Download Math 113/114 Lecture 3

Trigonometric Functions
Math 113/114 Lecture 3
Xi Chen 1
1 University
of Alberta
September 10, 2014
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Outline
1
Trigonometric Functions
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Basic Definition
Trigonometric functions cos x, sin x, tan x, cot x, sec x and
csc x are defined by
c
a
x
sin x =
a
c
cos x =
b
c
tan x =
a
b
cot x =
b
a
sec x =
c
b
csc x =
c
a
b
Pythagorean Theorem: a2 + b2 = c 2
We use Radian instead of Degree: 180◦ = π rad.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Basic Definition
Trigonometric functions cos x, sin x, tan x, cot x, sec x and
csc x are defined by
c
a
x
sin x =
a
c
cos x =
b
c
tan x =
a
b
cot x =
b
a
sec x =
c
b
csc x =
c
a
b
Pythagorean Theorem: a2 + b2 = c 2
We use Radian instead of Degree: 180◦ = π rad.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Special Angles
x
0
π
= 30◦
6
√
cos x
sin x
1
3
2
0
1
2
π
= 45◦
4
√
2
2
√
2
2
√
tan x
0
3
3
1
Xi Chen
π
= 60◦
3
π
= 90◦
2
1
2
0
√
3
2
√
3
Math 113/114 Lecture 3
1
ND
Trigonometric Functions
Basic Identities
By the definition,
sin x
, cot x
cos x
1
, csc x
sec x =
cos x
tan x =
cos x
sin x
1
=
sin x
=
Consequences of Pythagorean:
cos2 x + sin2 x = 1
sec2 x − tan2 x = 1
csc2 x − cot2 x = 1
| cos x|, | sin x| ≤ 1 and | sec x|, | csc x| ≥ 1 for all x.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Basic Identities
By the definition,
sin x
, cot x
cos x
1
, csc x
sec x =
cos x
tan x =
cos x
sin x
1
=
sin x
=
Consequences of Pythagorean:
cos2 x + sin2 x = 1
sec2 x − tan2 x = 1
csc2 x − cot2 x = 1
| cos x|, | sin x| ≤ 1 and | sec x|, | csc x| ≥ 1 for all x.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Basic Identities
By the definition,
sin x
, cot x
cos x
1
, csc x
sec x =
cos x
tan x =
cos x
sin x
1
=
sin x
=
Consequences of Pythagorean:
cos2 x + sin2 x = 1
sec2 x − tan2 x = 1
csc2 x − cot2 x = 1
| cos x|, | sin x| ≤ 1 and | sec x|, | csc x| ≥ 1 for all x.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Show that
cos4 x + sin4 x = 1 − 2(cos x sin x)2 for all x.
Proof.
cos2 x + sin2 x = 1 ⇒ (cos2 x + sin2 x)2 = 1
⇒ cos4 x + sin4 x + 2(cos x sin x)2 = 1
⇒ cos4 x + sin4 x = 1 − 2(cos x sin x)2
Example. Solve the equation sin(x) + cos2 (2x) = 2.
Solution. Since sin(x) ≤ 1 and cos2 (2x) ≤ 1,
sin(x) + cos2 (2x) = 2 if and only if sin(x) = cos2 (2x) = 1.
Therefore,
π
sin(x) = 1 ⇒ x = 2nπ +
2
2
and since cos (2(2nπ + π/2)) = 1, the solutions are
x = 2nπ + π/2 for n integers.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Show that
cos4 x + sin4 x = 1 − 2(cos x sin x)2 for all x.
Proof.
cos2 x + sin2 x = 1 ⇒ (cos2 x + sin2 x)2 = 1
⇒ cos4 x + sin4 x + 2(cos x sin x)2 = 1
⇒ cos4 x + sin4 x = 1 − 2(cos x sin x)2
Example. Solve the equation sin(x) + cos2 (2x) = 2.
Solution. Since sin(x) ≤ 1 and cos2 (2x) ≤ 1,
sin(x) + cos2 (2x) = 2 if and only if sin(x) = cos2 (2x) = 1.
Therefore,
π
sin(x) = 1 ⇒ x = 2nπ +
2
2
and since cos (2(2nπ + π/2)) = 1, the solutions are
x = 2nπ + π/2 for n integers.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Show that
cos4 x + sin4 x = 1 − 2(cos x sin x)2 for all x.
