inverse circular functions.pdf
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Inverse Circular Functions and Equations Involving Them
Institute of Mathematics, University of the Philippines
Lecture 26
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1 Inverse Circular Functions
2 Equations involving Inverse Circular Functions
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Recall: Circular functions are not one-to-one
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Recall: Circular functions are not one-to-one
We restrict each of their domains such that:
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Recall: Circular functions are not one-to-one
We restrict each of their domains such that:
the circular function is one-to-one on the restricted domainthe range of the restricted circular function is the same
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Inverse sine function
−1
1
−π−2π π 2π−
π
2−
3π
2
π
2
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Inverse sine function
−1
1
−π−2π π 2π−
π
2−
3π
2
π
2
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Inverse sine function
−1
1
−π−2π π 2π−
π
2−
3π
2
π
2
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Inverse sine function
−1
1
−π−2π π 2π−
π
2−
3π
2
π
2
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Inverse sine function
−1
1
−π−2π π 2π−
π
2−
3π
2
π
2
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Definition
Inverse sine function:
For any x ∈ [−1, 1],
y = sin−1 x ⇔ x = sin y, y ∈− π
2, π
2
π
2
−
π
2
−1 1
dom sin−1 : [−1, 1] ran sin−1 =−π
2, π
2
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1. sin−1
√ 3
2
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1. sin−1
√ 3
2
y = sin−1
√ 3
2
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1. sin−1
√ 3
2
y = sin−1
√ 3
2
⇒ sin y =
√ 3
2 , y ∈
− π
2, π
2
y = π
3
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1. sin−1
√ 3
2
y = sin−1
√ 3
2
⇒ sin y =
√ 3
2 , y ∈
− π
2, π
2
y = π
32. sin−1
−1
2
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1. sin−1
√ 3
2
y = sin−1
√ 3
2
⇒ sin y =
√ 3
2 , y ∈
− π
2, π
2
y = π
32. sin−1
−1
2
y = sin−1−1
2
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1. sin−1
√ 3
2
y = sin−1
√ 3
2
⇒ sin y =
√ 3
2 , y ∈
− π
2, π
2
y = π
32. sin−1
−1
2
y = sin−1−1
2 ⇒ sin y = −1
2, y ∈ −
π
2, π
2
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1. sin−1
√ 3
2
y = sin−1
√ 3
2
⇒ sin y =
√ 3
2 , y ∈
− π
2, π
2
y = π
32. sin−1
−1
2
y = sin−1
−1
2 ⇒ sin y = −1
2, y ∈ −
π
2, π
2 y = −π
6
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3. sin−1
−√
2
2
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3. sin−1
−√
2
2
= −π
4
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3. sin−1
−√
2
2
= −π
4
4. sin−1 0
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3. sin−1
−√
2
2
= −π
4
4. sin−1 0 = 0
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3. sin−1
−√
2
2
= −π
4
4. sin−1 0 = 0
5. sin−1(−1)
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3. sin−1
−√
2
2
= −π
4
4. sin−1 0 = 0
5. sin−1(−1) = −
π
2
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3. sin−1
−√
2
2
= −π
4
4. sin−1 0 = 0
5. sin−1(−1) = −
π
26. sin−1 2
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3. sin−1
−√
2
2
= −π
4
4. sin−1 0 = 0
5. sin−1(−1) = −
π
26. sin−1 2: undefined
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−1
1
−π−2π π 2π−
π
2−
3π
2
π
2
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−1
1
−π−2π π 2π−
π
2−
3π
2
π
2
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−1
1
−π−2π π 2π−
π
2−
3π
2
π
2
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−1
1
−π−2π π 2π−
π
2−
3π
2
π
2
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−1
1
−π−2π π 2π−
π
2−
3π
2
π
2
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Definition
Inverse cosine function:
For any x ∈ [−1, 1],
y = cos−1 x ⇔ x = cos y, y ∈ [0, π]
π
2
π
−1 1
dom cos−1 : [−1, 1] ran cos−1 = [0, π]
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1. cos−1
√ 3
2
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1. cos−1
√ 3
2
y = cos−1√ 3
2
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1. cos−1
√ 3
2
y = cos−1√ 3
2
⇒ cos y = √ 3
2 , y ∈ [0, π]
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1. cos−1
√ 3
2
y = cos−1√ 3
2
⇒ cos y = √ 3
2 , y ∈ [0, π]
y = π
6
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1. cos−1
√ 3
2
y = cos−1√ 3
2
⇒ cos y = √ 3
2 , y ∈ [0, π]
y = π
6
2. cos−1−√ 2
2
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1. cos−1
√ 3
2
y = cos−1√ 3
2
⇒ cos y = √ 3
2 , y ∈ [0, π]
y = π
6
2. cos−1−√ 2
2
= 3π
4
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√
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1. cos−1
√3
2
y = cos−1√ 3
2
⇒ cos y = √ 3
2 , y ∈ [0, π]
y = π
6
2. cos−1−√ 2
2
= 3π
4
3. cos−1(−1)
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√
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1. cos−1
√3
2
y = cos−1√ 3
2
⇒ cos y = √ 3
2 , y ∈ [0, π]
y = π
6
2. cos−1−√ 2
2
= 3π
4
3. cos−1(−1) = π
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√
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1. cos−1
√3
2
y = cos−1√ 3
2
⇒ cos y = √ 3
2 , y ∈ [0, π]
y = π
6
2. cos−1−√ 2
2
= 3π
4
3. cos−1(−1) = π
4. cos−1−
1
2
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√ 3
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1. cos−1
