Download 2017 Lecture note Trig 2

2017 MATH 1112: DR.YOU
CHAPTER 2. ANALYTIC TRIGONOMETRY
LECTURE 1-1 INVERSE TRIGONOMETRIC FUNCTIONS
INVERSE
The function
Inverse function
Domain of inverse
Range of inverse
𝑦 = sin 𝑥
𝑦 = sin−1 𝑥 or 𝑦 = arcsin 𝑥
[−1,1]
𝜋 𝜋
[− , ]
2 2
𝑦 = cos 𝑥
𝑦 = cos −1 𝑥 or 𝑦 = arccos 𝑥
[−1,1]
[0, 𝜋]
𝑦 = tan 𝑥
𝑦 = tan−1 𝑥 or 𝑦 = arctan 𝑥
(−∞, ∞)
𝑦 = csc 𝑥
𝑦 = csc −1 𝑥 or 𝑦 = arccsc 𝑥
(−∞, −1] ∪ [1, ∞)
𝑦 = sec 𝑥
𝑦 = sec −1 𝑥 or 𝑦 = arcsec 𝑥
(−∞, −1] ∪ [1, ∞)
𝑦 = cot 𝑥
𝑦 = cot −1 𝑥 or 𝑦 = arccot 𝑥
(−∞, ∞)
NOTE:
1
csc −1 𝑥 = sin−1 ( ),
𝑥
1
sec −1 𝑥 = cos−1 ( ),
𝑥
(0, 𝜋)
1
cot −1 𝑥 = tan−1 ( )
𝑥
Example 1
Your Turn 1
Find the exact value of
Find the exact value of
1
(A) sin−1 (2)
(B) sin−1 (−
√3
)
2
1
(A) Since 2 > 0, the value of angle 𝜃 in the unit circle
1
is in I quadrant such that sin 𝜃 = 2
1
2
sin−1 ( ) =
(B) Since −
√3
2
𝜋
6
< 0, the value of angle 𝜃 in unit circle
is in IV quadrant and negative such that sin 𝜃 =
−
√3
2
sin−1 (−
𝜋
√3
)=−
2
3
𝜋 𝜋
(− , )
2 2
𝜋 𝜋
𝑦≠0
[− , ] ,
2 2
𝜋
[0, 𝜋],
𝑦≠
2
(A) sin−1 (−
√2
)
2
(B) sin−1 (1)
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2017 MATH 1112: DR.YOU
Example 2
Your Turn 2
Find the exact value of
Find the exact value of
√2
(B) cos−1 (−
(A) cos−1 ( 2 )
(A) Since
√2
2
1
(A) cos−1 (− 2)
cos−1 (
√3
2
√2
.
2
𝜋
√2
)=
2
4
< 0, the value of angle 𝜃 in unit circle
is in II quadrant such that cos 𝜃 = −
cos−1 (−
√3
.
2
5𝜋
√3
)=
2
6
Example 3
Your Turn 3
Find the exact value of
Find the exact value of
(A) tan−1(1)
(B) tan−1(−√3)
(A) Since 1 > 0, the value of angle 𝜃 in the unit circle
is in I quadrant such that tan 𝜃 = 1.
𝜋
tan−1 (1) =
4
(B) Since −√3 < 0, the value of angle 𝜃 in unit circle
is in IV quadrant and negative such that sin 𝜃 =
−
√2
(B) cos−1 ( 2 )
> 0, the value of angle 𝜃 in unit circle is
in II quadrant such that cos 𝜃 =
(B) Since −
√3
)
2
√3
2
tan−1 (−√3) = −
𝜋
3
(A) tan−1(−1)
1
(B) tan−1 ( 3)
√
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2017 MATH 1112: DR.YOU
Example 4
Your Turn 4
Use a calculator to find exact value of the following in
Use a calculator to find exact value of the following in
degree. Round four decimal place.
radian. Round four decimal place.
4
12
(A) sin−1 (9)
(B) cos−1 (− 13)
(C) sin−1 (1.2)
(D) tan−1(4)
(A) sin−1 (9)
(B) cos−1 (−0.6)
(C) cos−1 (2)
(D) tan−1(6)
5
4
(A) sin−1 (9) ≈ 26.3978°
12
(B) cos−1 (− 13) ≈ 157.3801°
(C) sin−1 (1.2) is undefined: error since there are no
angles 𝜃 such that sin 𝜃 = 1.2.
(D) tan−1(4) = 75.9638°
Example 5
Your Turn 5
Use a calculator to find exact value of the following in
Use a calculator to find exact value of the following in
degree. Round four decimal place.
radian. Round four decimal place.
