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6.4: Values of the Trigonometric Functions
E. Kim
MTH 151
All notation and terminology is based on Swokowski, Cole. Algebra and
Trigonometry: with analytic geometry. Classic 12th Edition.
Accompanying handout:
I
I
Black-and-white: http://www.uwlax.edu/faculty/ekim/resources/unit-circle.pdf
Color: http://www.uwlax.edu/faculty/ekim/resources/unit-circle-color.pdf
1
Goal
√
−
− 21 ,
√ √
3
2
(0, 1)
√ 1
, 23
2
√
√ 2
, 22
2
2
, 22
2
√
− 23 , 12
√
3 1
,2
2
(−1, 0)
(1, 0)
√
√
− 23 , − 12
√
√ − 22 , − 22
√ − 12 , − 23
3
, − 12
2
√
√ 2
, − 22
2
(0, −1)
√ 1
, − 23
2
2
Goal
√
−
− 21 ,
√ √
3
2
(0, 1)
√ 1
, 23
2
√
√ 2
, 22
2
2
, 22
2
√
− 23 , 12
√
3 1
,2
2
(−1, 0)
(1, 0)
√
√
− 23 , − 12
√
√ − 22 , − 22
√ − 12 , − 23
3
, − 12
2
√
√ 2
, − 22
2
(0, −1)
√ 1
, − 23
2
2
Goal
√
−
− 21 ,
√ √
3
2
(0, 1)
√ 1
, 23
2
√
√ 2
, 22
2
2
, 22
2
√
− 23 , 12
√
3 1
,2
2
(−1, 0)
(1, 0)
√
√
− 23 , − 12
√
√ − 22 , − 22
√ − 12 , − 23
3
, − 12
2
√
√ 2
, − 22
2
(0, −1)
√ 1
, − 23
2
2
Why understand the unit circle?
For all special angles, we can compute sine, cosine, etc. by
knowing these values only for 30◦ , 45◦ , and 90◦ .
3
Why understand the unit circle?
For all special angles, we can compute sine, cosine, etc. by
knowing these values only for 30◦ , 45◦ , and 90◦ .
For the other special angles, just change the sign as appropriate.
The First Quadrant
(0, 1)
(1, 0)
4
The First Quadrant
(0, 1)
θ= π
= 30◦
6
√
3 1
, 2
2
(1, 0)
4
The First Quadrant
(0, 1)
θ= π
= 45◦
4
√
2
,
2
√
2
2
(1, 0)
4
The First Quadrant
θ= π
= 60◦
3
(0, 1)
1,
2
√ 3
2
(1, 0)
4
The First Quadrant
θ= π
= 60◦
3
(0, 1)
1,
2
√ 3
2
θ= π
= 45◦
4
√
2
,
2
√
2
2
θ= π
= 30◦
6
√
3 1
, 2
2
(1, 0)
4
What is a reference angle?
a a a a a a a aa a a aa a aaaaaa a a Take a nonquadrantal1 angle θ.
y
θ
x
1
Nonquadrantal means that θ is not a multiple of 90◦ .
5
What is a reference angle?
a a a a a a a aa a a aa a aaaaaa a a Take a nonquadrantal1 angle θ.
y
θ
θR , the reference angle
The acute angle made by
x
1
Nonquadrantal means that θ is not a multiple of 90◦ .
What is a reference angle?
a a a a a a a aa a a aa a aaaaaa a a Take a nonquadrantal1 angle θ.
y
θ
θR , the reference angle
The acute angle made by
x
1
I
terminal side of θ
Nonquadrantal means that θ is not a multiple of 90◦ .
What is a reference angle?
a a a a a a a aa a a aa a aaaaaa a a Take a nonquadrantal1 angle θ.
y
θ
θR , the reference angle
θR
The acute angle made by
x
1
I
terminal side of θ
I
x-axis
Nonquadrantal means that θ is not a multiple of 90◦ .