√3
2
y = cos−1√ 3
2
⇒ cos y = √ 3
2 , y ∈ [0, π]
y = π
6
2. cos−1−√ 2
2
= 3π
4
3. cos−1(−1) = π
4. cos−1−
1
2 =
2π
3
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√ 3
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1. cos−1
√3
2
y = cos−1√ 3
2
⇒ cos y = √ 3
2 , y ∈ [0, π]
y = π
6
2. cos−1−√ 2
2
= 3π
4
3. cos−1(−1) = π
4. cos−1−
1
2 =
2π
3
5. cos−1 0
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√ 3
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1. cos−1
√3
2
y = cos−1√ 3
2
⇒ cos y = √ 3
2 , y ∈ [0, π]
y = π
6
2. cos−1−√ 2
2
= 3π
4
3. cos−1(−1) = π
4. cos−1−
1
2 =
2π
3
5. cos−1 0 = π
2
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y,
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ ,
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
y = cot−1 x ⇔ x = cot y,
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
y = cot−1 x ⇔ x = cot y, x ∈ ,
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)
y = sec−1
x ⇔ x = sec y,
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)
y = sec−1
x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞),
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)
y = sec−1
x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈ [0, π] \ π
2
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)
y = sec−1
x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈ [0, π] \ π
2
y = csc−1 x ⇔ x = csc y,
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)
y = sec−1
x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈ [0, π] \ π
2
y = csc−1 x ⇔ x = csc y, x ∈ (−∞,−1] ∪ [1, +∞),
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)
y = sec−1
x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈ [0, π] \ π
2
y = csc−1 x ⇔ x = csc y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈−π
2, π
2
\ {0}
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Th h i i l f i d fi d i il l
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The other inverse circular functions are defined similarly:
y = tan−1 x ⇔ x = tan y, x ∈ , y ∈
−π
2, π
2
y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)
y = sec−1
x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈ [0, π] \ π
2
y = csc−1 x ⇔ x = csc y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈−π
2, π
2
\ {0}
The sin−1, cos−1, tan−1, . . . functions are also denoted
Arcsin, Arccos, Arctan, . . ..
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Inverse tangent function:
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x = π
2 x = −π
2
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Inverse tangent function:
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x = π
2 x = −π
2
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Inverse tangent function:
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y = π
2
y = −π
2
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Inverse tangent function:
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y = π
2
y = −π
2
dom tan−1 = ran tan−1 =−π
2, π
2
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Inverse cotangent function:
y = π
dom cot−1 = ran cot−1 = (0, π)
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Inverse secant function:
y = π
2
(1, 0)
(−1, π)
dom sec−1 = (−∞,−1] ∪ [1, +∞) ran sec−1 = [0, π] \π
2
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Inverse cosecant function:
(1, π2)
(−1,−π
2)
dom csc−1 = (−∞,−1] ∪ [1, +∞) ran csc−1 =−π
2, π
2
\ {0}
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π
2
sin−1 tan−1 csc−1
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sin , tan , csc
−π
2
0
cos−1, cot−1, sec−1
π
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π
2
sin−1 tan−1 csc−1
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sin , tan , csc
−π
2
0
cos−1, cot−1, sec−1
π
1 tan−1√ 3
3
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1 tan−1 csc−1
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sin , tan , csc
−π
2
0
cos−1, cot−1, sec−1
π
1 tan−1√ 3
3
= π
6
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−1
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sin , tan , csc
−π
2
0
cos−1, cot−1, sec−1
π
1 tan−1√ 3
3
= π
6
2 tan−1(−1)
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−1
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, ,
−π
2
0
cos−1, cot−1, sec−1
π
1 tan−1√ 3
3
= π
6
2 tan−1(−1) = −π
4
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−11 1 1
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, ,
−π
2
0
cos−1, cot−1, sec−1
π
1 tan−1√ 3
3
= π6
2 tan−1(−1) = −π
4
3 cot−1 −√ 3
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−11 1 1
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−π
2
0
cos−1, cot−1, sec−1
π
1 tan−1√ 3
3
= π6
2 tan−1(−1) = −π
4
3 cot−1 −√ 3 =
5π
6
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−11 1 1
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−π
2
0
cos−1, cot−1, sec−1
π
1 tan−1√ 3
3
= π6
2 tan−1(−1) = −π
4
3
cot−1 −√ 3 =
5π
6
4 Arccot 0
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−11 t 1 1
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−π
2
0
cos−1, cot−1, sec−1
π
1 tan−1√ 33
= π6
2 tan−1(−1) = −π
4
3
cot−1 −√ 3 =
5π
6
4 Arccot 0 = π
2
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−1−1 t−1 −1
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−π
2
0
cos 1, cot 1
, sec 1
π
1 tan−1√
33
= π6
2 tan−1(−1) = −π
4
3
cot−1 −√ 3 =
