos Asin B
cos Acos B − sin Asin B
sin ( A + B )
=
cos ( A + B )
)
3-3 Multiple Choice Homework
1.
2.
3.
4.
5.
sin Acos A + cos Asin A
cos Acos A − sin Asin A
sin ( A + A)
=
cos ( A + A)
B
C
D
E
A
3-4 Free Response Homework
= tan ( A + A)
1a.
tan A + tan A
1− tan Atan A
2 tan A
=
1− tan 2 A
=
b.
c.
d.
156
4 2
9
−527
625
−8
7 2
625
−
527
−
−9
4 2
527
336
9.
10.
2.
⎧π
⎪⎪ ± π n
θ = ⎨8
⎪ 3π ± π n
⎪⎩ 8
3.
⎧ π 3π ⎫
x=⎨ ,
⎬
⎩2 2 ⎭
4.
⎪⎧0° ± 90n
x=⎨
⎩⎪75° ± 90n
cot A + tan A
cos A sin A
=
+
sin A cos A
cos2 A + sin 2 A
=
sin Acos A
1
=
sin Acos A
2
=
2sin Acos A
= 2csc ( 2 A)
11.
cos2φ cot 2 φ + cos2φ
= cos2φ cot 2 φ +1
e.
f.
5.
6.
7.
8.
(
)(
= cos 2β
⎧π
⎪⎪ ± π n
x = ⎨12
⎪ 5π ± π n
⎪⎩ 12
(
)
= (cos φ − sin φ )(csc φ )
2
2
2
= cos2 φ csc2 φ −1
cos2 φ
=
−1
sin 2 φ
= cot 2 φ −1
⎧ π 5π 9π 13π
⎫
⎪⎪ 8 , 8 , 8 , 8 ,
⎪⎪
x=⎨
⎬
⎪− 3π , − 7π , − 11π , − 15π ⎪
⎪⎩ 8
8
8
8 ⎪⎭
x=
cos4 β − sin 4 β
= cos2 β − sin 2 β cos2 β + sin 2 β
12.
π π
± n
12 2
⎛ 1− tan 2 x cos2 x − sin 2 x ⎞
+
⎜
2sin xcos x ⎟⎠
⎝ 2 tan x
⎛
cos2x ⎞
= ⎜ cot 2x +
sin 2x ⎟⎠
⎝
1− tan 2 B
1+ tan 2 B
1− tan 2 B
=
sec2 B
1
tan 2 B
=
−
sec2 B sec2 B
= cos2 B − sin 2 B
= cos2B
= ( 2cot 2x )
2
= 4cot 2 2x
= 4 csc2 2x − 1
(
= 4csc2 2x − 4
156
)
2
2
)
13.
(1+ tanδ ) tan 2δ
2 tan δ
= (1+ tan δ )
1− tan 2 δ
2 tan δ (1+ tan δ )
=
(1− tanδ )(1+ tanδ )
=
2 tan δ
1− tan δ
3-4 Multiple Choice Homework
1.
2.
3.
4.
5.
6.
D
E
D
E
C
B
3a.
⎛1 ⎞
7
sin ⎜ x ⎟ =
⎝2 ⎠ 5 2
b.
⎛1 ⎞
−1
cos ⎜ x ⎟ =
⎝2 ⎠ 5 2
c.
⎛1 ⎞
tan ⎜ x ⎟ = −7
⎝2 ⎠
4a.
⎛1 ⎞
5
csc ⎜ x ⎟ = −
3
⎝2 ⎠
b.
⎛1 ⎞ 5
sec ⎜ x ⎟ =
⎝2 ⎠ 4
c.
⎛1 ⎞
4
cot ⎜ x ⎟ = −
3
⎝2 ⎠
5a.
3-5 Free Response Homework
1a.
b.
c.
2a.
b.
c.
⎛1 ⎞
1
sin ⎜ x ⎟ =
⎝2 ⎠
10
⎛1 ⎞
3
cos ⎜ x ⎟ =
⎝2 ⎠
10
⎛1 ⎞ 1
tan ⎜ x ⎟ =
⎝2 ⎠ 3
⎛1 ⎞
3
sin ⎜ x ⎟ =
⎝2 ⎠
10
⎛1 ⎞
1
cos ⎜ x ⎟ =
⎝2 ⎠
10
⎛1 ⎞
tan ⎜ x ⎟ = 3
⎝2 ⎠
156
⎛1 ⎞
csc ⎜ x ⎟ = − 10
⎝2 ⎠
b.