Formula for reference angle θR in every quadrant
y
x
6
Formula for reference angle θR in every quadrant
θ in Quadrant 1
y
θR = θ
θ
θR
x
6
Formula for reference angle θR in every quadrant
θ in Quadrant 1
y
θR = θ
θ
θR
x
6
Formula for reference angle θR in every quadrant
θ in Quadrant 1
y
θR = θ
θR
θ
x
6
Formula for reference angle θR in every quadrant
θ in Quadrant 1
y
θR = θ
θR
θ
x
6
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θR = π − θ
θR = 180◦ − θ
θR
θ
x
6
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θR = π − θ
θR = 180◦ − θ
θR
θ
x
6
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θR = π − θ
θR = 180◦ − θ
θ
θR
x
6
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θR = π − θ
θR = 180◦ − θ
θ
θR
x
6
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θ
θR = π − θ
θR = 180◦ − θ
θ in Quadrant 3
θR = θ − π
θR = θ − 180◦
θR
x
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θ
θR = π − θ
θR = 180◦ − θ
x
θ in Quadrant 3
θR = θ − π
θR = θ − 180◦
θR
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θ
θR = π − θ
θR = 180◦ − θ
x
θ in Quadrant 3
θR = θ − π
θR = θ − 180◦
θR
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θR = π − θ
θR = 180◦ − θ
θ
x
θ in Quadrant 3
θR = θ − π
θR = θ − 180◦
θR
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θR = π − θ
θR = 180◦ − θ
θ
x
θ in Quadrant 3
θR = θ − π
θR = θ − 180◦
θ in Quadrant 4
θR = 2π − θ
θR = 360◦ − θ
θR
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θR = π − θ
θR = 180◦ − θ
θ
x
θ in Quadrant 3
θR = θ − π
θR = θ − 180◦
θ in Quadrant 4
θR = 2π − θ
θR = 360◦ − θ
θR
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θR = π − θ
θR = 180◦ − θ
θ
x
θ in Quadrant 3
θR = θ − π
θR = θ − 180◦
θ in Quadrant 4
θR = 2π − θ
θR = 360◦ − θ
θR
Formula for reference angle θR in every quadrant
y
θ in Quadrant 1
θR = θ
θ in Quadrant 2
θR = π − θ
θR = 180◦ − θ
θ
θ in Quadrant 3
θR = θ − π
θR = θ − 180◦
θ in Quadrant 4
θR = 2π − θ
θR = 360◦ − θ
θR
x
Example: θ = 315◦
y
θ = 315◦
x
7
Example: θ = 315◦
y
θ = 315◦
x
θR
7
Example: θ = 315◦
y
θ = 315◦
x
θR
θR = 360◦ − 315◦
7
Example: θ = 315◦
y
θ = 315◦
x
θR = 45◦
θR = 360◦ − 315◦ = 45◦
7
Example: θ =
5π
6
y
θ=
5π
6
x
Example: θ =
5π
6
y
θ=
5π
6
θR
x
Example: θ =
5π
6
y
θ=
5π
6
θR
x
θR = π −
5π
6
Example: θ =
5π
6
y
θ=
θR =
5π
6
π
6
x
θR = π −
5π
6
=
π
6
Example: θ = 4. (Note this is four radians, not degrees!)
y
θ
x
9
Example: θ = 4. (Note this is four radians, not degrees!)
y
θ
x
θR
9
Example: θ = 4. (Note this is four radians, not degrees!)
y
θ
x
θR
θR = 4 − π
9
What if θ is not between 0◦ and 360◦ ?
10
What if θ is not between 0◦ and 360◦ ?
First, find the coterminal angle to θ between 0◦ and 360◦ .
10
What if θ is not between 0◦ and 360◦ ?
First, find the coterminal angle to θ between 0◦ and 360◦ .
Example: θ = −240◦ .
10
What if θ is not between 0◦ and 360◦ ?
First, find the coterminal angle to θ between 0◦ and 360◦ .
Example: θ = −240◦ .
θ = −240◦ is coterminal to −240◦ + 360◦
10
What if θ is not between 0◦ and 360◦ ?
First, find the coterminal angle to θ between 0◦ and 360◦ .
Example: θ = −240◦ .