5π
6
4 Arccot 0 = π
2
5 sec−1 −√ 2
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−1
cos−1 cot−1 sec−1
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−π
2
0
cos 1, cot 1
, sec 1
π
1 tan−1√
33
= π6
2 tan−1(−1) = −π
4
3
cot−1 −√ 3 =
5π
6
4 Arccot 0 = π
2
5 sec−1 −√ 2 =
3π
4
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−1
cos−1 cot−1 sec−1
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−π
2
0
cos , cot , sec
π
1 tan−1√
33
= π6
2 tan−1(−1) = −π
4
3
cot−1 −√ 3 =
5π
6
4 Arccot 0 = π
2
5 sec−1 −√ 2 =
3π
4
6 sec−1(−2)
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−1
cos−1 cot−1 sec−1
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−π
2
0
cos , cot , sec
π
1 tan−1√
33
= π6
2 tan−1(−1) = −π
4
3
cot−1 −√ 3 =
5π
6
4 Arccot 0 = π
2
5 sec−1 −√ 2 =
3π
4
6 sec−1(−2) = 2π
3
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−1
cos−1 cot−1 sec−1
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−π
2
0
cos , cot , sec
π
1 tan−1√
33
= π6
2 tan−1(−1) = −π
4
3
cot−1 −√ 3 =
5π
6
4 Arccot 0 = π
2
5 sec−1 −√ 2 =
3π
4
6 sec−1(−2) = 2π
3
7 Arccsc√
2
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−1
cos−1, cot−1, sec−1
7/24/2019 Inverse Circular Functions.pdf
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−π
2
0
cos , cot , sec
π
1 tan−1√
33
= π6
2 tan−1(−1) = −π
4
3
cot−1 −√ 3 =
5π
64 Arccot 0 =
π
2
5 sec−1 −√ 2 =
3π
4
6 sec−1(−2) = 2π
3
7 Arccsc√
2 = π
4
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
π
2
sin−1, tan−1, csc−1
cos−1, cot−1, sec−1
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−π
2
0
cos , cot , sec
π
1 tan−1√
33
= π6
2 tan−1(−1) = −π
4
3
cot−1 −√ 3 =
5π
64 Arccot 0 =
π
2
5 sec−1 −√ 2 =
3π
4
6 sec−1(−2) = 2π
3
7 Arccsc√
2 = π
4
8 csc−1
−2√ 3
3
(IMath, UP) Inverse Circular Functions Lec. 26 16 / 28
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Recall: (f −1 ◦ f )(x) = x for every x ∈ dom f
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(IMath UP) Inverse Circular Functions Lec 26 17 / 28
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Recall: (f −1 ◦ f )(x) = x for every x ∈ dom f
Thus
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Thus,
sin−1(sin x) = x for every x ∈ −π
2 , π
2
cos−1(cos x) = x for every x ∈ [0, π]
(IMath UP) Inverse Circular Functions Lec 26 17 / 28
Recall: (f −1 ◦ f )(x) = x for every x ∈ dom f
Thus,
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Thus,
sin−1(sin x) = x for every x ∈ −π
2 , π
2
cos−1(cos x) = x for every x ∈ [0, π]
tan−1(tan x) = x for every x ∈−π
2, π
2
cot−1
(cot x) = x for every x ∈ (0, π)sec−1(sec x) = x for every x ∈ [0, π] \
π
2
csc−1(csc x) = x for every x ∈
−π
2, π
2
\ {0}
(IMath UP) Inverse Circular Functions Lec 26 17 / 28
Recall: (f −1 ◦ f )(x) = x for every x ∈ dom f
Thus,
7/24/2019 Inverse Circular Functions.pdf
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,
sin−1(sin x) = x for every x ∈ −π
2 , π
2
cos−1(cos x) = x for every x ∈ [0, π]
tan−1(tan x) = x for every x ∈−π
2, π
2
cot−1
(cot x) = x for every x ∈ (0, π)sec−1(sec x) = x for every x ∈ [0, π] \
π
2
csc−1(csc x) = x for every x ∈
−π
2, π
2
\ {0}
Circ−1(Circ x) = x for every x in the restricted domain of Circ
(IMath UP) Inverse Circular Functions Lec 26 17 / 28
1 tan−1 tan π
7/24/2019 Inverse Circular Functions.pdf
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1 tan tan12
(IMath UP) Inverse Circular Functions Lec 26 18 / 28
1 tan−1 tan π
= π
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tan tan12 =
12
(IMath UP) Inverse Circular Functions Lec 26 18 / 28
1 tan−1 tan π
= π
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tan
tan
12
12
2 cos−1
cos 7π8
(IMath UP) Inverse Circular Functions Lec 26 18 / 28
1 tan−1 tan π
= π
7/24/2019 Inverse Circular Functions.pdf
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ta
ta
12 12
2 cos−1
cos 7π8
= 7π
8
(IMath UP) Inverse Circular Functions Lec 26 18 / 28
1 tan−1 tan π
12 =
π
12
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12 12
2 cos−1
cos 7π8
= 7π
8
3 sin−1
sin
5π
6
(IMath UP) Inverse Circular Functions Lec 26 18 / 28
1 tan−1 tan π
12 =
π
12
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12 12
2 cos−1
cos 7π8
= 7π
8
3 sin−1
sin
5π
6
= 5π
6
(IMath UP) Inverse Circular Functions Lec 26 18 / 28
1 tan−1 tan π
12 =
π
12
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12 12
2 cos−1
cos 7π8
= 7π
8
3 sin−1
sin
5π
6
= 5π
6 since
5π
6 /∈
−π
2, π
2
(IMath UP) Inverse Circular Functions Lec 26 18 / 28
1 tan−1 tan π
12 =
π
12
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12 12
2 cos−1
cos 7π8
= 7π
8
3 sin−1
sin
5π
6
= 5π
6 since
5π
6 /∈
−π
2, π
2
Rather, sin−1
sin 5π6
(IMath UP) Inverse Circular Functions Lec 26 18 / 28
1 tan−1 tan π
12 =
π
12
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12 12
2 cos−1
cos 7π8
= 7π
8
3 sin−1
sin
5π
6
= 5π
6 since
5π
6 /∈
−π
2, π
2
Rather, sin−1
sin 5π6
= sin−1
12
(IM th UP) I s Ci l F ti s L 26 18 / 28
1 tan−1 tan π
12 =
π
12
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12 12
2 cos−1
cos 7π8
= 7π
8
3 sin−1
sin
5π
6
= 5π
6 since
5π
6 /∈
−π
2, π
2
Rather, sin−1
sin 5π6
= sin−1
12
= π
6
(IM th UP) I Ci l F ti L 26 18 / 28
1 tan−1 tan π
12 =
π
12
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12 12
2 cos−1
cos 7π8
= 7π
8
3 sin−1
sin
5π
6
= 5π
6 since
5π
6 /∈
−π
2, π
2
Rather, sin−1
sin 5π6
= sin−1
12
= π
6
Note that π
6 ∈
−π
2, π
2
.