⎛1 ⎞
10
sec ⎜ x ⎟ =
3
⎝2 ⎠
c.
⎛1 ⎞
cot ⎜ x ⎟ = −3
⎝2 ⎠
6.
⎧π
⎪⎪ ± 4π n
x = ⎨3
⎪ 5π ± 4π n
⎪⎩ 3
7.
x=⎨
8.
x=⎨
9.
x = ±π ± 4π n
⎧ 8π
⎩ 3
⎧ 2π
⎩ 3
,
10π ⎫
⎬
3 ⎭
,
10π ⎫
⎬
3 ⎭
10.
11.
12.
13.
x=−
π
± 2π n
2
⎧ 2π
x = ⎨−
⎩
3
,
14.
4π ⎫
⎬
3 ⎭
⎛1 ⎞
2tan ⎜ x ⎟
⎝2 ⎠
⎛1 ⎞
1+ tan 2 ⎜ x ⎟
⎝2 ⎠
⎛1 ⎞
2 tan ⎜ x ⎟
⎝2 ⎠
=
⎛1 ⎞
sec2 ⎜ x ⎟
⎝2 ⎠
⎛1 ⎞
⎛1 ⎞
tan ⎜ x ⎟ + cot ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
1− cos x 1+ cos x
=
+
sin x
sin x
2
=
sin x
= 2csc x
⎛1 ⎞
2sin ⎜ x ⎟
⎛1 ⎞
⎝2 ⎠
=
cos2 ⎜ x ⎟
⎛1 ⎞
⎝2 ⎠
cos ⎜ x ⎟
⎝2 ⎠
⎛1 ⎞
⎛1 ⎞
= 2sin ⎜ x ⎟ cos ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
⎛1 ⎞
tan x tan ⎜ x ⎟
⎝2 ⎠
⎛1 ⎞
= sin 2 ⎜ x ⎟
⎝2 ⎠
= sin x
sin x (1− cos x )
=
cos x sin x
1− cos x
=
cos x
= sec x − 1
15.
16.
csc x − cot x
1
cos x
=
−
sin x sin x
1− cos x
=
sin x
⎛1 ⎞
= tan ⎜ x ⎟
⎝2 ⎠
⎛
1
1 ⎞
sin
x
+
cos
x
⎜⎝
2
2 ⎟⎠
2
1
1
1
1
= sin 2 x + 2sin xcos x + cos2 x
2
2
2
2
⎛1 ⎞
1
1
= sin 2 x + cos2 x + sin2 ⎜ x ⎟
2
2
⎝2 ⎠
= 1+ sin x
156
17.
⎛1 ⎞
⎛1 ⎞
cos ⎜ x ⎟ + sin ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
⎛1 ⎞
⎛1 ⎞
cos ⎜ x ⎟ − sin ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
⎛
⎛1 ⎞
⎛ 1 ⎞⎞
⎜ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎟
⎝2 ⎠
⎝ 2 ⎠⎠
⎝
=
⎛
⎛1 ⎞
⎛ 1 ⎞⎞
⎜ cos ⎜ x ⎟ − sin ⎜ x ⎟ ⎟
⎝2 ⎠
⎝ 2 ⎠⎠
⎝
18.