Proof.
cos2 x + sin2 x = 1 ⇒ (cos2 x + sin2 x)2 = 1
⇒ cos4 x + sin4 x + 2(cos x sin x)2 = 1
⇒ cos4 x + sin4 x = 1 − 2(cos x sin x)2
Example. Solve the equation sin(x) + cos2 (2x) = 2.
Solution. Since sin(x) ≤ 1 and cos2 (2x) ≤ 1,
sin(x) + cos2 (2x) = 2 if and only if sin(x) = cos2 (2x) = 1.
Therefore,
π
sin(x) = 1 ⇒ x = 2nπ +
2
2
and since cos (2(2nπ + π/2)) = 1, the solutions are
x = 2nπ + π/2 for n integers.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Show that
cos4 x + sin4 x = 1 − 2(cos x sin x)2 for all x.
Proof.
cos2 x + sin2 x = 1 ⇒ (cos2 x + sin2 x)2 = 1
⇒ cos4 x + sin4 x + 2(cos x sin x)2 = 1
⇒ cos4 x + sin4 x = 1 − 2(cos x sin x)2
Example. Solve the equation sin(x) + cos2 (2x) = 2.
Solution. Since sin(x) ≤ 1 and cos2 (2x) ≤ 1,
sin(x) + cos2 (2x) = 2 if and only if sin(x) = cos2 (2x) = 1.
Therefore,
π
sin(x) = 1 ⇒ x = 2nπ +
2
2
and since cos (2(2nπ + π/2)) = 1, the solutions are
x = 2nπ + π/2 for n integers.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Show that
cos4 x + sin4 x = 1 − 2(cos x sin x)2 for all x.
Proof.
cos2 x + sin2 x = 1 ⇒ (cos2 x + sin2 x)2 = 1
⇒ cos4 x + sin4 x + 2(cos x sin x)2 = 1
⇒ cos4 x + sin4 x = 1 − 2(cos x sin x)2
Example. Solve the equation sin(x) + cos2 (2x) = 2.
Solution. Since sin(x) ≤ 1 and cos2 (2x) ≤ 1,
sin(x) + cos2 (2x) = 2 if and only if sin(x) = cos2 (2x) = 1.
Therefore,
π
sin(x) = 1 ⇒ x = 2nπ +
2
2
and since cos (2(2nπ + π/2)) = 1, the solutions are
x = 2nπ + π/2 for n integers.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Show that
cos4 x + sin4 x = 1 − 2(cos x sin x)2 for all x.
Proof.
cos2 x + sin2 x = 1 ⇒ (cos2 x + sin2 x)2 = 1
⇒ cos4 x + sin4 x + 2(cos x sin x)2 = 1
⇒ cos4 x + sin4 x = 1 − 2(cos x sin x)2
Example. Solve the equation sin(x) + cos2 (2x) = 2.
Solution. Since sin(x) ≤ 1 and cos2 (2x) ≤ 1,
sin(x) + cos2 (2x) = 2 if and only if sin(x) = cos2 (2x) = 1.
Therefore,
π
sin(x) = 1 ⇒ x = 2nπ +
2
2
and since cos (2(2nπ + π/2)) = 1, the solutions are
x = 2nπ + π/2 for n integers.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Show that
cos4 x + sin4 x = 1 − 2(cos x sin x)2 for all x.
Proof.
cos2 x + sin2 x = 1 ⇒ (cos2 x + sin2 x)2 = 1
⇒ cos4 x + sin4 x + 2(cos x sin x)2 = 1
⇒ cos4 x + sin4 x = 1 − 2(cos x sin x)2
Example. Solve the equation sin(x) + cos2 (2x) = 2.
Solution. Since sin(x) ≤ 1 and cos2 (2x) ≤ 1,
sin(x) + cos2 (2x) = 2 if and only if sin(x) = cos2 (2x) = 1.
Therefore,
π
sin(x) = 1 ⇒ x = 2nπ +
2
2
and since cos (2(2nπ + π/2)) = 1, the solutions are
x = 2nπ + π/2 for n integers.