θ = −240◦ is coterminal to −240◦ + 360◦ = 120◦
10
What if θ is not between 0◦ and 360◦ ?
First, find the coterminal angle to θ between 0◦ and 360◦ .
Example: θ = −240◦ .
θ = −240◦ is coterminal to −240◦ + 360◦ = 120◦
To find θR , use θ = 120◦ .
10
Example: θ = −240◦ , but using 120◦
y
θ = 120◦
x
11
Example: θ = −240◦ , but using 120◦
y
θR
θ = 120◦
x
11
Example: θ = −240◦ , but using 120◦
y
θR
θ = 120◦
x
θR = 180◦ − 120◦
11
Example: θ = −240◦ , but using 120◦
y
θR = 60◦
θ = 120◦
x
θR = 180◦ − 120◦ = 60◦
11
Reference angles and exact values of sin, cos, and tan
− 12 ,
√
− 21 , −
3
2
√
3
2
, θ = 120◦
, θ = 240◦
√ 1
, 23 , θ
2
√ 1
, − 23 , θ
2
= 60◦
= 300◦
12
Reference angles and exact values of sin, cos, and tan
− 12 ,
√
− 21 , −
3
2
√
3
2
, θ = 120◦
, θ = 240◦
√ 1
, 23 , θ
2
√ 1
, − 23 , θ
2
= 60◦
= 300◦
12
Reference angles and exact values of sin, cos, and tan
− 12 ,
√
− 21 , −
3
2
√
3
2
, θ = 120◦
, θ = 240◦
√ 1
, 23 , θ
2
√ 1
, − 23 , θ
2
= 60◦
= 300◦
12
Reference angles and exact values of sin, cos, and tan
− 12 ,
√
− 21 , −
3
2
√
3
2
, θ = 120◦
, θ = 240◦
√ 1
, 23 , θ
2
√ 1
, − 23 , θ
2
= 60◦
= 300◦
12
Reference angles and exact values of sin, cos, and tan
− 12 ,
√
− 21 , −
3
2
√
3
2
, θ = 120◦
, θ = 240◦
√ 1
, 23 , θ
2
√ 1
, − 23 , θ
2
= 60◦
= 300◦
12
Reference angles and exact values of sin, cos, and tan
− 12 ,
√
− 21 , −
3
2
√
3
2
, θ = 120◦
, θ = 240◦
√ 1
, 23 , θ
2
√ 1
, − 23 , θ
2
= 60◦
= 300◦
12
Reference angles and exact values of sin, cos, and tan
− 12 ,
√
− 21 , −
3
2
√
3
2
, θ = 120◦
, θ = 240◦
√ 1
, 23 , θ
2
√ 1
, − 23 , θ
2
= 60◦
= 300◦
12
Reference angles and exact values of sin, cos, and tan
− 12 ,
√
− 21 , −
3
2
√
3
2
, θ = 120◦
, θ = 240◦
√ 1
, 23 , θ
2
√ 1
, − 23 , θ
2
= 60◦
= 300◦
12
Reference angles and exact values of sin, cos, and tan
− 12 ,
√
− 21 , −
3
2
√
3
2
, θ = 120◦
, θ = 240◦
√ 1
, 23 , θ
2
√ 1
, − 23 , θ
2
= 60◦
= 300◦
12
Reference angles and exact values of sin, cos, and tan
− 12 ,
√
− 21 , −
3
2
√
3
2
, θ = 120◦
, θ = 240◦
√ 1
, 23 , θ
2
√ 1
, − 23 , θ
2
= 60◦
= 300◦
12
If θ =
5π
6 ,
find sin θ, cos θ, and tan θ
y
θ=
5π
6
x
If θ =
5π
6 ,
find sin θ, cos θ, and tan θ
y
θ=
θR =
5π
6
π
6
x
θR = π −
5π
6
=
π
6
If θ =
5π
6 ,
find sin θ, cos θ, and tan θ
Reference angle is θR =
π
6
14
If θ =
5π
6 ,
find sin θ, cos θ, and tan θ
Reference angle is θR =
π
6
π
1
sin =
6
2
√
π
3
cos =
6
2
√
π
3
tan =
6
3
14
If θ =
5π
6 ,
find sin θ, cos θ, and tan θ
Reference angle is θR =
π
6
π
1