⎛
⎛1 ⎞
⎛ 1 ⎞⎞
⎜ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎟
⎝2 ⎠
⎝ 2 ⎠⎠
⎝
⎛
⎛1 ⎞
⎛ 1 ⎞⎞
⎜ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎟
⎝2 ⎠
⎝ 2 ⎠⎠
⎝
⎛1 ⎞
⎛1 ⎞
⎛1 ⎞
⎛1 ⎞
cos2 ⎜ x ⎟ + 2sin ⎜ x ⎟ cos ⎜ x ⎟ + sin 2 ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
⎝2 ⎠
⎝2 ⎠
=
⎛1 ⎞
⎛1 ⎞
cos2 ⎜ x ⎟ − sin 2 ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
=
⎛1 ⎞
⎛1 ⎞
cos ⎜ x ⎟ − sin ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
⎛1 ⎞
⎛1 ⎞
cos ⎜ x ⎟ + sin ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
⎛
⎛1 ⎞
⎛ 1 ⎞⎞
⎜ cos ⎜ x ⎟ − sin ⎜ x ⎟ ⎟
⎝2 ⎠
⎝ 2 ⎠⎠
⎝
⎛
⎛1 ⎞
⎛ 1 ⎞⎞
⎜ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎟
⎝2 ⎠
⎝ 2 ⎠⎠
⎝
⎛
⎛1 ⎞
⎛ 1 ⎞⎞
cos
x
+
sin
⎜
⎜⎝ 2 ⎟⎠
⎜⎝ 2 x ⎟⎠ ⎟
⎝
⎠
⎛
⎛1 ⎞
⎛ 1 ⎞⎞
cos
x
+
sin
⎜
⎜⎝ 2 ⎟⎠
⎜⎝ 2 x ⎟⎠ ⎟
⎝
⎠
⎛1 ⎞
⎛1 ⎞
cos2 ⎜ x ⎟ − sin 2 ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
=
⎛1 ⎞
⎛1 ⎞
⎛1 ⎞
⎛1 ⎞
cos2 ⎜ x ⎟ − 2sin ⎜ x ⎟ cos ⎜ x ⎟ + sin 2 ⎜ x ⎟
⎝2 ⎠
⎝2 ⎠
⎝2 ⎠
⎝2 ⎠
⎛1 ⎞
1+ sin2 ⎜ x ⎟
⎝2 ⎠
=
⎛1 ⎞
cos2 ⎜ x ⎟
⎝2 ⎠
1+ sin x
=
cos x
1
sin x
=
+
cos x cos x
= sec x + tan x
⎛1 ⎞
cos2 ⎜ x ⎟
⎝2 ⎠
=
⎛1 ⎞
1+ sin2 ⎜ x ⎟
⎝2 ⎠
cos x
=
1+ sin x
3-5 Multiple Choice Homework
1.
2.
3.
4.
5.
D
E
D
A
C
3-6 Free Response Homework
1.
156
⎧7.5°, 82.5°, 97.5°,
⎫
⎪
⎪
θ ∈⎨172.5°, 187.5°, 262.5°,⎬
⎪277.5°, 352.5°
⎪
⎩
⎭
2.
3.
⎧ −3π π
7π
, , −
⎪⎪
2
6
x ∈⎨ 2
⎪− 11π , π , 5π
⎪⎩ 6
6 6
⎫
⎪
⎬
⎪
⎪⎭
14.
⎧−5.553, − 3.871, 0.730,⎫
⎪
⎪
x ∈⎨
⎬
3π π
⎪2.412, − 2 , 2
⎪
⎩
⎭
15.
x ∈{−45°}
16.
x ∈{ ±4.373, ± 1.911}
17.
⎧19π
,
⎪⎪
18
A ∈⎨
⎪ 35π ,
⎪⎩ 18
18.
⎧0°, 75°, 90°, 165°, 180°,⎫
⎪
⎪
⎪255°, 270°, 345°, 360°, ⎪
x ∈⎨
⎬
⎪435°, 450°, 525°, 540°, ⎪
⎪615°, 630°, 705°, 720° ⎪
⎩
⎭
19.
⎪⎧ π
⎪⎫
x ∈⎨± , 0⎬
⎪⎩ 3
⎪⎭
20.
⎧ 7π 11π 19π
,
,
,
⎪⎪
18
18
18
x ∈⎨
⎪ 23π , 31π , 35π
⎪⎩ 18
18 18
21.
x ∈⎨
⎪⎧14°, 54°, 134°, ⎪⎫
θ ∈⎨
⎬
⎪⎩174°, 254°, 294° ⎪⎭
⎪⎧11π
4.
x ∈⎨
5.
x ∈⎨±
⎪⎩ 12
,
19π ⎪⎫
⎬
12 ⎪⎭
⎧⎪ 2π ⎫⎪
⎬
⎪⎩ 3 ⎪⎭
6.
⎧⎪30°, 90°, 150°, ⎫⎪
x ∈⎨
⎬
⎪⎩210°, 270°, 330° ⎪⎭
7.
⎧⎪ π π 7π 4π ⎫⎪
x ∈⎨ , ,
,
⎬
3 ⎪⎭
⎪⎩ 6 3 6
8.
x ∈⎨
9.