3
(A) csc−1 (2)
5
(B) sec−1 (− 3)
3
(A) We find angle 𝜃 such that csc 𝜃 = 2
2
2
⇒ 𝜃 = sin−1 ( )
3
3
3
2
csc−1 ( ) = sin−1 ( ) = 41.8103°
2
3
sin 𝜃 =
5
(B) We find angle 𝜃 such that sec 𝜃 = − 3
3
3
⇒ 𝜃 = cos−1 (− )
5
5
5
3
sec−1 (− ) = cos−1 (− ) = 126.8699°
3
5
cos 𝜃 = −
(A) csc−1 (1.2)
(B) cot−1 (3)
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2017 MATH 1112: DR.YOU
THE RELATION BETWEEN INVERSE FUNCTIONS
Functions
relation
sin−1(sin 𝜃) = 𝜃 ,
−
Sine function
Cosine function
𝜋
𝜋
≤𝜃≤
2
2
sin(sin−1 𝐴) = 𝐴,
−1 ≤ 𝐴 ≤ 1
cos−1(cos 𝜃) = 𝜃 ,
0≤𝜃≤𝜋
cos(cos−1 𝐴) = 𝐴,
−1 ≤ 𝐴 ≤ 1
tan−1 (tan 𝜃) = 𝜃,
−
tan(tan−1 𝐴) = 𝐴,
−∞ < 𝐴 < ∞
Tangent function
𝜋
𝜋
<𝜃<
2
2
Example 6
Your Turn 6
Find the exact value of
Find the exact value of
𝜋
2𝜋
(A) sin−1 (sin ( 3 ))
(C) sin−1 (sin (−
(B) sin−1 (sin ( 3 ))
2𝜋
))
3
𝜋
𝜋
(A) We find an angle 𝜃 such that − 2 ≤ 𝜃 ≤ 2 .
𝜋
𝜋
√3
sin−1 (sin ( )) = sin−1 ( ) =
3
2
3
𝜋
𝜋
(B) We find an angle 𝜃 such that − 2 ≤ 𝜃 ≤ 2 .
2𝜋
√3
√3
sin−1 (sin ( )) = sin−1 ( ) =
3
2
2
2𝜋
𝜋
sin−1 (sin ( )) =
3
3
𝜋
𝜋
(C) We find an angle 𝜃 such that − 2 ≤ 𝜃 ≤ 2 .
sin−1 (sin (−
2𝜋
𝜋
√3
)) = sin−1 (− ) = −
3
2
3
(Warning: not
5𝜋
3
𝜋
𝜋
since − 2 ≤ 𝜃 ≤ 2 )
4𝜋
(A) sin−1 (sin ( 3 ))
𝜋
(B) sin−1 (sin (− 6 ))
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Example 7
Your Turn 7
Find the exact value of
Find the exact value of
5𝜋
(A) tan−1 (tan ( 4 ))
𝜋
(B) tan−1 (tan (− 3 ))
𝜋
2𝜋
(A) tan−1 (tan ( 3 ))
(B) tan−1 (tan (−
11𝜋
))
6
𝜋
(A) We find an angle 𝜃 such that − 2 ≤ 𝜃 ≤ 2 .
5𝜋
𝜋
tan−1 (tan ( )) = tan−1 (1) = −
4
4
𝜋
𝜋
(B) We find an angle 𝜃 such that − 2 ≤ 𝜃 ≤ 2 .
𝜋
𝜋
tan−1 (tan (− )) = tan−1(−√3) = −
3
3
Example 8
Your Turn 8
Find the exact value of
Find the exact value of
𝜋
(A) cos−1 (cos (− 4 ))
3𝜋
(B) cos−1 (cos ( 4 ))
(A) We find an angle 𝜃 such that 0 ≤ 𝜃 ≤ 𝜋.
𝜋
4
cos−1 (cos (− )) = cos−1 (−
3𝜋
√2
)=
2
4
(B) We find an angle 𝜃 such that 0 ≤ 𝜃 ≤ 𝜋.
cos−1 (cos (
3𝜋
3𝜋
√2
)) = cos−1 (− ) =
4
2
4
5𝜋
(A) cos−1 (cos ( 4 ))
5𝜋
(B) cos−1 (cos ( 6 ))
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2017 MATH 1112: DR.YOU
Example 9
Your Turn 9
Find the exact value of
Find the exact value of
cos−1 (sin (
5𝜋
))
4
(A) sin−1 (cos (−
7𝜋
))
6
We find an angle 𝜃 such that 0 ≤ 𝜃 ≤ 𝜋.
5𝜋
3𝜋
√2
cos−1 (sin ( )) = cos −1 (− ) =
4
2
4
𝜋
(B) cos−1 (tan (− 4 ))
7𝜋
(C) cos−1 (sin ( 6 ))
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Example 10
Your Turn 10
Find the exact value of
Find the exact value of
5
5
(A) cos (tan−1 (12))
(B) tan (cos −1 (− 13))
5
𝜋
(A) Let 𝜃 = tan−1 (12). Then − 2 ≤ 𝜃 ≤
5
(A) cos (tan−1 (12))
3
(B) tan (cos −1 (− 5))
𝜋
2
We find a point in the terminal side of 𝜃
𝑥 = 12, 𝑦 = 5 in I quadrant
𝑥
𝑦
𝑟
12
5
13
5
12
cos (tan−1 ( )) = cos 𝜃 =
12
13
(C) cot (cos −1 (−
(B) Let 𝜃 = cos−1 (−
5
).