sin =
6
2
√
π
3
cos =
6
2
y
θ=
√
π
3
tan =
6
3
5π
6
x
If θ =
5π
6 ,
find sin θ, cos θ, and tan θ
Reference angle is θR =
π
6
π
1
sin =
6
2
√
π
3
cos =
6
2
y
(−, +)
θ=
√
π
3
tan =
6
3
5π
6
x
If θ =
5π
6 ,
find sin θ, cos θ, and tan θ
Reference angle is θR =
π
6
π
1
sin =
6
2
√
π
3
cos =
6
2
y
(−, +)
θ=
√
π
3
tan =
6
3
5π
6
x
5π
1
sin
=+
6
2
√
5π
3
cos
=−
6
2
√
5π
3
tan
=−
6
3
If θ = 315◦ , find sin θ, cos θ, and tan θ
y
θ = 315◦
x
15
If θ = 315◦ , find sin θ, cos θ, and tan θ
y
θ = 315◦
x
θR = 45◦
θR = 360◦ − 315◦ = 45◦
15
If θ = 315◦ , find sin θ, cos θ, and tan θ
Reference angle is θR = 45◦
16
If θ = 315◦ , find sin θ, cos θ, and tan θ
Reference angle is θR = 45◦
√
2
sin 45 =
2
◦
√
◦
cos 45 =
2
2
tan 45◦ = 1
16
If θ = 315◦ , find sin θ, cos θ, and tan θ
Reference angle is θR = 45◦
√
2
sin 45 =
2
◦
√
◦
cos 45 =
2
2
tan 45◦ = 1
y
θ = 315◦
x
16
If θ = 315◦ , find sin θ, cos θ, and tan θ
Reference angle is θR = 45◦
√
2
sin 45 =
2
◦
√
◦
cos 45 =
2
2
tan 45◦ = 1
y
θ = 315◦
x
(+, −)
If θ = 315◦ , find sin θ, cos θ, and tan θ
Reference angle is θR = 45◦
√
2
sin 45 =
2
◦
√
◦
cos 45 =
2
2
tan 45◦ = 1
y
θ = 315◦
x
(+, −)
√
2
sin 315 = −
2
◦
√
◦
cos 315 = +
2
2
tan 315◦ = −1
16
Summary of using reference angles
1. Find the reference angle θR for your angle θ.
2. Compute sin, cos, and tan for the reference angle θR .
3. Adjust the sign based on the quadrant of terminal side of θ.
Finding angles with a calculator
18
Inverse trigonometric functions
Problem
If θ is an acute angle and sin θ = 0.6635, what is θ?
19
Inverse trigonometric functions
Problem
If θ is an acute angle and sin θ = 0.6635, what is θ?
If sin θ = k, then θ = sin−1 k.
19
Inverse trigonometric functions
Problem
If θ is an acute angle and sin θ = 0.6635, what is θ?
If sin θ = k, then θ = sin−1 k.
θ = sin−1 (0.6635)
19
Inverse trigonometric functions
Problem
If θ is an acute angle and sin θ = 0.6635, what is θ?
If sin θ = k, then θ = sin−1 k.
θ = sin−1 (0.6635) ≈ 41.57◦ ≈ 0.7255
19
Inverse trigonometric functions
Problem
If θ is an acute angle and sin θ = 0.6635, what is θ?
If sin θ = k, then θ = sin−1 k.
θ = sin−1 (0.6635) ≈ 41.57◦ ≈ 0.7255
θ = sin−1 (0.6635)
19
Inverse trigonometric functions
Problem
If θ is an acute angle and sin θ = 0.6635, what is θ?
If sin θ = k, then θ = sin−1 k.
θ = sin−1 (0.6635) ≈ 41.57◦ ≈ 0.7255
θ = sin−1 (0.6635) ≈ degrees radians
19
Inverse trigonometric functions
Problem
If θ is an acute angle and sin θ = 0.6635, what is θ?
