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
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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 .
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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
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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
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