13
Then 0 ≤ 𝜃 ≤ 𝜋
We find a point in the terminal side of 𝜃
𝑥 = −5, 𝑟 = 13 in II quadrant
𝑥
𝑦
𝑟
−5
12
13
tan (cos−1 (−
5
12
)) = tan 𝜃 = −
13
5
7
))
25
3
4
(D) sin (tan−1 (− ))
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2017 MATH 1112: DR.YOU
LECTURE 2-2 TRIGONOMETRY IDENTITIES AND SUM/DIFFERENCE FORMULA
FUNDAMENTAL IDENTITIES
tan 𝜃 =
sin 𝜃
cos 𝜃
cot 𝜃 =
sin2 𝑥 + cos2 𝑥 = 1
1
cos 𝜃
=
tan 𝜃 sin 𝜃
csc 𝜃 =
1
sin 𝜃
tan2 𝑥 + 1 = sec 2 𝑥
sec 𝜃 =
cot 2 𝑥 + 1 = csc 2 𝑥
EVEN/ODD FUNCTIONS
Odd: Symmetry with respect to origin
Even: Symmetry with respect to 𝒚 -axis
sin(−𝜃) = − sin(𝜃)
csc(−𝜃) = − csc(𝜃)
tan(−𝜃) = − tan(𝜃)
cot(−𝜃) = − cot(𝜃)
cos(−𝜃) = cos(𝜃)
sec(−𝜃) = sec(−𝜃)
Example 1
Your Turn 1
Find the exact value of the expression.
Find the exact value of the expression.
(A) sin2(50°) + cos2(50°)
(A) cos2 (179°) + sin2(179°)
(B) tan2 (49°) − sec 2(49°)
(A) sin2(50°) + cos2(50°) = 1
(B) tan2 (49°) − sec 2(49°)
= sec
2 (49°)
− 1 − sec
2 (49°)
1
cos 𝜃
(B) sec 2(130°) − tan2 (130°)
= −1
(C) csc 2(68°) − cot 2(68°)
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2017 MATH 1112: DR.YOU
Example 2
Your Turn 2
Find the exact value of the expression.
Find the exact value of the expression.
(A) sin(−30°)
(B) cos(−30°)
(A) sin(−45°)
(B) cos(−120°)
(C) tan(−30°)
(D) sec(−60°)
1
(A) sin(−30°) = − sin(30°) = − 2
(B) cos(−30°) = cos(30°) =
√3
2
Sum and
sin(𝑎 + 𝑏) = sin 𝑎 cos 𝑏 + cos 𝑎 sin 𝑏
difference
sin(𝑎 − 𝑏) = sin 𝑎 cos 𝑏 − cos 𝑎 sin 𝑏
Formulas
cos(𝑎 + 𝑏) = cos 𝑎 cos 𝑏 − sin 𝑎 sin 𝑏
cos(𝑎 − 𝑏) = cos 𝑎 cos 𝑏 + sin 𝑎 sin 𝑏
tan(𝑎 + 𝑏) =
tan 𝑎 + tan 𝑏
1 − tan 𝑎 tan 𝑏
tan(𝑎 − 𝑏) =
tan 𝑎 − tan 𝑏
1 + tan 𝑎 tan 𝑏
Example 3
Your Turn 3
Find the exact value of each expression.
Find the exact value of each expression.
(A) sin(20°) cos(10°) + cos(20°) sin(10°)
(A) sin(40°) cos(50°) + cos(40°) sin(50°)
(B) sin(20°) cos(80°) − cos(20°) sin(80°)
sin(20°) cos(10°) + cos(20°) sin(10°)
= sin(20° + 10°) = sin(30°)
1
(B) sin(79°) cos(31°) − cos(79°) sin(31°)
=2
sin(20°) cos(80°) − cos(20°) sin(80°)
= sin(20° − 80°) = sin(−60°)
=−
√3
2
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2017 MATH 1112: DR.YOU
Example 4
Your Turn 4
Find the exact value of each expression.
Find the exact value of each expression.
(A) cos(70°) cos(20°) + sin(20°) sin(20°)
(B) cos(50°) cos(10°) − sin(50°) sin(10°)
3𝜋
𝜋
3𝜋
𝜋
𝜋
𝜋
𝜋
𝜋
(A) cos ( 8 ) cos ( 8 ) + sin ( 8 ) sin ( 8 )
cos(70°) cos(25°) + sin(70°) sin(25°)
= cos(70° − 25°) = cos(45°)
=
√2
2
(B) cos (12) cos ( 4 ) − sin (12) sin ( 4 )
cos(50°) cos(10°) − sin(50°) sin(10°)
= cos(50° + 10°) = cos(60°)
=
1
2
Example 5
Your Turn 5
Find the exact value of each expression.
Find the exact value of each expression.