(IM th UP) I Ci l F ti L 26 18 / 28
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In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ,
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(IM h UP) I Ci l F i L 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ)
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(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
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(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
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π
2
sin−1, tan−1, csc−1
−π
2
0
cos−1, cot−1, sec−1
π
(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
7/24/2019 Inverse Circular Functions.pdf
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In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
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π
2
sin−1, tan−1, csc−1
−π
2
0
cos−1, cot−1, sec−1
π
1 csc−1
csc 3π5
= csc−1
csc 2π
5
(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
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π
2
sin−1, tan−1, csc−1
−π
2
0
cos−1, cot−1, sec−1
π
1 csc−1
csc 3π5
= csc−1
csc 2π
5
= 2π
5
(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
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π
2
sin−1, tan−1, csc−1
−π
2
0
cos−1, cot−1, sec−1
π
1 csc−1
csc 3π5
= csc−1
csc 2π
5
= 2π
5
2 sec−1
sec 8π
7
(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
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π
2
sin−1, tan−1, csc−1
−π
2
0
cos−1, cot−1, sec−1
π
1 csc−1
csc 3π5
= csc−1
csc 2π
5
= 2π
5
2 sec−1
sec 8π7
= sec−1
sec 6π
7
(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
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π
2
sin−1, tan−1, csc−1
−π
2
0
cos−1, cot−1, sec−1
π
1 csc−1
csc 3π5
= csc−1
csc 2π
5
= 2π
5
2 sec−1
sec 8π7
= sec−1
sec 6π
7
= 6π
7
(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
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π
2
sin−1, tan−1, csc−1
−π
2
0
cos−1, cot−1, sec−1
π
1 csc−1
csc 3π5
= csc−1
csc 2π
5
= 2π
5
2 sec−1
sec 8π7
= sec−1
sec 6π
7
= 6π
7
3 tan−1 tan 8π
11
(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
π
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π
2
sin−1, tan−1, csc−1
−π
2
0
cos−1, cot−1, sec−1
π
1 csc−1
csc 3π5
= csc−1
csc 2π
5
= 2π
5
2 sec−1
sec 8π7
= sec−1
sec 6π
7
= 6π
7
3 tan−1 tan 8π
11 =
−3π
11
(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
π
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π
2
sin−1, tan−1, csc−1
−
π
2
0
cos−1, cot−1, sec−1
π
1 csc−1
csc 3π5
= csc−1
csc 2π
5
= 2π
5
2 sec−1
sec 8π7
= sec−1
sec 6π
7
= 6π
7
3 tan−1 tan 8π
11 =
−3π
114 cos−1
cos 19π10
(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
In general, to obtain Circ−1(Circ θ), find α in the restricted domain of
Circ such that Circ α = Circ θ, so that:
Circ−1(Circ θ) = Circ−1(Circ α) = α
π
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π
2
sin−1, tan−1, csc−1
−
π
2
0
cos−1, cot−1, sec−1
π
1 csc−1
csc 3π5
= csc−1
csc 2π
5
= 2π
5
2 sec−1
sec 8π7
= sec−1
sec 6π
7
= 6π
7
3 tan−1 tan 8π
11 =
−3π
114 cos−1
cos 19π10
= π
10
(IMath, UP) Inverse Circular Functions Lec. 26 19 / 28
Recall: (f ◦ f −1)(x) = x for every x ∈ dom f −1
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(IMath, UP) Inverse Circular Functions Lec. 26 20 / 28
Recall: (f ◦ f −1)(x) = x for every x ∈ dom f −1
Thus,
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sin(sin−1 x) = x for every x ∈ [−1, 1]
(IMath, UP) Inverse Circular Functions Lec. 26 20 / 28
Recall: (f ◦ f −1)(x) = x for every x ∈ dom f −1
Thus,
7/24/2019 Inverse Circular Functions.pdf
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sin(sin−1 x) = x for every x ∈ [−1, 1]
cos(cos−1 x) = x for every x ∈ [−1, 1]
(IMath, UP) Inverse Circular Functions Lec. 26 20 / 28
Recall: (f ◦ f −1)(x) = x for every x ∈ dom f −1
Thus,
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sin(sin−1 x) = x for every x ∈ [−1, 1]
cos(cos−1 x) = x for every x ∈ [−1, 1]
tan(tan−1 x) = x for every x ∈
cot(cot−1 x) = x for every x
∈
sec(sec−1 x) = x for every x ∈ (−∞,−1] ∪ [1, +∞)
csc(csc−1 x) = x for every x ∈ (−∞,−1] ∪ [1, +∞)
(IMath, UP) Inverse Circular Functions Lec. 26 20 / 28
Recall: (f ◦ f −1)(x) = x for every x ∈ dom f −1
Thus,
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sin(sin−1 x) = x for every x ∈ [−1, 1]
cos(cos−1 x) = x for every x ∈ [−1, 1]
tan(tan−1 x) = x for every x ∈
