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Algebra
MAA10

Sequences and Series

MAA11

Exponents and Logarithms

MAA12

Functions

Functions
MAA20

Polynomial Functions

MAA21

Rational Functions

MAA22

Exponential and Log Functions

MAA23

Trigonometric Functions

Geometry and Trigonometry
MAA30

Trigonometric Identities

MAA31

Solving Triangles

MAA32

Vectors

Calculus
MAA40

Limits and Continuity

MAA41

Derivatives

MAA42

Integrals

Statistics and Probability
MAA50

Descriptive Statistics

MAA51

Probability Concepts

MAA52

Probability Distributions

Advanced Topics
MAA60

Complex Numbers

MAA61

Further Calculus

MAA62

Matrices

Applications and Modeling
MAA70

Applications and Modeling

IA preparation
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IA Preparation (Math)

MAA81

IA Preparation (Physics)

Chapter MAA23

Trigonometric Functions

Functions

Angle Measurement

Angles can be measured in degrees or radians. Degrees are familiar from everyday use, with a full circle measuring 360°. Radians are preferred in calculus and advanced mathematics because they lead to simpler formulas.

Degree-Radian Conversion

180∘=π radians180^{\circ} = \pi \text{ radians}
1∘=π180 radians1^{\circ} = \frac{\pi}{180} \text{ radians}
1 radian=180∘π≈57.3∘1 \text{ radian} = \frac{180^{\circ}}{\pi} \approx 57.3^{\circ}
θrad=θdeg⋅π180\theta_{\text{rad}} = \theta_{\text{deg}} \cdot \frac{\pi}{180}
In radians, a full circle measures
2π2\pi
radians (approximately 6.28). Radians represent the ratio of arc length to radius, making them a natural unit for circular motion.
Problem: Convert
45∘45^{\circ}
to radians and
2π3\frac{2\pi}{3}
radians to degrees.
1
For
45∘45^{\circ}
to radians:
45∘=45⋅π180=π4 radians45^{\circ} = 45 \cdot \frac{\pi}{180} = \frac{\pi}{4} \text{ radians}
2
For
2π3\frac{2\pi}{3}
radians to degrees:
2π3=2π3⋅180∘π=360∘3=120∘\frac{2\pi}{3} = \frac{2\pi}{3} \cdot \frac{180^{\circ}}{\pi} = \frac{360^{\circ}}{3} = 120^{\circ}

The Unit Circle

The unit circle is a circle with radius 1 centered at the origin of the coordinate plane. It's a fundamental tool for understanding trigonometric functions.
For any angle
θ\theta
, we can define:
•
cos⁡θ\cos\theta
= x-coordinate of the point on the unit circle at angle
θ\theta
•
sin⁡θ\sin\theta
= y-coordinate of the point on the unit circle at angle
θ\theta
•
tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}
= slope of the line from origin to the point on the unit circle

Special Angles on the Unit Circle

Angle (degrees)Angle (radians)
cos⁡θ\cos\theta
sin⁡θ\sin\theta
tan⁡θ\tan\theta
0∘0^{\circ}
00
11
00
00
30∘30^{\circ}
π6\frac{\pi}{6}
32\frac{\sqrt{3}}{2}
12\frac{1}{2}
13\frac{1}{\sqrt{3}}
45∘45^{\circ}
π4\frac{\pi}{4}
22\frac{\sqrt{2}}{2}
22\frac{\sqrt{2}}{2}
11
60∘60^{\circ}
π3\frac{\pi}{3}
12\frac{1}{2}
32\frac{\sqrt{3}}{2}
3\sqrt{3}
90∘90^{\circ}
π2\frac{\pi}{2}
00
11
undefined\text{undefined}
180∘180^{\circ}
π\pi
−1-1
00
00
270∘270^{\circ}
3π2\frac{3\pi}{2}
00
−1-1
undefined\text{undefined}
360∘360^{\circ}
2π2\pi
11
00
00

The Six Trigonometric Functions

While sine and cosine are the most fundamental trigonometric functions, there are six trigonometric functions in total. Each has its own properties and uses.

Definitions of Trigonometric Functions

sin⁡θ=oppositehypotenuse\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}
cos⁡θ=adjacenthypotenuse\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}
tan⁡θ=sin⁡θcos⁡θ=oppositeadjacent\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\text{opposite}}{\text{adjacent}}
cot⁡θ=cos⁡θsin⁡θ=adjacentopposite\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\text{adjacent}}{\text{opposite}}
sec⁡θ=1cos⁡θ=hypotenuseadjacent\sec\theta = \frac{1}{\cos\theta} = \frac{\text{hypotenuse}}{\text{adjacent}}
csc⁡θ=1sin⁡θ=hypotenuseopposite\csc\theta = \frac{1}{\sin\theta} = \frac{\text{hypotenuse}}{\text{opposite}}
The domains of these functions are restricted by their definitions:
•
sin⁡θ\sin\theta
and
cos⁡θ\cos\theta
are defined for all real values of
θ\theta
.
•
tan⁡θ\tan\theta
and
sec⁡θ\sec\theta
are undefined when
cos⁡θ=0\cos\theta = 0
, which occurs at
θ=π2+nπ\theta = \frac{\pi}{2} + n\pi
where
nn
is an integer.
•
cot⁡θ\cot\theta
and
csc⁡θ\csc\theta
are undefined when
sin⁡θ=0\sin\theta = 0
, which occurs at
θ=nπ\theta = n\pi
where
nn
is an integer.