tan(20°) + tan(25°)
1 − tan(20°) tan(25°)
(A)
tan(40°)−tan(10°)
1+tan(40°) tan(10°)
(B)
5𝜋
𝜋
)+tan( )
12
4
5𝜋
𝜋
1−tan( ) tan( )
12
4
tan(20°) + tan(25°)
1 − tan(20°) tan(25°)
= tan(20° + 25°) = tan(45°)
=1
tan(
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2017 MATH 1112: DR.YOU
Example 6
Your Turn 6
Find the exact value of
Find the exact value of
sin(30° + 45°)
sin(60° − 45°)
sin(30° + 45°)
= sin(30°) cos(45°) + cos(30°) sin(45°)
=
1 √2 √3 √2
∙
+
∙
2 2
2 2
=
√2 + √6
4
Example 7
Your Turn 7
Find the exact value of
Find the exact value of
cos(135° + 60°)
cos(45° − 30°)
cos(135° + 60°)
= cos(135°) cos(60°) − sin(135°) sin(60°)
= (−
=
√2 1 √2 √3
∙
)∙ −
2
2
2 2
−√2 + √6
4
Example 8
Your Turn 8
Find the exact value of
Find the exact value of
tan(30° − 45°)
tan(30° − 45°) =
=
=
tan(30°) − tan(45°)
1 + tan(30°) tan(45°)
√3
−1
3
√3
1+ ∙1
3
√3−3
√3
= 3+
−12−6√3
6
=
(√3−3)(3−√3)
(3+√3)(3−√3)
= −2 − √3
tan(45° − 60°)
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2017 MATH 1112: DR.YOU
Example 9
Your Turn 9
Find the exact value of
Find the exact value of
sin(105°)
sin(75°)
sin(105°)
= sin(60° + 45°)
= sin(60°) cos(45°) + cos(60°) sin(45°)
=
√3 √2 1 √2
∙
+ ∙
2 2
2 2
=
√6 + √2
4
Example 10
Your Turn 10
Find the exact value of
Find the exact value of
cos(285°)
cos(285°)
= cos(240° + 45°)
= cos(240°) cos(45°) − sin(240°) sin(45°)
1 √2
√3 √2
= (− ) ∙
− (− ) ∙
2
2
2
2
=
−√2 + √6
4
cos (
7𝜋
)
12
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2017 MATH 1112: DR.YOU
Example 11
Your Turn 11
Find the exact value of
Find the exact value of
tan(165°)
tan(15°)
tan(15°)
= tan(45° − 30°)
=
tan(45°) − tan(30°)
1 + tan(45°) tan(30°)
=
√3
3
√3
1+ ∙1
3
=
3−√3
3+√3
=
(3−√3)(3−√3)
=
12+6√3
6
1−
(3+√3)(3−√3)
= 2 + √3
Your Turn
Your Turn
Find the exact value of
Find the exact value of
17𝜋
sin (
)
12
cos (−
5𝜋
)
12
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2017 MATH 1112: DR.YOU
Example 12
Your Turn 12
Simplify the expression by using sum and difference
Simplify the expression by using sum and difference
formula.
formula.
sin(7𝑥) cos(3𝑥) + cos(7𝑥) sin(3𝑥)
(A) cos(2𝑥) cos(5𝑥) + sin(2𝑥) sin(5𝑥)
sin(7𝑥) cos(3𝑥) + cos(7𝑥) sin(3𝑥)
(B) sin(5𝑥) cos(2𝑥) − cos(5𝑥) sin(2𝑥)
= sin(7𝑥 + 3𝑥)
= sin(10𝑥)
Example 13
Your Turn 13
Find the exact value of sin(𝛼 + 𝛽) when
Find the exact value of sin(𝛼 − 𝛽) when
3
sin 𝛼 = 5 , 0 < 𝛼 <
𝜋
2
12
𝜋
and cos 𝛽 = 13 , − 2 < 𝛽 < 0
We find points on terminal side which is corresponding to
angles 𝛼 and 𝛽.
𝑥
𝑦
𝑟
𝛼:I
+4
+3
5
𝛽 : IV
+12
−5
13
quadrant
sin(𝛼 + 𝛽)
= sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽
3 12
4
5
= ( ) ( ) + ( ) (− )
5 13
5
13
=
16
65
4
cos 𝛼 = 5 , 0 < 𝛼 <
𝜋
2
3
𝜋
and sin 𝛽 = − 5 , − 2 < 𝛽 < 0.
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2017 MATH 1112: DR.YOU
Your Turn
Your Turn
Find the exact value of cos(𝛼 + 𝛽) when
Find the exact value of cos(𝛼 − 𝛽) when
4
tan 𝛼 = − 3 ,
𝜋
2
1
𝜋
< 𝛼 < 𝜋 and cos 𝛽 = 2 , 0 < 𝛽 < 2 .
Example 14
5
𝜋
Your Turn 14
𝜋
2
Verify the identity sin ( + 𝑥) = cos 𝑥
3
sin 𝛼 = − 13 , − 2 < 𝛼 < 0 and cos 𝛽 = 5 , 0 < 𝛽 <
Verify the identity: cos(𝜋 − 𝑥) = − cos 𝑥
𝜋
2
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Example 15
Your Turn 15
Find the exact value of
Find the exact value of
1
3
sin [cos−1 ( ) + sin−1 (− )]
2
5
1
4
5
cos [tan−1 ( ) + cos −1 ( )]
3
13
3
Let 𝛼 = cos −1 (2) and 𝛽 = sin−1 (− 5).
We find points on terminal side which is corresponding to
angles 𝛼 and 𝛽.