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Periodic Behavior
Basic Relations:
π
) = − sin x
2
π
sin(−x) = − sin x, sin(x + ) = cos x
2
cos(−x) = cos(x), cos(x +
Example. Find cos(x + π), sin(x + π) and tan(x + π).
π
π
π
) + ) = − sin(x + ) = − cos(x)
2
2
2
π
π
π
sin(x + π) = sin((x + ) + ) = cos(x + ) = − sin(x)
2
2
2
sin(x + π)
− sin(x)
tan(x + π) =
=
= tan(x).
cos(x + π)
− cos(x)
cos(x + π) = cos((x +
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Periodic Behavior
Basic Relations:
π
) = − sin x
2
π
sin(−x) = − sin x, sin(x + ) = cos x
2
cos(−x) = cos(x), cos(x +
Example. Find cos(x + π), sin(x + π) and tan(x + π).
π
π
π
) + ) = − sin(x + ) = − cos(x)
2
2
2
π
π
π
sin(x + π) = sin((x + ) + ) = cos(x + ) = − sin(x)
2
2
2
sin(x + π)
− sin(x)
tan(x + π) =
=
= tan(x).
cos(x + π)
− cos(x)
cos(x + π) = cos((x +
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Periodic Behavior
Basic Relations:
π
) = − sin x
2
π
sin(−x) = − sin x, sin(x + ) = cos x
2
cos(−x) = cos(x), cos(x +
Example. Find cos(x + π), sin(x + π) and tan(x + π).
π
π
π
) + ) = − sin(x + ) = − cos(x)
2
2
2
π
π
π
sin(x + π) = sin((x + ) + ) = cos(x + ) = − sin(x)
2
2
2
sin(x + π)
− sin(x)
tan(x + π) =
=
= tan(x).
cos(x + π)
− cos(x)
cos(x + π) = cos((x +
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Find cos(x + 2π) and sin(x + 2π).
cos(x + 2π) = cos((x + π) + π) = − cos(x + π) = cos(x)
sin(x + 2π) = sin((x + π) + π) = − sin(x + π) = sin(x).
Example. Find cos(2015π − x) and sin(2015π − x).
cos(2015π − x) = cos((2014π − x) + π) = (−1) cos(2014π − x)
= (−1)2 cos(2013π − x) = ... = (−1)2015 cos(−x) = − cos x
sin(2015π − x) = sin((2014π − x) + π) = (−1) sin(2014π − x)
= (−1)2 sin(2013π − x) = ... = (−1)2015 sin(−x) = sin x
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Find cos(x + 2π) and sin(x + 2π).
cos(x + 2π) = cos((x + π) + π) = − cos(x + π) = cos(x)
sin(x + 2π) = sin((x + π) + π) = − sin(x + π) = sin(x).
Example. Find cos(2015π − x) and sin(2015π − x).
cos(2015π − x) = cos((2014π − x) + π) = (−1) cos(2014π − x)
= (−1)2 cos(2013π − x) = ... = (−1)2015 cos(−x) = − cos x
sin(2015π − x) = sin((2014π − x) + π) = (−1) sin(2014π − x)
= (−1)2 sin(2013π − x) = ... = (−1)2015 sin(−x) = sin x
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Find cos(x + 2π) and sin(x + 2π).
cos(x + 2π) = cos((x + π) + π) = − cos(x + π) = cos(x)
sin(x + 2π) = sin((x + π) + π) = − sin(x + π) = sin(x).
Example. Find cos(2015π − x) and sin(2015π − x).
cos(2015π − x) = cos((2014π − x) + π) = (−1) cos(2014π − x)
= (−1)2 cos(2013π − x) = ... = (−1)2015 cos(−x) = − cos x
sin(2015π − x) = sin((2014π − x) + π) = (−1) sin(2014π − x)
= (−1)2 sin(2013π − x) = ... = (−1)2015 sin(−x) = sin x
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Find cos(x + 2π) and sin(x + 2π).
cos(x + 2π) = cos((x + π) + π) = − cos(x + π) = cos(x)
sin(x + 2π) = sin((x + π) + π) = − sin(x + π) = sin(x).