Graphs of Trigonometric Functions

Trigonometric functions are periodic, meaning their values repeat after a certain interval.

Key Properties of Trigonometric Graphs

Sine Function
• Domain: All real numbers
• Range:
[−1,1][-1, 1]

• Period:
2π2\pi

• x-intercepts:
nπn\pi
, where
nn
is an integer
Cosine Function
• Domain: All real numbers
• Range:
[−1,1][-1, 1]

• Period:
2π2\pi

• x-intercepts:
π2+nπ\frac{\pi}{2} + n\pi
, where
nn
is an integer
Tangent Function
• Domain: All real numbers except
π2+nπ\frac{\pi}{2} + n\pi

• Range: All real numbers
• Period:
π\pi

• x-intercepts:
nπn\pi
, where
nn
is an integer

Trigonometric Function Transformations

Like other functions, trigonometric functions can undergo various transformations. The general form for a transformed sine or cosine function is:
f(x)=Asin⁡(B(x−C))+Dorf(x)=Acos⁡(B(x−C))+Df(x) = A\sin(B(x - C)) + D \quad \text{or} \quad f(x) = A\cos(B(x - C)) + D
Each parameter affects the graph in a specific way:
Transformation Parameters:
A: Amplitude
The absolute value of A determines the amplitude (half the distance between maximum and minimum values). If A is negative, the function is reflected across the
xx
-axis.
B: Period
The period of the function is
2π∣B∣\frac{2\pi}{|B|}
. A larger
∣B∣|B|
means a shorter period (faster oscillation).
C: Phase Shift
The function is shifted horizontally by C units. Positive C shifts to the right; negative C shifts to the left.
D: Vertical Shift
The function is shifted vertically by D units. Positive D shifts upward; negative D shifts downward.
Problem: Find the amplitude, period, phase shift, and vertical shift of
f(x)=3sin⁡(2(x−π4))−1f(x) = 3\sin(2(x - \frac{\pi}{4})) - 1
1
Compare
f(x)=3sin⁡(2(x−π4))−1f(x) = 3\sin(2(x - \frac{\pi}{4})) - 1
with
f(x)=Asin⁡(B(x−C))+Df(x) = A\sin(B(x - C)) + D
2
Identify the parameters:
•
A=3A = 3
⟹ Amplitude = |A| = 3
•
B=2B = 2
⟹ Period =
2π∣B∣=2π2=π\frac{2\pi}{|B|} = \frac{2\pi}{2} = \pi
•
C=π4C = \frac{\pi}{4}
⟹ Phase shift = C =
π4\frac{\pi}{4}
units to the right
•
D=−1D = -1
⟹ Vertical shift = 1 unit downward
3
Therefore, compared to
y=sin⁡(x)y = \sin(x)
:
• The function is stretched vertically by a factor of 3
• The function completes a full cycle in
π\pi
units (twice as fast)
• The function is shifted
π4\frac{\pi}{4}
units to the right
• The function is shifted 1 unit downward

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles when given values of trigonometric ratios. Because trigonometric functions are not one-to-one over their entire domains, we restrict their domains to define their inverses.

Inverse Trigonometric Functions

Arcsine (inverse sine)
y=arcsin⁡x  ⟺  sin⁡y=xy = \arcsin x \iff \sin y = x
Domain:
[−1,1][-1, 1]

Range:
[−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]
Arccosine (inverse cosine)
y=arccos⁡x  ⟺  cos⁡y=xy = \arccos x \iff \cos y = x
Domain:
[−1,1][-1, 1]

Range:
[0,π][0, \pi]
Arctangent (inverse tangent)
y=arctan⁡x  ⟺  tan⁡y=xy = \arctan x \iff \tan y = x
Domain: All real numbers
Range:
(−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2})
Problem: Find all solutions to
sin⁡x=12\sin x = \frac{1}{2}
in
[0,2π)[0, 2\pi)
1
First, find the principal solution using the inverse sine:
x=arcsin⁡(12)=π6x = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}
2
Since sine has a period of
2π2\pi
and is positive in the first and second quadrants, we also have:
x=π−π6=5π6x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}
3
To find all solutions in
[0,2π)[0, 2\pi)
, we need to add the period
2π2\pi
to these values:
x=π6+2πk or x=5π6+2πkx = \frac{\pi}{6} + 2\pi k \text{ or } x = \frac{5\pi}{6} + 2\pi k
, where
kk
is an integer
4
For
k=0k = 0
, we get
x=π6x = \frac{\pi}{6}
and
x=5π6x = \frac{5\pi}{6}