𝑥
𝑦
𝑟
𝛼:I
+1
+√3
2
𝛽 : IV
+4
−3
5
quadrant
1
3
sin [cos−1 ( ) + sin−1 (− )]
2
5
= sin(𝛼 + 𝛽)
= sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽
=(
=
1
3
√3 4
) ( ) + ( ) (− )
2
5
2
5
4√3 − 3
10
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2017 MATH 1112: DR.YOU
Your Turn
Your Turn
Find the exact value of
Find the exact value of
4
5
cos [tan−1 ( ) + cos−1 ( )]
3
13
3
4
sin [sin−1 ( ) + cos−1 (− )]
5
5
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2017 MATH 1112: DR.YOU
LECTURE 2-3 HALF/DOUBLE FORMULA
sin(2𝜃) = 2 sin 𝜃 cos 𝜃
Double angle
Formulas
tan(2𝜃) =
cos(2𝜃) = cos 2 𝜃 − sin2 𝜃
Half angle
cos(2𝜃) = 2 cos2 𝜃 − 1
𝜃
1 − cos 𝜃
sin ( ) = ±√
2
2
Formulas
2 tan 𝜃
1 − tan2 𝜃
cos(2𝜃) = 1 − 2 sin2 𝜃
𝜃
1 − cos 𝜃
tan ( ) =
2
sin 𝜃
𝜃
1 + cos 𝜃
cos ( ) = ±√
2
2
𝜃
(the sign is determined by the quadrant of 2 )
Example 1
Your Turn 1
Find the exact value of cos(2𝜃) if
Find the exact value of cos(2𝜃) if
sin(𝜃) =
5
13
3
(A) cos(𝜃) = − 5
cos(2𝜃)
= 1 − 2 sin2 𝜃
3
5 2
= 1 − 2( )
13
=
(B) sin(𝜃) = 7
119
169
Example 2
Your Turn 2
Find the exact value of sin(2𝜃) if
Find the exact value of sin(2𝑥) if
3
sin(𝜃) = ,
5
𝜋
<𝜃<𝜋
2
cos(𝜃) =
We find a point on terminal side which is corresponding
to angle 𝜃.
quadrant
𝜃 : II
𝑥
𝑦
𝑟
−4
+3
5
3
4
24
sin(2𝜃) = 2 sin(𝜃) cos(𝜃) = 2 ( ) (− ) = −
5
5
25
12
,
13
0<𝜃<
𝜋
2
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2017 MATH 1112: DR.YOU
Example 3
Your Turn 3
Find the exact value of 2 sin(15°) cos(15°)
Find the exact value of 2 sin(22.5°) cos(22.5°)
2 sin(15°) cos(15°)
= sin(2 ∙ 15°)
= sin(30°)
1
=2
Example 4
Your Turn 4
Find the exact value of cos2 (75°) − sin2(75°)
Find the exact value of
(A) cos2 (105°) − sin2(105°)
cos2 (75°) − sin2(75°)
= cos(2 ∙ 75°)
(B) 2 cos2 (15°) − 1
= cos(150°)
=
√3
2
(C) 1 − 2 sin2(22.5°)
Example 5
Your Turn 5
Find the exact value of sin(−22.5°) by half-angle
Find the exact value of cos(15°) by half-angle formula.
formula.
Since −22.5° is in IV quadrant, the value of sine is
negative;
−45°
sin(−22.5°) = sin (
)
2
=
1 − cos(−45°)
1 − √2⁄2
√
= −√
2
2
𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞
−
⏟
√2 − √2
2 − √2
= −√
=−
4
2
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2017 MATH 1112: DR.YOU
Your Turn
Your Turn
Find the exact value of sin(75°) by half-angle formula.
Find the exact value of cos(112.5°) by half-angle
formula.
Example 6
Your Turn 6
𝜋
8
Find the exact value of sin ( ) by half-angle formula.
𝜋
12
Find the exact value of cos ( ) by half-angle formula.
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2017 MATH 1112: DR.YOU
Example 7
Your Turn 7
Find the exact value of tan(15°) by half-angle formula.
Find the exact value of tan(22.5°) by half-angle formula.
30°
tan(15°) = tan ( )
2
1 − cos(30°)
=
sin(30°)
√3
1− 2
=
1
2
= 2 − √3
Sum to product
Formulas
Product to Sum
Formulas
𝑎+𝑏
𝑎−𝑏
) cos (
)
2
2
𝑎−𝑏
𝑎+𝑏
sin 𝑎 − sin 𝑏 = 2 sin (
) cos (
)
2
2
1
sin 𝑎 sin 𝑏 = {cos(𝑎 − 𝑏) − cos(𝑎 + 𝑏)}