Example. Find cos(2015π − x) and sin(2015π − x).
cos(2015π − x) = cos((2014π − x) + π) = (−1) cos(2014π − x)
= (−1)2 cos(2013π − x) = ... = (−1)2015 cos(−x) = − cos x
sin(2015π − x) = sin((2014π − x) + π) = (−1) sin(2014π − x)
= (−1)2 sin(2013π − x) = ... = (−1)2015 sin(−x) = sin x
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Example
Example. Find
sin(
2015π
2015π
) and cos(
)
3
3
2015π
π
π
) = sin(672π − ) = (−1)672 sin(− )
3
3√
3
π
3
= − sin( ) = −
3
2
π
π
2015π
) = cos(672π − ) = (−1)672 cos(− )
cos(
3
3
3
π
1
= cos( ) =
3
2
sin(
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Example
Example. Find
sin(
2015π
2015π
) and cos(
)
3
3
2015π
π
π
) = sin(672π − ) = (−1)672 sin(− )
3
3√
3
π
3
= − sin( ) = −
3
2
π
π
2015π
) = cos(672π − ) = (−1)672 cos(− )
cos(
3
3
3
π
1
= cos( ) =
3
2
sin(
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Addition Formula
Derive
cos(x + y) = cos(x) cos(y ) − sin(x) sin(y )
from law of cosine:
c
x
y
cos(x + y) =
a
c 2 + e2 − (a + d)2
2ce
b
c = b sec x, e = b sec y
e
d
a = b tan x, d = b tan y
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Derive a formula for cos(x − y ).
cos(x − y ) = cos(x + (−y)) = cos(x) cos(−y) − sin(x) sin(−y)
= cos(x) cos(y) + sin(x) sin(y)
Example. Derive formulas sin(x + y) and sin(x − y ).
π
)
2
π
π
= − cos(x) cos(y + ) + sin(x) sin(y + )
2
2
= cos(x) sin(y ) + sin(x) cos(y )
sin(x + y) = − cos(x + y +
sin(x − y) = sin(x + (−y))
= cos(x) sin(−y ) + sin(x) cos(−y)
= sin(x) cos(y ) − cos(x) sin(y).
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Derive a formula for cos(x − y ).
cos(x − y ) = cos(x + (−y)) = cos(x) cos(−y) − sin(x) sin(−y)
= cos(x) cos(y) + sin(x) sin(y)
Example. Derive formulas sin(x + y) and sin(x − y ).
π
)
2
π
π
= − cos(x) cos(y + ) + sin(x) sin(y + )
2
2
= cos(x) sin(y ) + sin(x) cos(y )
sin(x + y) = − cos(x + y +
sin(x − y) = sin(x + (−y))
= cos(x) sin(−y ) + sin(x) cos(−y)
= sin(x) cos(y ) − cos(x) sin(y).
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Derive a formula for cos(x − y ).
cos(x − y ) = cos(x + (−y)) = cos(x) cos(−y) − sin(x) sin(−y)
= cos(x) cos(y) + sin(x) sin(y)
Example. Derive formulas sin(x + y) and sin(x − y ).
π
)
2
π
π
= − cos(x) cos(y + ) + sin(x) sin(y + )
2
2
= cos(x) sin(y ) + sin(x) cos(y )
sin(x + y) = − cos(x + y +
sin(x − y) = sin(x + (−y))
= cos(x) sin(−y ) + sin(x) cos(−y)
= sin(x) cos(y ) − cos(x) sin(y).
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Derive a formula for cos(x − y ).
cos(x − y ) = cos(x + (−y)) = cos(x) cos(−y) − sin(x) sin(−y)
= cos(x) cos(y) + sin(x) sin(y)
Example. Derive formulas sin(x + y) and sin(x − y ).
π
)
2
π
π
= − cos(x) cos(y + ) + sin(x) sin(y + )
2
2
= cos(x) sin(y ) + sin(x) cos(y )
sin(x + y) = − cos(x + y +
sin(x − y) = sin(x + (−y))
= cos(x) sin(−y ) + sin(x) cos(−y)
= sin(x) cos(y ) − cos(x) sin(y).
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Derive formulas cos(2x) and sin(2x).
Let x = y. Then
cos(2x) = cos(x + x) = cos2 x − sin2 x
= 2 cos2 x − 1 = 1 − 2 sin2 x
sin(2x) = sin(x + x) = cos x sin x + sin x cos x = 2 cos x sin x
Example. Derive a formula for cos(3x).
