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Published by chirulai99, 2020-10-28 21:09:27

Trigonometric Formulas

Calculus

Keywords: Calculus

Trigonometric Formula Sheet

Definition of the Trig Functions

Right Triangle Definition Unit Circle Definition
Assume θ can be any angle.
Assume that:
y
0 < θ < π or 0◦ < θ < 90◦
2

(x, y)

hypotenuse opposite 1 x
y
θ
adjacent θ

x

sin θ = opp csc θ = hyp sin θ = y csc θ = 1
hyp opp 1 y

cos θ = adj sec θ = hyp cos θ = x sec θ = 1
hyp adj 1 x

tan θ = opp cot θ = adj tan θ = y cot θ = x
adj opp x y

Domains of the Trig Functions

sin θ, ∀ θ ∈ (−∞, ∞) csc θ, ∀ θ = nπ, where n ∈ Z

cos θ, ∀ θ ∈ (−∞, ∞) sec θ, ∀θ= n + 1 π, where n ∈ Z
2

tan θ, ∀θ= n + 1 π, where n ∈ Z cot θ, ∀ θ = nπ, where n ∈ Z
2

Ranges of the Trig Functions

−1 ≤ sin θ ≤ 1 csc θ ≥ 1 and csc θ ≤ −1
−1 ≤ cos θ ≤ 1 sec θ ≥ 1 and sec θ ≤ −1
−∞ ≤ tan 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.

sin(ωθ) ⇒ T = 2π csc(ωθ) ⇒ T = 2π
ω ω

cos(ωθ) ⇒ T = 2π sec(ωθ) ⇒ T = 2π
ω ω

tan(ωθ) ⇒ T = π cot(ωθ) ⇒ T = π
ω ω

1

Identities and Formulas

Tangent and Cotangent Identities Half Angle Formulas

tan θ = sin θ cot θ = cos θ sin θ = ± 1 − cos(2θ)
cos θ sin θ cos θ = ± 2

Reciprocal Identities 1 + cos(2θ)
2
1 1
sin θ = csc θ csc θ = sin θ 1 − cos(2θ)
1 + cos(2θ)
1 1 tan θ = ±
sec cos
cos θ = θ sec θ = θ

1 1 Sum and Difference Formulas
cot tan
tan θ = θ cot θ = θ

sin(α ± β) = sin α cos β ± cos α sin β

Pythagorean Identities cos(α ± β) = cos α cos β ∓ sin α sin β
sin2 θ + cos2 θ = 1
tan2 θ + 1 = sec2 θ tan(α ± β) = tan α ± tan β
1 + cot2 θ = csc2 θ 1 ∓ tan α tan β

Even and Odd Formulas Product to Sum Formulas

sin(−θ) = − sin θ csc(−θ) = − csc θ sin α sin β = 1 [cos(α − β) − cos(α + β)]
cos(−θ) = cos θ sec(−θ) = sec θ 2
tan(−θ) = − tan θ cot(−θ) = − cot θ
cos α cos β = 1 [cos(α − β) + cos(α + β)]
Periodic Formulas csc(θ + 2πn) = csc θ 2
If n is an integer sec(θ + 2πn) = sec θ
cot(θ + πn) = cot θ sin α cos β = 1 [sin(α + β) + sin(α − β)]
sin(θ + 2πn) = sin θ 2
cos(θ + 2πn) = cos θ
tan(θ + πn) = tan θ cos α sin β = 1 [sin(α + β) − sin(α − β)]
2

Sum to Product Formulas

Double Angle Formulas sin α + sin β = 2 sin α+β cos α−β
2 2

sin(2θ) = 2 sin θ cos θ sin α − sin β = 2 cos α+β sin α−β
2 2

cos(2θ) = cos2 θ − sin2 θ cos α + cos β = 2 cos α+β cos α−β
= 2 cos2 θ − 1 2 2
= 1 − 2 sin2 θ
cos α − cos β = −2 sin α+β sin α−β
2 2
2 tan θ
tan(2θ) = 1 − tan2 θ Cofunction Formulas