2
1
sin 𝑎 cos 𝑏 = {sin(𝑎 + 𝑏) + sin(𝑎 − 𝑏)}
2
sin 𝑎 + sin 𝑏 = 2 sin (
𝑎+𝑏
𝑎−𝑏
cos 𝑎 + cos 𝑏 = 2 cos (
) cos (
)
2
2
𝑎+𝑏
𝑎−𝑏
cos 𝑎 − cos 𝑏 = −2 sin (
) sin (
)
2
2
1
cos 𝑎 cos 𝑏 = {cos(𝑎 − 𝑏) + cos(𝑎 + 𝑏)}
2
Example 8
Your Turn 8
Find the exact value of sin(75°) + sin(15°)
Find the exact value of cos(225°) − cos(195°)
sin(75°) + sin(15°)
= 2 sin (
75° + 15°
75° − 15°
) cos (
)
2
2
= 2 sin(45°) cos(30°)
= 2(
=
√2
2
√2 1
)( )
2
2
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2017 MATH 1112: DR.YOU
Example 9
Your Turn 9
Find the exact value of sin(195°) ∙ cos(75°)
Find the exact value of cos(285°) ∙ cos(195°)
sin(195°) ∙ cos(75°)
1
= {sin(195° + 75°) + sin(195° − 75°)}
2
1
= {sin(270°) + sin(120°)}
2
1
= 2 {−1 +
=
√3
}
2
−2+√3
4
Example 10
Your Turn 10
Simplify the expression by using sum to product formula
Simplify the expression by using sum to product formula
cos(5𝑥) − cos(𝑥)
sin(2𝑥) − sin(4𝑥)
sin(2𝑥) − sin(4𝑥)
= 2 sin (
2𝑥 − 4𝑥
2𝑥 + 4𝑥
) cos (
)
2
2
= 2 sin(−𝑥) cos(3𝑥)
= 2 sin(𝑥) cos(3𝑥)
Your Turn
Your Turn
Simplify the expression by using sum to product formula
Simplify the expression by using sum to product formula
sin(𝑥) + sin(3𝑥)
cos(𝑥) + cos(5𝑥)
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2017 MATH 1112: DR.YOU
Example 11
Your Turn 11
Simplify the expression by using product to sum formula
Simplify the expression by using product to sum formula
cos(10𝑥) ∙ sin(6𝑥)
sin(6𝑥) ∙ cos(2𝑥)
cos(10𝑥) ∙ sin(6𝑥)
= sin(6𝑥) ∙ cos(10𝑥)
1
= {sin(6𝑥 + 10𝑥) + sin(6𝑥 − 10𝑥)}
2
1
= {sin(16𝑥) + sin(−4𝑥)}
2
1
= {sin(16𝑥) − sin(4𝑥)}
2
Your Turn
Your Turn
Simplify the expression by using product to sum formula
Simplify the expression by using product to sum formula
cos(3𝑥) ∙ cos(5𝑥)
sin(4𝑥) ∙ sin(2𝑥)
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2017 MATH 1112: DR.YOU
LECTURE 2-4 IDENTITIES
Fundamental
Identities
Half angle
Formulas
tan 𝜃 =
sin 𝜃
cos 𝜃
cot 𝜃 =
cos 𝜃
sin 𝜃
cot 𝜃 =
1
tan 𝜃
csc 𝜃 =
sin2 𝜃 + cos 2 𝜃 = 1
tan2 𝜃 + 1 = sec 2 𝜃
𝜃
1 − cos 𝜃
sin ( ) = ±√
2
2
𝜃
1 + cos 𝜃
cos ( ) = ±√
2
2
1
sin 𝜃
sec 𝜃 =
1
cos 𝜃
cot 2 𝜃 + 1 = csc 2 𝜃
𝜃
1 − cos 𝜃
tan ( ) =
2
sin 𝜃
𝜃
Formulas
( the sign is determined by the quadrant of 2 )
2 tan 𝜃
sin(2𝜃) = 2 sin 𝜃 cos 𝜃
tan(2𝜃) =
1 − tan2 𝜃
Sum and
sin(𝑎 + 𝑏) = sin 𝑎 cos 𝑏 + cos 𝑎 sin 𝑏
difference
sin(𝑎 − 𝑏) = sin 𝑎 cos 𝑏 − cos 𝑎 sin 𝑏
Formulas
cos(𝑎 + 𝑏) = cos 𝑎 cos 𝑏 − sin 𝑎 sin 𝑏
Double angle
cos(2𝜃) = cos 2 𝜃 − sin2 𝜃
cos(2𝜃) = 2 cos2 𝜃 − 1
cos(𝑎 − 𝑏) = cos 𝑎 cos 𝑏 + sin 𝑎 sin 𝑏
Sum to product
Formulas
Product to Sum
Formulas
𝑎+𝑏
𝑎−𝑏
) cos (
)
2
2
𝑎−𝑏
𝑎+𝑏
sin 𝑎 − sin 𝑏 = 2 sin (
) cos (
)
2
2
1
sin 𝑎 sin 𝑏 = {cos(𝑎 − 𝑏) − cos(𝑎 + 𝑏)}
2
1
sin 𝑎 cos 𝑏 = {sin(𝑎 + 𝑏) + sin(𝑎 − 𝑏)}
2
sin 𝑎 + sin 𝑏 = 2 sin (
cos(2𝜃) = 1 − 2 sin2 𝜃
tan(𝑎 + 𝑏) =
tan 𝑎 + tan 𝑏
1 − tan 𝑎 tan 𝑏
tan(𝑎 − 𝑏) =
tan 𝑎 − tan 𝑏
1 + tan 𝑎 tan 𝑏
𝑎+𝑏
𝑎−𝑏
cos 𝑎 + cos 𝑏 = 2 cos (
) cos (
)
2
2
𝑎+𝑏
𝑎−𝑏
cos 𝑎 − cos 𝑏 = −2 sin (
) sin (
)