cot(cot−1 x) = x for every x
∈
sec(sec−1 x) = x for every x ∈ (−∞,−1] ∪ [1, +∞)
csc(csc−1 x) = x for every x ∈ (−∞,−1] ∪ [1, +∞)
Circ(Circ−1 x) = x for every x in the range of Circ
(IMath, UP) Inverse Circular Functions Lec. 26 20 / 28
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1 cos
cos−1 23
= 2
3
(IMath, UP) Inverse Circular Functions Lec. 26 21 / 28
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1 cos
cos−1 23
= 2
3
2 tan(tan−1(−4))
(IMath, UP) Inverse Circular Functions Lec. 26 21 / 28
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1 cos
cos−1 23
= 2
3
2 tan(tan−1(−4)) = −4
(IMath, UP) Inverse Circular Functions Lec. 26 21 / 28
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1 cos
cos−1 23
= 2
3
2 tan(tan−1(−4)) = −4
3 cossin−1√ 32
(IMath, UP) Inverse Circular Functions Lec. 26 21 / 28
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1 cos
cos−1 23
= 2
3
2 tan(tan−1(−4)) = −4
3 cossin−1√ 32 = cos
π
3
(IMath, UP) Inverse Circular Functions Lec. 26 21 / 28
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1 cos
cos−1 23
= 2
3
2 tan(tan−1(−4)) = −4
3 cossin−1√ 32 = cos
π
3 = 12
(IMath, UP) Inverse Circular Functions Lec. 26 21 / 28
Evaluate: sin
cos−1−3
5
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(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos
−1 3
5.
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−5
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos
−1 3
5. Then, cos θ = 3
5,
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−5 −5
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos
−1 3
5. Then, cos θ = 3
5, θ
∈ π2 , π.
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−5 −5 ∈ 2
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos
−1 3
5. Then, cos θ = 3
5, θ
∈ π2 , π.
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−5 −5 ∈ 2
Since θ lies in QII,
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos
−1 −3
5. Then, cos θ =
−3
5, θ
∈ π2 , π.
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−5 −5 ∈ 2
Since θ lies in QII, sin θ =
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos−1 −3
5. Then, cos θ =
−3
5, θ
∈ π2 , π.
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5 5 ∈ 2
Since θ lies in QII, sin θ =√
1− cos2 θ
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos−1 −3
5. Then, cos θ =
−3
5, θ
∈ π2 , π.
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∈ Since θ lies in QII, sin θ =
√ 1− cos2 θ =
1− 9
25
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos−1 −
3
5. Then, cos θ =
−3
5
, θ∈
π
2
, π.
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∈ Since θ lies in QII, sin θ =
√ 1− cos2 θ =
1− 9
25 =
1625
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos−1 −
3
5. Then, cos θ =
−3
5
, θ∈
π
2
, π.
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∈ Since θ lies in QII, sin θ =
√ 1− cos2 θ =
1− 9
25 =
1625 = 4
5
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos−1 −
3
5. Then, cos θ =
−3
5
, θ∈
π
2
, π.
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∈ Since θ lies in QII, sin θ =
√ 1− cos2 θ =
1− 9
25 =
1625 = 4
5
Alternative Solution:
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos−1 −
3
5. Then, cos θ =
−3
5
, θ ∈
π
2
, π.
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∈ Since θ lies in QII, sin θ =
√ 1− cos2 θ =
1− 9
25 =
1625 = 4
5
Alternative Solution: Take x =
−3, r = 5.
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos−1 −
3
5. Then, cos θ =
−3
5
, θ ∈
π
2
, π.
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Since θ lies in QII, sin θ =
√ 1− cos2 θ =
1− 9
25 =
1625 = 4
5
Alternative Solution: Take x =
−3, r = 5.
y =
52 − (−3)2
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos−1 −
3
5. Then, cos θ =
−3
5
, θ ∈
π
2
, π.
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Since θ lies in QII, sin θ =
√ 1− cos2 θ =
1− 9
25 =
1625 = 4
5
Alternative Solution: Take x =
−3, r = 5.
y =
52 − (−3)2 =√
16
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
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Evaluate: sin
cos−1−3
5
Let θ = cos−1 −
3
5. Then, cos θ =
−3
5
, θ ∈
π
2
, π.
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Since θ lies in QII, sin θ =
√ 1− cos2 θ =
1− 9
25 =
1625 = 4
5
Alternative Solution: Take x =
−3, r = 5.
y =
52 − (−3)2 =√
16 = 4
sin θ
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: sin
cos−1−3
5
Let θ = cos−1 −
3
5. Then, cos θ =
−3
5
, θ ∈
π
2
, π.