Example. Derive a formula for tan(x + y).
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Derive formulas cos(2x) and sin(2x).
Let x = y. Then
cos(2x) = cos(x + x) = cos2 x − sin2 x
= 2 cos2 x − 1 = 1 − 2 sin2 x
sin(2x) = sin(x + x) = cos x sin x + sin x cos x = 2 cos x sin x
Example. Derive a formula for cos(3x).
Example. Derive a formula for tan(x + y).
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Derive formulas cos(2x) and sin(2x).
Let x = y. Then
cos(2x) = cos(x + x) = cos2 x − sin2 x
= 2 cos2 x − 1 = 1 − 2 sin2 x
sin(2x) = sin(x + x) = cos x sin x + sin x cos x = 2 cos x sin x
Example. Derive a formula for cos(3x).
Example. Derive a formula for tan(x + y).
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Examples
Example. Derive formulas cos(2x) and sin(2x).
Let x = y. Then
cos(2x) = cos(x + x) = cos2 x − sin2 x
= 2 cos2 x − 1 = 1 − 2 sin2 x
sin(2x) = sin(x + x) = cos x sin x + sin x cos x = 2 cos x sin x
Example. Derive a formula for cos(3x).
Example. Derive a formula for tan(x + y).
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Trigonometric Equations and Inequalities
The solutions for cos x = a are x = 2nπ ± θ if cos θ = a.
The solutions for sin x = a are x = 2nπ + θ and
x = 2nπ + π − θ if sin θ = a
The solutions for tan x = a are x = nπ + θ if tan θ = a.
The solutions for cos x > a are 2nπ − θ < x < 2nπ + θ if
cos θ = a and θ ∈ [0, π].
The solutions for sin x > a are 2nπ + θ < x < 2nπ + π − θ if
sin θ = a and θ ∈ [−π/2, π/2].
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Trigonometric Equations and Inequalities
The solutions for cos x = a are x = 2nπ ± θ if cos θ = a.
The solutions for sin x = a are x = 2nπ + θ and
x = 2nπ + π − θ if sin θ = a
The solutions for tan x = a are x = nπ + θ if tan θ = a.
The solutions for cos x > a are 2nπ − θ < x < 2nπ + θ if
cos θ = a and θ ∈ [0, π].
The solutions for sin x > a are 2nπ + θ < x < 2nπ + π − θ if
sin θ = a and θ ∈ [−π/2, π/2].
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Trigonometric Equations and Inequalities
The solutions for cos x = a are x = 2nπ ± θ if cos θ = a.
The solutions for sin x = a are x = 2nπ + θ and
x = 2nπ + π − θ if sin θ = a
The solutions for tan x = a are x = nπ + θ if tan θ = a.
The solutions for cos x > a are 2nπ − θ < x < 2nπ + θ if
cos θ = a and θ ∈ [0, π].
The solutions for sin x > a are 2nπ + θ < x < 2nπ + π − θ if
sin θ = a and θ ∈ [−π/2, π/2].
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Trigonometric Equations and Inequalities
The solutions for cos x = a are x = 2nπ ± θ if cos θ = a.
The solutions for sin x = a are x = 2nπ + θ and
x = 2nπ + π − θ if sin θ = a
The solutions for tan x = a are x = nπ + θ if tan θ = a.
The solutions for cos x > a are 2nπ − θ < x < 2nπ + θ if
cos θ = a and θ ∈ [0, π].
The solutions for sin x > a are 2nπ + θ < x < 2nπ + π − θ if
sin θ = a and θ ∈ [−π/2, π/2].
Xi Chen
Math 113/114 Lecture 3
Trigonometric Functions
Trigonometric Equations and Inequalities
The solutions for cos x = a are x = 2nπ ± θ if cos θ = a.
The solutions for sin x = a are x = 2nπ + θ and
x = 2nπ + π − θ if sin θ = a
The solutions for tan x = a are x = nπ + θ if tan θ = a.
The solutions for cos x > a are 2nπ − θ < x < 2nπ + θ if
cos θ = a and θ ∈ [0, π].
The solutions for sin x > a are 2nπ + θ < x < 2nπ + π − θ if
sin θ = a and θ ∈ [−π/2, π/2].
Xi Chen
Math 113/114 Lecture 3