2
2
1
cos 𝑎 cos 𝑏 = {cos(𝑎 − 𝑏) + cos(𝑎 + 𝑏)}
2
Example 1
Your Turn 1
Establish the identity
Establish the identity
cos 𝜃 (tan 𝜃 + cot 𝜃) = csc 𝜃
csc 𝜃 ∙ cos 𝜃 = cot 𝜃
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2017 MATH 1112: DR.YOU
Example 2
Your Turn 2
Establish the identity
Establish the identity
cos2 𝜃 (1 + tan2 𝜃) = 1
(sec 𝜃 − 1)(sec 𝜃 + 1) = tan2 𝜃
Example 3
Your Turn 3
Establish the identity
Establish the identity
9 sec2 𝜃 − 5 tan2 𝜃 = 5 + 4 sec2 𝜃
3 sin2 𝜃 + 4 cos2 𝜃 = 3 + cos2 𝜃
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2017 MATH 1112: DR.YOU
Example 4
Your Turn 4
Establish the identity
Establish the identity
1 − sin 𝜃
= (sec 𝜃 + tan 𝜃)2
1 + sin 𝜃
1−
Example 5
Your Turn 5
Establish the identity
Establish the identity
1 − sin 𝜃
cos 𝜃
=
cos 𝜃
1 − sin 𝜃
sin2 𝜃
1 + cos 𝜃
= cos 𝜃
cos 𝜃
= sec 𝜃 − tan 𝜃
1 + sin 𝜃
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2017 MATH 1112: DR.YOU
REVIEW PROBLEMS FOR EXAM 2
1. Find the exact value of the following expression in radian.
(A) cos−1 (−
√3
)
2
(B) sin−1 (−
(C) tan−1 (√3)
√2
)
2
2. Find the exact value of the expression in radian without using a calculator.
14𝜋
))
3
9𝜋
(A) sin−1 (sin (
9𝜋
(B) cos−1 (cos ( 4 ))
(C) sin−1 (sin ( 8 ))
3. Find the exact value of the following expression.
3
12
(A) cot (sin−1 (− 5))
1
(C) sec (tan−1 (2))
(B) tan (cos −1 (− 13))
3
5
4. Find the exact value of the following expression: sin [sin−1 (5) − cos −1 (− 13)]
3
5. Given cos 𝑎 = − 5 ,
𝜋
2
5
≤ 𝑎 ≤ 𝜋 and sin 𝑏 = − 13 ,
24
3𝜋
2
≤ 𝑏 ≤ 2𝜋, find sin(𝑎 + 𝑏).
5
6. Given cos 𝑎 = − 25 with 𝑎 in quadrant II and sin 𝑏 = − 13 with 𝑏 in qudrant IV, find cos(𝑎 + 𝑏)
5𝜋
2𝜋
5𝜋
2𝜋
7. Find the exact value of sin ( 18 ) cos ( 18 ) − cos ( 18 ) sin ( 18 ).
8. Write sin(3𝑥) ∙ cos(4𝑥) + cos(3𝑥) ∙ sin(4𝑥) in terms of a single trigonometric function.
9. Write sin(2𝑥) cos(3𝑥) − cos(2𝑥) sin(3𝑥) in terms of a single trigonometric function.
10. Using a Sum or Difference Identity, find the exact value cos(105°)
4
5
𝜋
2
11. Find the exact value of cos(2𝜃) given sin 𝜃 = and 0 < 𝜃 < .
𝜋
12. Using a half-angle Identity, find the exact value sin (− 12)
13. Use a half-angle identity to evaluate tan(22.5°).
14. Find the exact value of
5𝜋
𝜋
3𝜋
(A) cos ( 12 ) cos (12)
15. Simplify:
(sin 𝜃+cos 𝜃)2 −1
sin 𝜃 cos 𝜃
16. Establish the identity:
𝜋
(B) sin ( 8 ) cos ( 8 )
1−sin 𝜃
cos 𝜃
cos 𝜃
+ 1−sin 𝜃 = 2 sec 𝜃.
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2017 MATH 1112: DR.YOU
SOLUTIONS:
√3
)
2
with
√2
)
2
with − 2 < 𝑥 < 0. Since sin 𝜃 = −
1. (A) Let 𝜃 = cos−1 (−
𝜋
2
< 𝑥 < 𝜋. Since cos 𝜃 = −
√3
2
and 𝜃 is in the second quadrant, 𝜃 =
5𝜋
6
by using
the unit circle.
(B) Let 𝜃 = sin−1 (−
𝜋
√2
2
and 𝜃 is negative in the fourth quadrant, 𝜃 =
𝜋
− 4 by using the unit circle.
𝜋
(C) Let 𝜃 = tan−1(√3) with 0 < 𝑥 < 2 . Since tan 𝜃 = √3 and 𝜃 is in the first quadrant, 𝜃 =
𝜋
3
by using the unit
circle.
2. (A) Since
angle
𝜋
.
3
14𝜋
3
𝜋
2
𝜋
2
is not in the interval − < 𝑥 < , we may not simply cancel the sine and sine inverse function. The
14𝜋 2𝜋
~
3
3
lies in quadrant II. Reflect this angle over the 𝑦-axis into quadrant I and its corresponding angle is
Then
14𝜋
2𝜋
𝜋
sin−1 (sin (
)) = sin−1 (sin ( )) =
3
3
3
(B) Since
9𝜋
4
The angle
9𝜋 𝜋
~
4
4
(C) Since
9𝜋
8
angle
9𝜋
8
𝜋
𝜋
is not in the interval − 2 < 𝑥 < 2 , we may not simply cancel the cosine and cosine inverse function.
