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2
(x, y)
y
θ
x
Trigonometric Formula Sheet
Definition of the Trig Functions
Right Triangle Definition
Assume that:
Unit Circle Definition
Assume θ can be any angle.
0 < θ <
π
or 0
◦
< θ < 90
◦
y
hypotenuse
θ
adjacent
opposite
x
sin θ =
opp
hyp
csc θ =
hyp
opp
sin θ =
y
csc θ =
y
cos θ =
adj
hyp
sec θ =
hyp
adj
cos θ =
x
sec θ =
x
tan θ =
opp
adj
cot θ =
adj
opp
tan θ =
y
x
cot θ =
x
y
sin θ, ∀ θ ∈ (−∞ , ∞)
cos θ, ∀ θ ∈ (−∞ , ∞)
Domains of the Trig Functions
csc θ, ∀ θ /= nπ, where n ∈ Z
sec θ, ∀ θ n +
π, where n ∈ Z
tan θ, ∀ θ
n +
π, where n ∈ Z
cot θ, ∀ θ nπ, where n ∈ Z
Ranges of the Trig Functions
− 1 ≤ sin θ ≤ 1
− 1 ≤ cos θ ≤ 1
−∞ ≤ tan θ ≤ ∞
csc θ ≥ 1 and csc θ ≤
− 1 sec θ ≥ 1 and sec θ
−∞ ≤ cot θ ≤ ∞
Periods of the Trig Functions
The period of a function is the number, T, such that f ( θ +T ) = f ( θ ).
So, if ω is a fixed number and θ is any angle we have the following periods.
2 π sin( ωθ ) ⇒ T =
ω
2 π cos( ωθ ) ⇒ T =
ω
π
2 π csc( ωθ ) ⇒ T =
ω
2 π sec( ωθ ) ⇒ T =
ω
π
tan( ωθ ) ⇒ T =
ω
cot( ωθ ) ⇒ T =
ω
r
Identities and Formulas
Tangent and Cotangent Identities Half Angle Formulas
sin θ
tan θ =
cos θ
cos θ
cot θ =
sin θ
sin θ = ±
1 − cos(2 θ )
Reciprocal Identities
cos θ = ±
r
1 + cos(2 θ )
s
1 − cos(2 θ )
Pythagorean Identities
sin
2
θ + cos
2
θ =
1 tan
2
θ + 1 =
sec
2
θ 1 + cot
2
θ
= csc
2
θ
Even and Odd Formulas
Sum and Difference Formulas
sin( α ± β ) = sin α cos β ± cos α sin β
cos( α ± β ) = cos α cos β ∓ sin α sin β
tan( α β ) =
tan α ± tan β
1 ∓ tan α tan β
Product to Sum Formulas
sin(− θ ) = − sin θ
cos(− θ ) = cos θ
tan(− θ ) = − tan
θ
Periodic Formulas
If n is an integer
sin( θ + 2 πn ) = sin θ
cos( θ + 2 πn ) = cos θ
tan( θ + πn ) = tan θ
csc(− θ ) = − csc
θ sec(− θ ) = sec θ
cot(− θ ) = − cot
θ
csc( θ + 2 πn ) = csc θ
sec( θ + 2 πn ) = sec θ
cot( θ + πn ) = cot θ
sin α sin β =
[cos( α − β ) − cos( α + β )]
cos α cos β =
[cos( α − β ) + cos( α + β )]
sin α cos β =
[sin( α + β ) + sin( α − β )]
cos α sin β =
[sin( α + β ) − sin( α − β )]
Sum to Product Formulas
sin α + sin β = 2 sin
α + β
cos
α −
β
sin(2 θ ) = 2 sin θ cos θ
cos(2 θ ) = cos
2
θ − sin
2
θ
sin α − sin β = 2 cos
α + β
sin
α − β
cos α + cos β = 2 cos
α + β
cos
α − β
= 2 cos
2
θ −
= 1 − 2 sin
2
θ
2 tan θ
cos α − cos β = − 2
sin
α + β
sin
α − β
tan(2 θ ) =
1 − tan
2
θ
Degrees to Radians Formulas
If x is an angle in degrees and t is an angle in
radians then:
Cofunction Formulas
sin
π
− θ = cos θ
csc
π
− θ = sec θ
cos
π
− θ = sin θ
sec
π
− θ = csc θ
π t
πx
⇒ t =
◦
t
and x =
tan
π
− θ = cot
θ
tan θ =
1 + cos(2 θ )
Double Angle Formulas
◦
x
◦
π
sin θ =
csc θ
csc θ =
sin θ
cos θ =
sec θ
sec θ =
cos θ
tan θ =
cot θ
cot θ =
tan θ
13
√
90 ◦, π
2
22
13
√
√ 2 √ 2 √ 2 √ 2
120 ◦, 2 π
3
60 ◦, π
3
√
3 1
135 ◦, 3 π 45 ◦, π
22
4
4
22
150 ◦, 5 π
6
30 ◦, π
6
180 ◦, π
0 ◦, 2π
210 ◦, 7 π
6
330 ◦, 11 π
6
√ 3
1
√ 3
2 2
225 ◦, 5 π
4
315 ◦, 7 π
4
1
2 2
240 ◦, 4 π
3
300 ◦, 5 π
3
√ 2 √ 2
√ 2 √ 2
1
√ 3 √ 3
22
270 ◦, 3 π
1
2
2
2
Unit Circle
For any ordered pair on the unit circle ( x, y ) : cos θ = x and sin θ = y
Example
cos (
7 π
) = −
sin (
7 π
) = −
2
2
2
2
2
β
a c
γ
α
Inverse Trig Functions
Definition
θ = sin
− 1
( x ) is equivalent to x = sin θ
Inverse Properties
These properties hold for x in the domain and θ in
the range
θ = cos
− 1
( x ) is equivalent to x = cos θ
θ = tan
− 1
( x ) is equivalent to x = tan θ
Domain and Range
sin(sin
− 1