If sin θ = k, then θ = sin−1 k.
θ = sin−1 (0.6635) ≈ 41.57◦ ≈ 0.7255
θ = sin−1 (0.6635) ≈ degrees radians
If cos θ = k, then θ = cos−1 k.
If tan θ = k, then θ = tan−1 k.
19
What about csc, sec, and cot?
Use the reciprocal formulas:
1
1
sec θ =
csc θ =
sin θ
cos θ
cot θ =
1
tan θ
20
What about csc, sec, and cot?
Use the reciprocal formulas:
1
1
sec θ =
csc θ =
sin θ
cos θ
cot θ =
1
tan θ
Example: csc θ = 2
20
What about csc, sec, and cot?
Use the reciprocal formulas:
1
1
sec θ =
csc θ =
sin θ
cos θ
cot θ =
1
tan θ
Example: csc θ = 2
Convert to
1
=2
sin θ
20
What about csc, sec, and cot?
Use the reciprocal formulas:
1
1
sec θ =
csc θ =
sin θ
cos θ
cot θ =
1
tan θ
Example: csc θ = 2
Convert to
1
=2
sin θ
Take reciprocal of both sides
sin θ =
1
2
20
What about csc, sec, and cot?
Use the reciprocal formulas:
1
1
sec θ =
csc θ =
sin θ
cos θ
cot θ =
1
tan θ
Example: csc θ = 2
Convert to
1
=2
sin θ
Take reciprocal of both sides
sin θ =
1
2
Use inverse sine function (also called arcsin or asin)
1
θ = sin−1
2
20
What about csc, sec, and cot?
Use the reciprocal formulas:
1
1
sec θ =
csc θ =
sin θ
cos θ
cot θ =
1
tan θ
Example: csc θ = 2
Convert to
1
=2
sin θ
Take reciprocal of both sides
sin θ =
1
2
Use inverse sine function (also called arcsin or asin)
1
θ = sin−1
= 30◦
2
20
An example of each inverse trig function
sin θ = 0.5
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
21
An example of each inverse trig function
sin θ = 0.5
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
cos θ = 0.5
θ = cos−1 ( 12 ) = 60◦ ≈ 1.0472
21
An example of each inverse trig function
sin θ = 0.5
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
cos θ = 0.5
θ = cos−1 ( 12 ) = 60◦ ≈ 1.0472
tan θ = 0.5
θ = tan−1 ( 21 ) ≈ 26.57◦ ≈ 0.4636
21
An example of each inverse trig function
sin θ = 0.5
csc θ = 2
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
cos θ = 0.5
θ = cos−1 ( 12 ) = 60◦ ≈ 1.0472
tan θ = 0.5
θ = tan−1 ( 21 ) ≈ 26.57◦ ≈ 0.4636
21
An example of each inverse trig function
sin θ = 0.5
csc θ = 2
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
cos θ = 0.5
sec θ = 2
θ = cos−1 ( 12 ) = 60◦ ≈ 1.0472
θ = cos−1 ( 21 ) = 60◦ ≈ 1.0472
tan θ = 0.5
θ = tan−1 ( 21 ) ≈ 26.57◦ ≈ 0.4636
21
An example of each inverse trig function
sin θ = 0.5
csc θ = 2
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
cos θ = 0.5
sec θ = 2
θ = cos−1 ( 12 ) = 60◦ ≈ 1.0472
θ = cos−1 ( 21 ) = 60◦ ≈ 1.0472
tan θ = 0.5
cot θ = 2
θ = tan−1 ( 21 ) ≈ 26.57◦ ≈ 0.4636
θ = tan−1 ( 12 ) ≈ 26.57◦ ≈ 0.4636
21
An example of each inverse trig function
sin θ = 0.5
csc θ = 2
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
θ = sin−1 ( 12 ) = 30◦ ≈ 0.5236
cos θ = 0.5
sec θ = 2
θ = cos−1 ( 12 ) = 60◦ ≈ 1.0472
θ = cos−1 ( 21 ) = 60◦ ≈ 1.0472
tan θ = 0.5
cot θ = 2
θ = tan−1 ( 21 ) ≈ 26.57◦ ≈ 0.4636
θ = tan−1 ( 12 ) ≈ 26.57◦ ≈ 0.4636
Column on right copies column on left
because of reciprocal identites
21
Since f (x) = sin(x) is periodic, what is sin−1 k giving you?