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Since θ lies in QII, sin θ =
√ 1− cos2 θ =
1− 9
25 =
1625 = 4
5
Alternative Solution: Take x =
−3, r = 5.
y =
52 − (−3)2 =√
16 = 4
sin θ = 45
(IMath, UP) Inverse Circular Functions Lec. 26 22 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
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(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√ 5, α ∈
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√
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√ 5, α ∈ −π
2 , 0
, tan β =
2
3 , β ∈ 0,
π
2
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√
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√ 5, α ∈ −π
2 , 0
, tan β =
2
3 , β ∈ 0,
π
2
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√tan(α− β ) =
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2 , 0
, tan β =
2
3 , β ∈ 0,
π
2
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tan(α− β ) = tan α− tan β
1 + tan α tan β
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2 , 0
, tan β =
2
3 , β ∈ 0,
π
2
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tan(α− β ) = tan α− tan β
1 + tan α tan β
For α: take r
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2 , 0
, tan β =
2
3 , β ∈ 0,
π
2
β
7/24/2019 Inverse Circular Functions.pdf
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tan(α− β ) = tan α− tan β
1 + tan α tan β
For α: take r = √ 5,
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2 , 0
, tan β =
2
3 , β ∈ 0,
π
2
t t β
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tan(α− β ) = tan α− tan β
1 + tan α tan β
For α: take r =
√ 5, y = −1.
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2 , 0
, tan β =
2
3 , β ∈ 0,
π
2
t t β
7/24/2019 Inverse Circular Functions.pdf
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tan(α− β ) = tan α− tan β
1 + tan α tan β
For α: take r =
√ 5, y = −1.
⇒ x =√
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2 , 0,
tan β =
2
3, β ∈ 0,
π
2
t t β
7/24/2019 Inverse Circular Functions.pdf
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tan(α− β ) = tan α− tan β
1 + tan α tan β
For α: take r =
√ 5, y = −1.
⇒ x =√
5
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2 , 0,
tan β =
2
3, β ∈ 0,
π
2
tan α tan β
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tan(α− β ) = tan α− tan β
1 + tan α tan β
For α: take r =
√ 5, y = −1.
⇒ x =√
5− 1
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2, 0, tan β = 2
3, β ∈ 0, π
2
tan α tan β
7/24/2019 Inverse Circular Functions.pdf
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tan(α− β ) = tan α− tan β
1 + tan α tan β
For α: take r =
√ 5, y = −1.
⇒ x =√
5− 1 = 2
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2, 0, tan β = 2
3, β ∈ 0, π
2
tan α− tan β
7/24/2019 Inverse Circular Functions.pdf
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tan(α− β ) = tan α− tan β
1 + tan α tan β
For α
: take r =
√ 5
, y = −1
.
⇒ x =√
5− 1 = 2 ⇒ tan α =
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2, 0, tan β = 2
3, β ∈ 0, π
2
tan α− tan β
7/24/2019 Inverse Circular Functions.pdf
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tan(α− β ) = tan α tan β
1 + tan α tan β
For α
: take r =
√ 5
, y = −1
.
⇒ x =√
5− 1 = 2 ⇒ tan α = − 12
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2, 0, tan β = 2
3, β ∈ 0, π
2
tan α− tan β
7/24/2019 Inverse Circular Functions.pdf
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tan(α− β ) = tan α tan β
1 + tan α tan β
For α: take r =√
5, y = −
1.
⇒ x =√
5− 1 = 2 ⇒ tan α = − 12
tan(α
−β ) =
−12 − 2
3
1 +−1
2 2
3
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Evaluate: tan
csc−1(−√ 5)− tan−1 23
Let α = csc−1(−√ 5), β = tan−1 2
3 .
csc α = −√
5, α ∈ −π
2, 0, tan β = 2
3, β ∈ 0, π
2
( )tan α− tan β
7/24/2019 Inverse Circular Functions.pdf
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tan(α− β ) = tan α tan β
1 + tan α tan β
For α: take r =√
5, y = −
1.