( x )) = x
cos(cos
− 1
( x )) = x
tan(tan
− 1
( x )) =
x
sin
− 1
(sin( θ )) = θ
cos
− 1
(cos( θ )) = θ
tan
− 1
(tan( θ )) = θ
Function
θ = sin
− 1
( x )
θ = cos
− 1
( x )
θ = tan
− 1
( x )
Domain
− 1 ≤ x ≤ 1
− 1 ≤ x ≤ 1
−∞ ≤ x ≤ ∞
Range
π π
≤ θ ≤
0 ≤ θ ≤
π π π
— < θ <
Other Notations
sin
− 1
( x ) = arcsin( x )
cos
− 1
( x ) = arccos( x )
tan
− 1
( x ) = arctan( x )
Law of Sines, Cosines, and Tangents
b
Law of Sines Law of Tangents
sin α sin β sin
γ
a − b
tan
1
( α − β )
a b
c
Law of Cosines
a +
b
b − c
tan
1
( α + β )
tan
1
( β − γ )
a
2
= b
2
2
− 2 bc cos
α
b + c
2
tan
1
( β + γ )
b
2
= a
2
2
− 2 ac cos
β
a − c
tan
1
( α − γ )
c
2
= a
2
2
− 2 ab cos γ
a + c
− a = i
a, a ≥ 0
Complex Numbers
i =
− 1 i
2
= − 1 i
3
= − i i
4
( a + bi )( a − bi ) = a
2
b
2
( a + bi ) + ( c + di ) = a + c + ( b +
d ) i ( a + bi ) − ( c + di ) = a − c + ( b −
d ) i ( a + bi )( c + di ) = ac − bd + ( ad +
bc ) i
| a + bi | =
a
2
2
Complex Modulus
( a + bi ) = a − bi Complex Conjugate
( a + bi )( a + bi ) = | a + bi |
2
DeMoivre’s Theorem
Let z = r (cos θ + i sin θ ), and let n be a positive integer.
Then:
Example: Let z = 1 − i , find z
6
z
n
= r
n
(cos nθ + i sin nθ ).
Solution: First write z in polar form.
r =
2
2
θ = arg ( z ) = tan
− 1
π
Polar Form: z =
2 cos −
π
π
Applying DeMoivre’s Theorem gives :
z
6
6
cos 6 · −
π
π
3
cos −
3 π
3 π
= 8(0 + i (1))
= 8 i
n
− 1
θ = arg ( z ) = tan
4
4
4
4
◦
◦
◦
◦
Finding the nth roots of a number using DeMoivre’s Theorem
Example: Find all the complex fourth roots of 4. That is, find all the complex solutions of
x
4
We are asked to find all complex fourth roots of 4.
These are all the solutions (including the complex values) of the equation x
4
For any positive integer n , a nonzero complex number z has exactly n distinct n th roots.
More specifically, if z is written in the trigonometric form r (cos θ + i sin θ ), the n th roots of
z are given by the following formula.
(∗) r
1
cos
◦
k
+ i sin
◦
k
, for k = 0 , 1 , 2 , ..., n − 1.
n n n n
Remember from the previous example we need to write 4 in trigonometric form by using:
r = ( a )
2
2
and θ = arg ( z ) = tan
− 1
b
a
So we have the complex number a + ib = 4 + i 0.
Therefore a = 4 and b = 0
So r =
2
2
= 4 and
Finally our trigonometric form is 4 = 4(cos 0
◦
◦
Using the formula (∗) above with n = 4, we can find the fourth roots of 4(cos 0
◦
◦
For k = 0 , 4 cos + + i sin + =
2 (cos(
◦
) + i sin(
◦
For k = 1 , 4
1
cos
◦
◦
◦
◦
2 (cos(
◦
) + i sin(
◦
2 i
For k = 2 , 4
1
cos
◦
◦
◦
◦
2 (cos(
◦
) + i sin(
◦
For k = 3 , 4
1
cos
◦
◦
◦
◦
2 (cos(
◦
) + i sin(
◦
2 i
Thus all of the complex roots of x
4
= 4 are:
2i , −
2i.
More Conic Sections
Hyperbola
Standard Form for Horizontal Transverse Axis :
( x h )
2
a
2
( y k )
2
b
2
Standard Form for V ertical Transverse Axis :
( y k )
2
a
2
( x h )
2
b
2
Where ( h , k )= center
a =distance between center and either vertex
Foci can be found by using b
2
= c
2
− a
2
Where c is the distance between
center and either focus. ( b > 0 )
Parabola
Vertical axis: y = a ( x − h )
2
+ k
Horizontal axis: x = a ( y − k )
2
+ h
Where ( h , k )= vertex
a =scaling factor
f (x) = sin(x)
1
√ 3
2
√ 2
2
1
2
0 πππ
643
π
2
2π3π5π
3 4 6
π
7π5π4π
64 3
3π
2
5π7π 11π
3 4 6
2π
−
1
2
— 2
√ 2
— 2
-
√ 3
Example : sin
5π
√
4
= −
2
2
f (x) = cos(x)
1
√ 3
2
√ 2
2
1
2
0 πππ
643
π
2
2π3π5π
346
π
7π5π4π
643
3π
2
5π7π 11π
346
2π
−
1
2
— 2
√ 2
— 2
-
√ 3
Example : cos
7π
√
6
= −
3
2
f ( x )
x
f ( x )
x