Inverse Sine
If you put sin−1 k into your calculator, the answer will be an angle
I
in [− π2 , π2 ] using radian mode
I
in [−90◦ , 90◦ ] in degree mode
Since f (x) = sin(x) is periodic, what is sin−1 k giving you?
Inverse Sine
If you put sin−1 k into your calculator, the answer will be an angle
I
in [− π2 , π2 ] using radian mode
I
in [−90◦ , 90◦ ] in degree mode
Inverse Cosine
If you put cos−1 k into your calculator, the answer will be an angle
I
in [0, π] using radian mode
I
in [0, 180◦ ] in degree mode
Since f (x) = sin(x) is periodic, what is sin−1 k giving you?
Inverse Sine
If you put sin−1 k into your calculator, the answer will be an angle
I
in [− π2 , π2 ] using radian mode
I
in [−90◦ , 90◦ ] in degree mode
Inverse Cosine
If you put cos−1 k into your calculator, the answer will be an angle
I
in [0, π] using radian mode
I
in [0, 180◦ ] in degree mode
Inverse Tangent
If you put tan−1 k into your calculator, the answer will be an angle
I
in (− π2 , π2 ) using radian mode
I
in (−90◦ , 90◦ ) in degree mode
Examples with negative values
sin θ = −0.5
θ = sin−1 (− 21 ) = −30◦ ≈ −0.5236
Examples with negative values
sin θ = −0.5
θ = sin−1 (− 21 ) = −30◦ ≈ −0.5236
cos θ = −0.5
θ = cos−1 (− 21 ) = 120◦ ≈ 2.0944
Examples with negative values
sin θ = −0.5
θ = sin−1 (− 21 ) = −30◦ ≈ −0.5236
cos θ = −0.5
θ = cos−1 (− 21 ) = 120◦ ≈ 2.0944
tan θ = −0.5
θ = tan−1 (− 12 ) ≈ −26.57◦ ≈ −0.4636
Examples with negative values
sin θ = −0.5
θ = sin−1 (− 21 ) = −30◦ ≈ −0.5236
cos θ = −0.5
θ = cos−1 (− 21 ) = 120◦ ≈ 2.0944
tan θ = −0.5
θ = tan−1 (− 12 ) ≈ −26.57◦ ≈ −0.4636
If the calculator doesn’t give you the angle θ you wanted...
...use reference angles to find the angle you want!
23
Find θ such that tan θ = −0.4623 and 0◦ ≤ θ < 360◦
Find θ such that tan θ = −0.4623 and 0◦ ≤ θ < 360◦
Putting tan−1 (−0.4623) in the calculator (in degree mode).
Find θ such that tan θ = −0.4623 and 0◦ ≤ θ < 360◦
Putting tan−1 (−0.4623) in the calculator (in degree mode).
I
Get ≈ −24.8◦ .
Find θ such that tan θ = −0.4623 and 0◦ ≤ θ < 360◦
Putting tan−1 (−0.4623) in the calculator (in degree mode).
I
Get ≈ −24.8◦ .
Problem: We wanted θ such that 0◦ ≤ θ < 360◦ .
Find θ such that tan θ = −0.4623 and 0◦ ≤ θ < 360◦
Putting tan−1 (−0.4623) in the calculator (in degree mode).
I
Get ≈ −24.8◦ .
Problem: We wanted θ such that 0◦ ≤ θ < 360◦ .
y
x
−24.8◦
Find θ such that tan θ = −0.4623 and 0◦ ≤ θ < 360◦
Putting tan−1 (−0.4623) in the calculator (in degree mode).
I
Get ≈ −24.8◦ .