9𝜋
4
𝜋
4
lies in quadrant I. Then cos−1 (cos ( )) = cos−1 (cos ( )) =
𝜋
2
𝜋
4
𝜋
2
is not in the interval − < 𝑥 < , we may not simply cancel the sine and sine inverse function. The
lies in quadrant III. Reflect this angle over the 𝑦-axis into quadrant IV and its corresponding angle is
𝜋
8
negative − . Then
9𝜋
𝜋
sin−1 (sin ( )) = −
8
8
3
5
𝜋
2
3. (A) Let 𝜃 = sin−1 (− ) with − < 𝑥 <
𝜋
2
: 𝜃 lies in quadrant IV ⇒ 𝑦 = −3, 𝑥 = 4 and 𝑟 = 5
3
𝑥
4
4
cot (sin−1 (− )) = =
=−
5
𝑦 −3
3
12
(B) Let 𝜃 = cos−1 (− 13) with 0 < 𝑥 < 𝜋 : 𝜃 lies in quadrant II ⇒ 𝑦 = 5, 𝑥 = −12 and 𝑟 = 13
tan (cos−1 (−
1
2
𝜋
2
(C) Let 𝜃 = tan−1 ( ) with − < 𝑥 <
𝜋
2
12
𝑦
5
)) = = −
13
𝑥
12
: 𝜃 lies in quadrant I ⇒ 𝑦 = 1, 𝑥 = 2 and 𝑟 = √12 + 22 = √5
1
𝑟 √5
sec (tan−1 ( )) = =
2
𝑥
2
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2017 MATH 1112: DR.YOU
3
5
4. Let 𝑎 = sin−1 (5) and 𝑏 = cos−1 (− 13)
quadrant
𝑥
𝑦
𝑟
𝑎: I quadrant
4
3
5
𝑏: II quadrant
−5
12
13
3
5
3
5
4 12
33
sin [sin−1 ( ) − cos−1 (− )] = sin(𝑎 + 𝑏) = sin 𝑎 cos 𝑏 + cos 𝑎 sin 𝑏 = ( ) (− ) + ( ) ( ) =
5
13
5
13
5 13
65
5. First, find the corresponding terminal points for each angle;
quadrant
𝑥
𝑦
𝑟
𝑎: II quadrant
−3
4
5
𝑏: IV quadrant
12
−5
13
4 12
3
5
33
sin(𝑎 + 𝑏) = sin 𝑎 cos 𝑏 + cos 𝑎 sin 𝑏 = ( ) ( ) + (− ) (− ) =
5 13
5
13
65
6. First, find the corresponding terminal points for each angle;
quadrant
𝑥
𝑦
𝑟
𝑎: II quadrant
−24
7
25
𝑏: IV quadrant
−5
12
13
cos(𝑎 + 𝑏) = cos 𝑎 cos 𝑏 − sin 𝑎 sin 𝑏 = (−
5𝜋
18
2𝜋
18
5𝜋
18
2𝜋
18
5𝜋
18
7. sin ( ) cos ( ) − cos ( ) sin ( ) = sin (
−
2𝜋
)
18
24
5
7 12
36
) (− ) − ( ) ( ) =
25
13
25 13
325
𝜋
6
= sin ( ) =
1
2
8. sin(3𝑥) ∙ cos(4𝑥) + cos(3𝑥) ∙ sin(4𝑥) = sin(3𝑥 + 4𝑥) = sin(7𝑥)
9. sin(2𝑥) cos(3𝑥) − cos(2𝑥) sin(3𝑥) = sin(2𝑥 − 3𝑥) = sin(−𝑥) = − sin(𝑥)
1
√2
√3
√2
10. cos(60° + 45°) = cos(60°) cos(45°) − sin(60°) sin(45°) = (2) ( 2 ) − ( 2 ) ( 2 ) =
4 2
32
7
11. cos(2𝜃) = 1 − 2 sin2 𝜃 = 1 − 2 (5) = 1 − 25 = − 25
−𝜋⁄6
)
2
𝜋
𝜋
12. sin (− 12) = sin (
(it is negative since − 6 lies in quadrant IV)
𝜋
6
1−cos(− )
= −√
2
45°
)
2
13. tan(22.5°) = tan (
=
1−
= −√
√3
2
2
1−cos(45°)
sin(45°)
= −√
=
√2
2
√2
2
1−
2−√3
4
=
=−
2−√2
√2
√2−√3
2
= √2 − 1
√2−√6
4
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2017 MATH 1112: DR.YOU
5𝜋
1
𝜋
4𝜋
6𝜋
1 1
1
14. (A) cos ( 12 ) cos (12) = 2 {cos (12) + cos (12)} = 2 {2 + 0} = 4
3𝜋
1
𝜋
4𝜋
2𝜋
1
(B) sin ( 8 ) cos (8 ) = 2 {sin ( 8 ) + sin ( 8 )} = 2 {1 +
15.
√2
2
}=
2+√2
4
(sin 𝜃+cos 𝜃)2 −1
sin 𝜃 cos 𝜃
=
=
sin2 𝜃+2 sin 𝜃 cos 𝜃+cos2 𝜃−1
sin 𝜃 cos 𝜃
2 sin 𝜃 cos 𝜃
sin 𝜃 cos 𝜃
by FOIL
since sin2 𝜃 + cos2 𝜃 = 1
=2
16.
1−sin 𝜃
cos 𝜃
+
cos 𝜃
1−sin 𝜃
=
(1−sin 𝜃)2
cos 𝜃(1−sin 𝜃)
=
1−2sin 𝜃+sin2 𝜃
cos 𝜃(1−sin 𝜃)
=
1−2 sin 𝜃+sin2 𝜃+cos2 𝜃
cos 𝜃(1−sin 𝜃)
2−2 sin 𝜃
= cos 𝜃(1−sin 𝜃)
2(1−sin 𝜃)
+
(cos 𝜃)2
cos 𝜃(1−sin 𝜃)
cos2 𝜃
+ cos 𝜃(1−sin 𝜃)
: find the least common denominator
: FOIL
: Make it one fraction and simplify
: since sin2 𝜃 + cos2 𝜃 = 1
= cos 𝜃(1−sin 𝜃) : Factor the numerator and simplify
2
= cos 𝜃
= 2 sec 𝜃
77