Degrees to Radians Formulas sin π − θ = cos θ cos π − θ = sin θ
If x is an angle in degrees and t is an angle in 2 2
radians then:
csc π − θ = sec θ sec π − θ = csc θ
2 2
180◦t
π = t ⇒ t = πx and x = π tan π − θ = cot θ cot π − θ = tan θ
180◦ x 180◦ 2 2

2

Unit Circle

(0, 1)

√ 90◦, π √
1 3 2 1 3
(− 2 , 2 ) ( 2 , 2 )

√√ √√
2 2 2 2
(− 2 , 2 ) 60◦, ( 2 , 2 )

120◦, 2π π
3 3

√ 135◦, 3π 45◦, π √
3 1 4 4 3 1
(− 2 , 2 ) ( 2 , 2 )

150◦, 5π 30◦, π
6 6

(−1, 0) 180◦, π 0◦, 2π (1, 0)

210◦, 7π 330◦, 11π
6 6

√ √
3 1 3 1
(− 2 , − 2 ) 225◦, 5π 315◦, 7π ( 2 , − 2 )
4 4

√√ 240◦, 4π 300◦, 5π √√
3 3
2 2 2 2
(− 2 , − 2 ) ( 2 , − 2 )

√ √
1 3 1 3
(− 2 , − 2 ) 270◦, 3π ( 2 , − 2 )
2

(0, −1)

F or any ordered pair on the unit circle (x, y) : cos θ = x and sin θ = y

Example


7π 3 7π 1
cos ( 6 ) = − 2 sin ( 6 ) = − 2

3

Inverse Trig Functions

Definition Inverse Properties
θ = sin−1(x) is equivalent to x = sin θ These properties hold for x in the domain and θ in
the range

θ = cos−1(x) is equivalent to x = cos θ sin(sin−1(x)) = x sin−1(sin(θ)) = θ
θ = tan−1(x) is equivalent to x = tan θ cos(cos−1(x)) = x cos−1(cos(θ)) = θ

Domain and Range tan(tan−1(x)) = x tan−1(tan(θ)) = θ

Function Domain Range
θ = sin−1(x) −1 ≤ x ≤ 1
θ = cos−1(x) −1 ≤ x ≤ 1 − π ≤ θ ≤ π Other Notations
θ = tan−1(x) −∞ ≤ x ≤ ∞ 2 2 sin−1(x) = arcsin(x)
cos−1(x) = arccos(x)
0≤θ≤π tan−1(x) = arctan(x)

− π < θ < π
2 2

Law of Sines, Cosines, and Tangents

β c
a

γ α
b

Law of Sines Law of Tangents

sin α = sin β = sin γ a− b = tan 1 (α − β)
a b c a+ b tan 2 + β)

1 (α
2

Law of Cosines 1
2
b − c = tan (β − γ)
b + c tan + γ)
a2 = b2 + c2 − 2bc cos α 1 (β
2

b2 = a2 + c2 − 2ac cos β a − c tan 1 (α − γ)
a + c tan 2 + γ)
=
1 (α
c2 = a2 + b2 − 2ab cos γ 2

4

Complex Numbers

√ i2 = −1 i3 = −i i4 = 1
i = −1

√√ (a + bi)(a − bi) = a2 + b2
−a = i a, a ≥ 0

(a + bi) + (c + di) = a + c + (b + d)i √
|a + bi| = a2 + b2 Complex Modulus

(a + bi) − (c + di) = a − c + (b − d)i (a + bi) = a − bi Complex Conjugate

(a + bi)(c + di) = ac − bd + (ad + bc)i (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:
zn = rn(cos nθ + i sin nθ).

Example: Let z = 1 − i, find z6.

Solution: First write z in polar form.


r = (1)2 + (−1)2 = 2

θ = arg(z) = tan−1 −1 = − π
1 4

Polar Form: z = 2 cos − π + i sin − π
4 4

Applying DeMoivre’s Theorem gives :

z6 = √6 cos 6 · − π + i sin 6 · − π
2 4 4

= 23 cos − 3π + i sin − 3π
2 2

= 8(0 + i(1))

= 8i

5

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
x4 = 4.