⇒ x =√
5− 1 = 2 ⇒ tan α = − 12
tan(α
−β ) =
−12 − 2
3
1 +−1
2 2
3 =
−7
(IMath, UP) Inverse Circular Functions Lec. 26 23 / 28
Solve for x: cos−1x
4
+ tan−1
−√ 3
3
= cot−1 0
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 24 / 28
Solve for x: cos−1x
4
+ tan−1
−√ 3
3
= cot−1 0
cos−1 x
4
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 24 / 28
Solve for x: cos−1x
4
+ tan−1
−√ 3
3
= cot−1 0
cos−1 x
4 + −
π
6 =
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 24 / 28
Solve for x: cos−1x
4
+ tan−1
−√ 3
3
= cot−1 0
cos−1 x
4 + −
π
6 =
π
2
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 24 / 28
Solve for x: cos−1x
4
+ tan−1
−√ 3
3
= cot−1 0
cos−1 x
4 + −
π
6 =
π
2cos−1
x
4
=
π
6 +
π
2
7/24/2019 Inverse Circular Functions.pdf
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4
6 2
(IMath, UP) Inverse Circular Functions Lec. 26 24 / 28
Solve for x: cos−1x
4
+ tan−1
−√ 3
3
= cot−1 0
cos−1 x
4 + −
π
6 =
π
2cos−1
x
4
=
π
6 +
π
2
7/24/2019 Inverse Circular Functions.pdf
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4
6 2
cos−1
x
4 =
2π
3
(IMath, UP) Inverse Circular Functions Lec. 26 24 / 28
Solve for x: cos−1x
4
+ tan−1
−√ 3
3
= cot−1 0
cos−1 x
4 + −
π
6 =
π
2cos−1
x
4
=
π
6 +
π
2
7/24/2019 Inverse Circular Functions.pdf
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4
6 2
cos−1
x
4 =
2π
3
x4
= cos 2π3
(IMath, UP) Inverse Circular Functions Lec. 26 24 / 28
Solve for x: cos−1x
4
+ tan−1
−√ 3
3
= cot−1 0
cos−1 x
4 + −
π
6 =
π
2cos−1
x
4
=
π
6 +
π
2
7/24/2019 Inverse Circular Functions.pdf
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4
6 2
cos−1
x
4 =
2π
3
x4
= cos 2π3
x = 4
−1
2
(IMath, UP) Inverse Circular Functions Lec. 26 24 / 28
Solve for x: cos−1x
4
+ tan−1
−√ 3
3
= cot−1 0
cos−1 x
4 + −
π
6 =
π
2cos−1
x
4
=
π
6 +
π
2
7/24/2019 Inverse Circular Functions.pdf
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4
6 2
cos−1
x
4 =
2π
3
x4
= cos 2π3
x = 4
−1
2
x = −2
(IMath, UP) Inverse Circular Functions Lec. 26 24 / 28
Solve for x: sec−1(−√ 2)− tan−1(x− 2) = csc−1(−2)
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 25 / 28
Solve for x: sec−1(−√ 2)− tan−1(x− 2) = csc−1(−2)
3π
4
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 25 / 28
Solve for x: sec−1(−√ 2)− tan−1(x− 2) = csc−1(−2)
3π
4 −tan−1(x
−2) =
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 25 / 28
Solve for x: sec−1(−√ 2)− tan−1(x− 2) = csc−1(−2)
3π
4 −tan−1(x
−2) =
− π
6
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 25 / 28
Solve for x: sec−1(−√ 2)− tan−1(x− 2) = csc−1(−2)
3π
4 −tan−1(x
−2) =
− π
63π
4 +
π
6 = tan−1(x − 2)
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 25 / 28
Solve for x: sec−1(−√ 2)− tan−1(x− 2) = csc−1(−2)
3π
4 −tan−1(x
−2) =
− π
63π
4 +
π
6 = tan−1(x − 2)
7/24/2019 Inverse Circular Functions.pdf
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11π
12 = tan−1(x − 2)
(IMath, UP) Inverse Circular Functions Lec. 26 25 / 28
Solve for x: sec−1(−√ 2)− tan−1(x− 2) = csc−1(−2)
3π
4 −tan−1(x
−2) =
− π
63π
4 +
π
6 = tan−1(x − 2)
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11π
12 = tan−1(x − 2)
But ran tan−1 =−π
2, π
2
.
(IMath, UP) Inverse Circular Functions Lec. 26 25 / 28
Solve for x: sec−1(−√ 2)− tan−1(x− 2) = csc−1(−2)
3π
4 −tan−1(x
−2) =
−
π
63π
4 +
π
6 = tan−1(x − 2)
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11π
12 = tan−1(x − 2)
But ran tan−1 =−π
2, π
2
.
No solution
(IMath, UP) Inverse Circular Functions Lec. 26 25 / 28
Solve for x: sin−1(x)− cos
−1(x) = π
6
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(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1(x)− cos
−1(x) = π
6
Let α = sin−1 x, β = cos−1 x.
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(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1(x)− cos
−1(x) = π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x
,
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(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1(x)− cos
−1(x) = π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x
, α ∈ −
π
2 ,
π
2,
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(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1(x)− cos
−1(x) = π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x,
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(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1(x)− cos
−1(x) = π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
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(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1(x)− cos
−1(x) = π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
α = π
6 + β
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(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1(x)− cos
−1(x) = π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
α = π
6 + β π
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sin α = sinπ
6 + β
(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
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Solve for x: sin−1(x)− cos
−1(x) = π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
α = π
6 + β π
β
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sin α = sinπ
6 + β
sin α = sin π
6 cos β + cos π
6 sin β
x =
(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
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Solve for x: sin−1(x)− cos
−1(x) = π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
α = π
6 + β
i iπ
+ β
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sin α = sin
6 + β
sin α = sin π
6 cos β + cos π
6 sin β
x = 1
2(x)
(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1(x)− cos
−1(x) = π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
α = π
6 + β
i iπ
+ β
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sin α = sin
6 + β
sin α = sin π
6 cos β + cos π
6 sin β
x = 1
2(x) +
√ 3
2
(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1
(x)− cos−1(x) =
π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
α = π
6 + β
si siπ
+ β
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sin α = sin
6 + β
sin α = sin π
6 cos β + cos π
6 sin β
x = 1
2(x) +
√ 3
2 sin β
(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1
(x)− cos−1
(x) = π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
α = π
6 + β
sin α sinπ
+ β
7/24/2019 Inverse Circular Functions.pdf
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sin α = sin
6 + β
sin α = sin π
6 cos β + cos π
6 sin β
x = 1
2(x) +
√ 3
2 sin β
Since β ∈ [0, π],
(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1
(x)− cos−1
(x) =
π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
α = π
6 + β
sin α = sinπ
+ β
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sin α = sin
6 + β
sin α = sin π
6 cos β + cos π
6 sin β
x = 1
2(x) +
√ 3
2 sin β
Since β ∈ [0, π], sin β =
(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1
(x)− cos−1
(x) =
π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
α = π
6 + β
sin α = sinπ
+ β
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sin α = sin
6 + β
sin α = sin
π
6 cos β + cos
π
6 sin β
x = 1
2(x) +
√ 3
2 sin β
Since β ∈ [0, π], sin β =
1− cos2 β
(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
Solve for x: sin−1
(x)− cos−1
(x) =
π
6
Let α = sin−1 x, β = cos−1 x.