θ ∈{25°, 295°}
10.
θ ∈{−165°, − 15°}
11.
⎧⎪ π ⎫⎪
x ∈ ⎨− ⎬
⎪⎩ 4 ⎪⎭
12.
13.
⎧
⎪
⎪
⎪
x=⎨
⎪
⎪
⎪
⎩
,⎪
⎧⎪ 5π
7π 17π 19π ⎫⎪
,
,
⎬
12 ⎪⎭
⎪⎩ 12 12 12
,
t ∈{90°, 210°, 330°}
156
π
± 2π n
6
5π
± 2π n
6
π
± 2π n
2
⎧⎪ π
23π
,
18
43π
,
18
5π ⎫⎪
⎬
⎪⎩12 12 ⎪⎭
,
31π
,
18
47π
18
⎫
⎪⎪
⎬
⎪
⎪⎭
⎫
⎪⎪
⎬
⎪
⎪⎭
⎪⎧ 2π ⎪⎫
⎬
⎪⎩ 3 ⎪⎭
22.
x ∈⎨
23.
x ∈{0.554, 2.125, 3.696, 5.266}
31.
24.
⎪⎧ π 3π ⎪⎫
x ∈⎨ ,
⎬
⎩⎪ 2 2 ⎭⎪
25.
x ∈⎨
26.
⎪⎧ 2π ⎪⎫
a ∈⎨ ⎬
⎪⎩ 3 ⎪⎭
27.
⎧ ±15π ±13π ±11π ±9π ⎫
,
,
,
,⎪
⎪⎪
⎪
8
8
8
8
x ∈⎨
⎬
⎪ ±7π , ±5π , ±3π , ±π
⎪
⎪⎩ 8
⎪⎭
8
8
8
⎪⎧ 5π
⎪⎩ 6
,
11π ⎪⎫
⎬
6 ⎪⎭
28.
x ∈ ⎡⎣ −2π , 2π )
29.
⎧⎪−63.435°, − 26.565°,⎫⎪
θ ∈⎨
⎬
⎪⎩116.525°, 153.435° ⎪⎭
30.
θ ∈{±67.5°, 112.5°, 247.5°}
⎧ 3π 7π 11π 15π 19π ⎫
,
,
,
,⎪
⎪⎪ ,
⎪
16
16
16
16
16
x ∈⎨
⎬
⎪ 23π , 27π , 31π
⎪
⎪⎩ 16
⎪⎭
16 16
π
± 2π n
2
32.
A= −
33.
θ ∈{±60°, 0°, 180°}
34.
θ ∈{ ±60°}
35.
⎧11π
,
⎪
⎪ 36
x ∈⎨
⎪ 43π ,
⎪
⎩ 36
36.
⎧ 11π
7π π ⎫
, − ,
,⎪
⎪⎪−
⎪
12
12
12
x ∈⎨
⎬
⎪ 5π , ± π , ± 3π ⎪
⎪⎩ 12
4
4 ⎪⎭
37.
No solutions
35π 59π 19π ⎫
,
,
,
36 36 36 ⎪⎪
⎬
67π
⎪
⎪
36
⎭
3-6 Multiple Choice Homework
1.
2.
3.
4.
5.
156
C
C
C
D
A
Trigonometric Identities Practice Test Answer Key
Page 1
1.
33
65
2.
−
7
25
3.
−7
4.
65
63
5.
−
120
119
6.
−
7.
1
1
tan x + cot x
2
2
1
1
sin x cos x
2 +
2
=
1
1
cos x sin x
2
2
1
1
sin 2 x + cos 2 x
2
2
=
1
1
sin x cos x
2
2
1
=
1
1
sin x cos x
2
2
1
1
= csc x sec x
2
2
8.
⎧⎪ 3π
π ⎫⎪
, − ⎬
4 ⎭⎪
⎩⎪ 4
9.
x ∈⎨
10.
x ∈⎨
⎧ 7π
11π 19π 23π 31π 35π ⎫
,
,
,
,
⎬
18 18 18 18 ⎪⎭
⎪
⎩ 18 18
,
156
2
5
1+ tan 2 x
1− tan 2 x
sin 2 x
1+
cos 2 x
=
sin 2 x
1−
cos 2 x
sin 2 x
1+
2
cos 2 x
cos
x
=
i
sin 2 x cos 2 x
1−
cos 2 x
cos 2 x + sin 2 x
=
cos 2 x − sin 2 x
1
=
2
cos x − sin 2 x
1
=
cos 2x
= sec 2x
Page 2
1.