We are asked to find all complex fourth roots of 4.
These are all the solutions (including the complex values) of the equation x4 = 4.

For any positive integer n , a nonzero complex number z has exactly n distinct nth roots.
More specifically, if z is written in the trigonometric form r(cos θ + i sin θ), the nth roots of
z are given by the following formula.

(∗) r1 cos θ + 360◦k + i sin θ + 360◦k , f or k = 0, 1, 2, ..., n − 1.
n n n n n

Remember from the previous example we need to write 4 in trigonometric form by using:

r = (a)2 + (b)2 and θ = arg(z) = tan−1 b .
a

So we have the complex number a + ib = 4 + i0.

Therefore a = 4 and b = 0

So r = (4)2 + (0)2 = 4 and

θ = arg(z) = tan−1 0 =0
4
Finally our trigonometric form is 4 = 4(cos 0◦ + i sin 0◦)

Using the formula (∗) above with n = 4, we can find the fourth roots of 4(cos 0◦ + i sin 0◦)

• For k = 0, 41 cos 0◦ + 360◦ ∗ 0 + i sin 0◦ + 360◦ ∗ 0 √√
4 4 4 4 4 = 2 (cos(0◦) + i sin(0◦)) = 2

• For k = 1, 41 cos 0◦ + 360◦ ∗ 1 + i sin 0◦ + 360◦ ∗ 1 = √ (cos(90◦) + i sin(90◦)) = √
4 44 44 2 2i

• For k = 2, 41 cos 0◦ + 360◦ ∗ 2 + i sin 0◦ + 360◦ ∗ 2 = √ (cos(180◦) + i sin(180◦)) = √
4 4 4 4 4 2 −2

• For k = 3, 41 cos 0◦ + 360◦ ∗ 3 + i sin 0◦ + 360◦ ∗ 3 √√
4 4 4 4 4 = 2 (cos(270◦) + i sin(270◦)) = − 2i

Thus all of the complex roots of x4 = 4 are:
√√ √ √
2, 2i, − 2, − 2i .

6

Formulas for the Conic Sections

Circle
StandardF orm : (x − h)2 + (y − k)2 = r2

W here (h, k) = center and r = radius

Ellipse

Standard F orm f or Horizontal M ajor Axis :

(x − h)2 + (y − k)2 = 1
a2 b2

Standard F orm f or V ertical M ajor Axis :

(x − h)2 + (y − k)2 = 1
b2 a2

Where (h, k)= center

2a=length of major axis

2b=length of minor axis

(0 < b < a)

Foci can be found by using c2 = a2 − b2

Where c= foci length

7

More Conic Sections

Hyperbola

Standard F orm f or Horizontal T ransverse Axis :

(x − h)2 − (y − k)2 = 1
a2 b2

Standard F orm f or V ertical T ransverse Axis :

(y − k)2 − (x − h)2 = 1
a2 b2

Where (h, k)= center
a=distance between center and either vertex

Foci can be found by using b2 = c2 − a2
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

8

f (x)

f (x) = sin(x)

1


3

√2
2
2

1
2

x0 πππ π 2π 3π 5π π 7π 5π 4π 3π 5π 7π 11π 2π
643 2 346 643 2 346

− 1
2


2
− √2

− 3
2

-1

5π  √
2
Example : sin  = − 2

4

f (x)

f (x) = cos(x)

1


3

√2
2
2

1
2

x0 πππ π 2π 3π 5π π 7π 5π 4π 3π 5π 7π 11π 2π
643 2 346 643 2 346

− 1
2


2
− √2

− 3
2

-1

 7π  √
 3
Example : cos  = − 2

6

9

− π f (x) π
2 2

f (x) = tan x


3

1


3
3

−π − 5π − 3π − 2π − π − π − π 0 πππ 2π 3π 5π x
6 4 3 3 4 6 643 346
√ π

− 3
3

−1


−3

10


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