Then sin α = x, α ∈ −
π
2, π
2, cos β = x, β
∈ [0, π].
α = π
6 + β
sin α = sinπ
+ β
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sin α = sin
6 + β
sin α = sin
π
6 cos β + cos
π
6 sin β
x = 1
2(x) +
√ 3
2 sin β
Since β ∈ [0, π], sin β =
1− cos2 β = √ 1− x2
(IMath, UP) Inverse Circular Functions Lec. 26 26 / 28
x = 1
2(x) +
√ 3
2
1− x2
7/24/2019 Inverse Circular Functions.pdf
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(IMath, UP) Inverse Circular Functions Lec. 26 27 / 28
x = 1
2(x) +
√ 32
1− x2
2x = x +√
3
1− x2
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(IMath, UP) Inverse Circular Functions Lec. 26 27 / 28
x = 1
2(x) +
√ 32
1− x2
2x = x +√
3
1− x2
x = √ 3 1− x
2
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(IMath, UP) Inverse Circular Functions Lec. 26 27 / 28
x = 1
2(x) +
√ 32
1− x2
2x = x +√
3
1− x2
x = √ 3 1− x
2
x2 = 3(1− x2)
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(IMath, UP) Inverse Circular Functions Lec. 26 27 / 28
x = 1
2(x) +
√ 32
1− x2
2x = x +√
3
1− x2
x = √ 3 1− x
2
x2 = 3(1− x2)
x2 = 3− 3x2
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(IMath, UP) Inverse Circular Functions Lec. 26 27 / 28
x = 1
2(x) +
√ 32
1− x2
2x = x +√
3
1− x2
x = √ 3 1− x
2
x2 = 3(1− x2)
x2 = 3− 3x2
7/24/2019 Inverse Circular Functions.pdf
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4x2 = 3
(IMath, UP) Inverse Circular Functions Lec. 26 27 / 28
x = 1
2(x) +
√ 32
1− x2
2x = x +√
3
1− x2
x = √ 3 1− x
2
x2 = 3(1− x2)
x2 = 3− 3x2
7/24/2019 Inverse Circular Functions.pdf
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4x2 = 3
x2 = 34
(IMath, UP) Inverse Circular Functions Lec. 26 27 / 28
x = 1
2(x) +
√ 32
1− x2
2x = x +√
3
1− x2
x = √ 3 1− x
2
x2 = 3(1− x2)
x2 = 3− 3x2
2
7/24/2019 Inverse Circular Functions.pdf
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4x2 = 3
x2 = 34
x = ±√
3
2
(IMath, UP) Inverse Circular Functions Lec. 26 27 / 28
Check:
x = √ 32
: sin−1 √ 32 − cos−1 √ 3
2
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(IMath, UP) Inverse Circular Functions Lec. 26 28 / 28
Check:
x = √ 32
: sin−1 √ 32 − cos−1 √ 3
2 = π
3 −
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(IMath, UP) Inverse Circular Functions Lec. 26 28 / 28
Check:
x = √ 32
: sin−1 √ 32 − cos−1 √ 3
2 = π
3 − π
6
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(IMath, UP) Inverse Circular Functions Lec. 26 28 / 28
Check:
x = √ 32
: sin−1 √ 32 − cos−1 √ 3
2 = π
3 − π
6 = π
6
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(IMath, UP) Inverse Circular Functions Lec. 26 28 / 28
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Check:
x = √ 32
: sin−1 √ 32 − cos−1 √ 3
2 = π
3 − π
6 = π
6
x = −√
3
2 : sin−1
−√
3
2
− cos−1
−√
3
2
7/24/2019 Inverse Circular Functions.pdf
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2
2
2
(IMath, UP) Inverse Circular Functions Lec. 26 28 / 28
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Check:
x = √ 32
: sin−1 √ 32 − cos−1 √ 3
2 = π
3 − π
6 = π
6
x = −√
3
2 : sin−1
−√
3
2
− cos−1
−√
3
2
= −π
3 − 5π
6
7/24/2019 Inverse Circular Functions.pdf
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2
2
2
3 6
(IMath, UP) Inverse Circular Functions Lec. 26 28 / 28
Check:
x = √ 32
: sin−1 √ 32 − cos−1 √ 3
2 = π
3 − π
6 = π
6
x = −√
3
2 : sin−1
−√
3
2
− cos−1
−√
3
2
= −π
3 − 5π
6 = −7π
6
7/24/2019 Inverse Circular Functions.pdf
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2
2
2
3 6 6
(IMath, UP) Inverse Circular Functions Lec. 26 28 / 28
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