−
56
65
2.
−
24
25
3.
4.
−
65
16
5.
−
119
120
6.
7.
1
2 tan x
2
1
1+ tan 2 x
2
1
2 tan x
2
=
1
sec 2 x
2
1
2sin x
2
=
1
1
cos x sec 2 x
2
2
1
2sin x
2
=
1
1
1
cos x sec x sec x
2
2
2
1
2sin x
2
=
1
sec x
2
1
1
= 2sin x cos x
2
2
⎛ 1 ⎞
= 2sin ⎜ 2 i x ⎟
⎝ 2 ⎠
8.
−
7
73
1
5
1 − tan 2 β
1 + tan 2 β
sin 2 β
1−
cos 2 β
=
sin 2 β
1+
cos 2 β
cos 2 β − sin 2 β
cos 2 β
=
cos 2 β + sin 2 β
cos 2 β
cos 2 β − sin 2 β
cos 2 β
cos 2 β
=
i
cos 2 β + sin 2 β cos 2 β
cos 2 β
= cos 2 β − sin 2 β
= cos 2 β
= sin x
156
9.
⎧ 5π
9π ⎫
x ∈ ⎨− , − ⎬
4 ⎭
⎩ 4
10.
⎧ π
5π
13π ⎫
⎪⎪− 18 , − 18 , − 18 , ⎪⎪
x ∈⎨
⎬
17π
25π
29π
⎪−
⎪
, −
, −
⎪⎩ 18
18
18 ⎪⎭
Page 3
1.
63
65
2.
−
4.
65
33
5.
24
7
7.
25
7
=
=
=
)(
) (
(
)(
)
1
6.
1
1
tan x + cot x
2
2
sin x
1 + cos x
=
+
1 + cos x
sin x
sin x sin x + 1 + cos x 1 + cos x
=
sin x 1 + cos x
(
5
3.
5 2
)
sin 2 x + 1 + 2cos x + cos 2 x
(
sin x 1 + cos x
2 + 2cos x
(
sin x 1 + cos x
(
2 1 + cos x
(
)
sin x 1 + cos x
)
)
)
2
sin x
= 2csc x
=
8.
(1+ tanδ )( tan 2δ )
⎛ 2 tanδ ⎞
= (1+ tanδ ) ⎜
⎝ 1− tan 2 δ ⎟⎠
=
=
2 tanδ (1+ tanδ )
(1− tanδ )(1+ tanδ )
2 tanδ
1− tanδ
156
⎧ 5π
π⎫
, − ⎬
4 ⎭⎪
⎩⎪ 4
9.
x ∈⎨−
10.
⎧ 25π
29π
37π ⎫
, −
, −
, ⎪
⎪⎪−
⎪
18
18
18
x ∈⎨
⎬
⎪− 41π , − 49π , − 53π ⎪
⎪⎩ 18
18
18 ⎪⎭
Page 4
1.
−
16
65
2.
−
4.
−
65
56
5.
7
24
7.
25
24
3.
1
5
6.
−
7
5 2
( ) ( )
sin ( A + B ) − sin ( A − B )
sin A + B + sin A − B
sin Acos B + cos Asin B + sin Acos B − cos Asin B
sin Acos B + cos Asin B − sin Acos B + cos Asin B
2sin Acos B
=
2cos Asin B
sin A cos B
=
i
cos A sin B
= tan Acot B
=
8.
(
)
(
)
cos a + b cos b + sin a + b sin b
= cos b ⎡⎣cos acos b − sin asin b ⎤⎦ + sin b ⎡⎣sin acos b + cos asin b ⎤⎦
= cos acos 2 b − sin asin bcos b + sin asin bcos b + cos asin 2 b
(
= cos a cos 2 b + sin 2 b
)
= cos a
⎧⎪ 3π
9.
x ∈⎨
10.
x ∈⎨
⎩⎪ 4
⎧ 7π
,
7π ⎫⎪
⎬
4 ⎭⎪
11π
π
5π
13π
17π ⎫
, − , − , −
, −
⎬
18
18
18
18 ⎪⎭
⎪⎩ 18 18
,
156