Problem: We wanted θ such that 0◦ ≤ θ < 360◦ .
y
Solution: tan is
180◦ -periodic:
x
−24.8◦
Find θ such that tan θ = −0.4623 and 0◦ ≤ θ < 360◦
Putting tan−1 (−0.4623) in the calculator (in degree mode).
I
Get ≈ −24.8◦ .
Problem: We wanted θ such that 0◦ ≤ θ < 360◦ .
y
Solution: tan is
180◦ -periodic:
θ
I
x
−24.8◦
Add 180◦ to −24.8◦
I
θ = 155.2◦
Find θ such that tan θ = −0.4623 and 0◦ ≤ θ < 360◦
Putting tan−1 (−0.4623) in the calculator (in degree mode).
I
Get ≈ −24.8◦ .
Problem: We wanted θ such that 0◦ ≤ θ < 360◦ .
y
Solution: tan is
180◦ -periodic:
θ
I
x
−24.8◦
Add 180◦ to −24.8◦
I
I
θ = 155.2◦
Add 180◦ to 155.2◦
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θ = 335.2◦
Find θ such that cos θ = −0.3842 and 0 ≤ θ < 2π
Find θ such that cos θ = −0.3842 and 0 ≤ θ < 2π
Putting cos−1 (−0.3842) in the calculator (in radian mode).
Find θ such that cos θ = −0.3842 and 0 ≤ θ < 2π
Putting cos−1 (−0.3842) in the calculator (in radian mode).
I
Get ≈ 1.9651.
Find θ such that cos θ = −0.3842 and 0 ≤ θ < 2π
Putting cos−1 (−0.3842) in the calculator (in radian mode).
I
Get ≈ 1.9651.
I
Since 1.9651 is between 0 and π, reference angle is
≈ π − 1.9651 ≈ 1.1765
y
x
Find θ such that cos θ = −0.3842 and 0 ≤ θ < 2π
Putting cos−1 (−0.3842) in the calculator (in radian mode).
I
Get ≈ 1.9651.
I
Since 1.9651 is between 0 and π, reference angle is
≈ π − 1.9651 ≈ 1.1765
y
θR
θ ≈ 1.9651 is a solution.
θ
x
Find θ such that cos θ = −0.3842 and 0 ≤ θ < 2π
Putting cos−1 (−0.3842) in the calculator (in radian mode).
I
Get ≈ 1.9651.
I
Since 1.9651 is between 0 and π, reference angle is
≈ π − 1.9651 ≈ 1.1765
y
θR
θ ≈ 1.9651 is a solution.
θ
x
Find another θ with the same
x value?
Find θ such that cos θ = −0.3842 and 0 ≤ θ < 2π
Putting cos−1 (−0.3842) in the calculator (in radian mode).
I
Get ≈ 1.9651.
I
Since 1.9651 is between 0 and π, reference angle is
≈ π − 1.9651 ≈ 1.1765
y
θ ≈ 1.9651 is a solution.
θR
θ
x
θR
Find another θ with the same
x value?
Find θ such that cos θ = −0.3842 and 0 ≤ θ < 2π
Putting cos−1 (−0.3842) in the calculator (in radian mode).
I
Get ≈ 1.9651.
I
Since 1.9651 is between 0 and π, reference angle is
≈ π − 1.9651 ≈ 1.1765
y
θ
θ ≈ 1.9651 is a solution.
x
θR
Find another θ with the same
x value?
Find θ such that cos θ = −0.3842 and 0 ≤ θ < 2π
Putting cos−1 (−0.3842) in the calculator (in radian mode).
I
Get ≈ 1.9651.
I
Since 1.9651 is between 0 and π, reference angle is
≈ π − 1.9651 ≈ 1.1765
y
θ
θ ≈ 1.9651 is a solution.
x
Find another θ with the same
x value?
θR
θ ≈ π + 1.1765 ≈ 4.3180 is a
second solution
Main idea
Use the symmetry in the circle with ± to get sin, cos, tan
Main idea
Use the symmetry in the circle with ± to get sin, cos, tan
The angles which have related x and y value have the same
reference angle!