For
k=1k = 1
, we get
x=π6+2π=13π6x = \frac{\pi}{6} + 2\pi = \frac{13\pi}{6}
and
x=5π6+2π=17π6x = \frac{5\pi}{6} + 2\pi = \frac{17\pi}{6}
5
Therefore, in the interval
[0,2π)[0, 2\pi)
, the solutions are
x=π6x = \frac{\pi}{6}
and
x=5π6x = \frac{5\pi}{6}

Applications of Trigonometric Functions

Trigonometric functions have numerous applications in various fields, including physics, engineering, and computer science.
Simple Harmonic Motion
The position of an object in simple harmonic motion (like a mass on a spring) can be described by:
x(t)=Acos⁡(ωt+ϕ)x(t) = A\cos(\omega t + \phi)
where A is the amplitude,
ω\omega
is the angular frequency, and
ϕ\phi
is the phase angle.
Wave Functions
Waves (including sound and light) can be modeled using trigonometric functions:
y(x,t)=Asin⁡(kx−ωt)y(x,t) = A\sin(kx - \omega t)
where k is the wave number and
ω\omega
is the angular frequency.
Navigation and Surveying
Trigonometric functions are essential for calculating distances and angles in navigation, surveying, and GPS systems.
Problem: A Ferris wheel has a radius of 20 meters and rotates once every 40 seconds. If a passenger starts at the bottom of the wheel at
t=0t = 0
, find their height above the ground at time
tt
1
Let's set up a coordinate system where the center of the wheel is at (0, 20) and the ground level is at y = 0
2
The angle
θ\theta
changes with time:
θ(t)=2πt40=πt20\theta(t) = \frac{2\pi t}{40} = \frac{\pi t}{20}
radians
3
The position of the passenger is:
x(t)=20sin⁡(πt20)x(t) = 20\sin(\frac{\pi t}{20})
y(t)=20−20cos⁡(πt20)y(t) = 20 - 20\cos(\frac{\pi t}{20})
4
The height above ground is given by y(t):
h(t)=20−20cos⁡(πt20)=20(1−cos⁡(πt20))h(t) = 20 - 20\cos(\frac{\pi t}{20}) = 20(1 - \cos(\frac{\pi t}{20}))
5
This function has:
• Minimum value: 0 meters (at t = 0, 40, 80, ...)
• Maximum value: 40 meters (at t = 20, 60, 100, ...)
• Period: 40 seconds (one full rotation)

Solving Trigonometric Equations

Solving trigonometric equations often requires finding all angles that satisfy a given condition. This typically involves using inverse trigonometric functions and understanding the periodicity of trigonometric functions.
General Approach:
1. Isolate the trigonometric function
Rearrange the equation to isolate a single trigonometric function.
2. Find the principal solution
Use inverse trigonometric functions to find the initial solutions.
3. Find all solutions in the given interval
Use periodicity and symmetry to find all solutions within the specified interval.
4. Check your solutions
Substitute values back into the original equation to verify your solutions.
Problem: Solve
2sin⁡2x−sin⁡x−1=02\sin^2 x - \sin x - 1 = 0
for
0≤x<2π0 \leq x < 2\pi
1
Let
u=sin⁡xu = \sin x
. Then our equation becomes:
2u2−u−1=02u^2 - u - 1 = 0
2
Solve this quadratic equation:
(2u+1)(u−1)=0(2u + 1)(u - 1) = 0
u=−12 or u=1u = -\frac{1}{2} \text{ or } u = 1
3
Now we need to find all
xx
values such that:
sin⁡x=−12 or sin⁡x=1\sin x = -\frac{1}{2} \text{ or } \sin x = 1
4
For
sin⁡x=−12\sin x = -\frac{1}{2}
:
• Principal solution:
x=arcsin⁡(−12)=−π6x = \arcsin(-\frac{1}{2}) = -\frac{\pi}{6}
• Since
sin⁡\sin
is negative in the third and fourth quadrants, we also have
x=−π6+2π=11π6x = -\frac{\pi}{6} + 2\pi = \frac{11\pi}{6}
and
x=π+π6=7π6x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}
5
For
sin⁡x=1\sin x = 1
:
•
x=arcsin⁡(1)=π2x = \arcsin(1) = \frac{\pi}{2}
•
x=π2+2π=5π2x = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}
(outside our interval)
6
Therefore, in the interval
[0,2π)[0, 2\pi)
, the solutions are
x=π2,7π6,11π6x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}