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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
MAA80

IA Preparation (Math)

MAA81

IA Preparation (Physics)

Chapter MAA30

Trigonometric Identities

Geometry and Trigonometry

The Unit Circle

The unit circle is a key visual representation when dealing with solving trigonometric equations. It is defined as a circle with its origin at the center of the
xyxy
plane and has a radius of
11
.
If we take any point on the circle, we can recognize an angle is formed between a line that connects the point to the origin of the
xyxy
plane, and with the horizontal. If we analyze this line using basic trigonometry, we can split the coordinates of the points into
xx
- and
yy
-components. These components are defined as
x=cos⁡θx=\cos{\theta}
and
y=sin⁡θy=\sin{\theta}
.
Using the Pythagorean theorem, we know that the sum of the squares of the two components equals the square of the hypotenuse. Since the radius of the unit circle is
11
, we derive the fundamental identity:

Pythagorean Identity

cos⁡2θ+sin⁡2θ=1\cos^2{\theta}+\sin^2{\theta}=1
From this identity, we can rearrange terms to derive two additional relationships:
sin⁡2θ=1−cos⁡2θ\sin^2{\theta}=1-\cos^2{\theta}
cos⁡2θ=1−sin⁡2θ\cos^2{\theta}=1-\sin^2{\theta}

Derivation of
tan⁡θ\tan{\theta}

The tangent of an angle
θ\theta
is defined as the ratio of the opposite side to the adjacent side in a right triangle. On the unit circle, the opposite side corresponds to the
yy
-coordinate (
sin⁡θ\sin{\theta}
), and the adjacent side corresponds to the
xx
-coordinate (
cos⁡θ\cos{\theta}
). Thus, we derive:
tan⁡θ=sin⁡θcos⁡θ\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}
This relationship is valid for all angles where
cos⁡θ≠0\cos{\theta} \neq 0
. It is particularly useful when simplifying trigonometric expressions.

Fundamental Trigonometric Identities

Reciprocal Identities

Reciprocal identities are fundamental relationships between trigonometric functions:
sin⁡θ=1csc⁡θ\sin{\theta} = \frac{1}{\csc{\theta}}
cos⁡θ=1sec⁡θ\cos{\theta} = \frac{1}{\sec{\theta}}
tan⁡θ=1cot⁡θ\tan{\theta} = \frac{1}{\cot{\theta}}

Quotient Identities

Quotient identities express tangent and cotangent in terms of sine and cosine:
tan⁡θ=sin⁡θcos⁡θ\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}
cot⁡θ=cos⁡θsin⁡θ\cot{\theta} = \frac{\cos{\theta}}{\sin{\theta}}

Working with Trigonometric Expressions

Problem:
sin⁡θcos⁡θ+cot⁡θ\frac{\sin{\theta}}{\cos{\theta}} + \cot{\theta}
1
Using the quotient identities:
sin⁡θcos⁡θ=tan⁡θ\frac{\sin{\theta}}{\cos{\theta}} = \tan{\theta}
and
cot⁡θ=cos⁡θsin⁡θ\cot{\theta} = \frac{\cos{\theta}}{\sin{\theta}}
2
Substituting these into the expression:
tan⁡θ+cos⁡θsin⁡θ\tan{\theta} + \frac{\cos{\theta}}{\sin{\theta}}
3
Combine the terms under a common denominator:
sin⁡2θ+cos⁡2θsin⁡θcos⁡θ\frac{\sin^2{\theta} + \cos^2{\theta}}{\sin{\theta}\cos{\theta}}
4
Using the Pythagorean identity
sin⁡2θ+cos⁡2θ=1\sin^2{\theta} + \cos^2{\theta} = 1
, the expression simplifies to:
1sin⁡θcos⁡θ\frac{1}{\sin{\theta}\cos{\theta}}

Sum and Difference Identities

These identities allow us to compute the sine, cosine, and tangent of the sum or difference of two angles. They are derived geometrically or using the unit circle.

Sum and Difference Identities

sin⁡(a±b)=sin⁡acos⁡b±cos⁡asin⁡b\sin{(a\pm b)} = \sin{a}\cos{b} \pm \cos{a}\sin{b}
cos⁡(a±b)=cos⁡acos⁡b∓sin⁡asin⁡b\cos{(a\pm b)} = \cos{a}\cos{b} \mp \sin{a}\sin{b}
tan⁡(a±b)=tan⁡a±tan⁡b1∓tan⁡atan⁡b\tan{(a\pm b)} = \frac{\tan{a} \pm \tan{b}}{1 \mp \tan{a}\tan{b}}
Derivation of
sin⁡(a+b)\sin{(a+b)}
1
Consider two angles
aa
and
bb
on the unit circle. The coordinates of a point for angle
aa
are
(cos⁡a,sin⁡a)(\cos{a}, \sin{a})
, and for angle
bb
, they are
(cos⁡b,sin⁡b)(\cos{b}, \sin{b})
2
The angle
a+ba+b
corresponds to rotating the point for
aa
by an additional angle
bb
3
Using geometric projections, the y-coordinate of the rotated point gives:
sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin{(a+b)} = \sin{a}\cos{b} + \cos{a}\sin{b}
Derivation of
cos⁡(a+b)\cos{(a+b)}
1
Similarly, the x-coordinate of the rotated point gives:
cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos{(a+b)} = \cos{a}\cos{b} - \sin{a}\sin{b}
Note: The difference identities are derived in the same way, but with the angle
bb
subtracted instead of added.

Double Angle Identities

Double angle identities allow us to express trigonometric functions of
2θ2\theta
in terms of
sin⁡θ\sin{\theta}
and
cos⁡θ\cos{\theta}
. These identities are derived using the sum identities.

Double Angle Formulas

sin⁡2θ=2sin⁡θcos⁡θ\sin{2\theta} = 2\sin{\theta}\cos{\theta}
cos⁡2θ=cos⁡2θ−sin⁡2θ\cos{2\theta} = \cos^2{\theta} - \sin^2{\theta}
tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan{2\theta} = \frac{2\tan{\theta}}{1-\tan^2{\theta}}
Alternative forms for
cos⁡2θ\cos{2\theta}
:
cos⁡2θ=2cos⁡2θ−1\cos{2\theta} = 2\cos^2{\theta} - 1
cos⁡2θ=1−2sin⁡2θ\cos{2\theta} = 1 - 2\sin^2{\theta}
Derivation of
sin⁡2θ\sin{2\theta}
1
Start with the sum identity for sine:
sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin{(a+b)} = \sin{a}\cos{b} + \cos{a}\sin{b}
2
Set
a=b=θa = b = \theta
. Substituting gives:
sin⁡(2θ)=sin⁡θcos⁡θ+cos⁡θsin⁡θ\sin{(2\theta)} = \sin{\theta}\cos{\theta} + \cos{\theta}\sin{\theta}
3
Combine like terms to get:
sin⁡2θ=2sin⁡θcos⁡θ\sin{2\theta} = 2\sin{\theta}\cos{\theta}
Derivation of
cos⁡2θ\cos{2\theta}
1
Start with the sum identity for cosine:
cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos{(a+b)} = \cos{a}\cos{b} - \sin{a}\sin{b}
2
Set
a=b=θa = b = \theta
. Substituting gives:
cos⁡(2θ)=cos⁡θcos⁡θ−sin⁡θsin⁡θ\cos{(2\theta)} = \cos{\theta}\cos{\theta} - \sin{\theta}\sin{\theta}
3
Simplify to get:
cos⁡2θ=cos⁡2θ−sin⁡2θ\cos{2\theta} = \cos^2{\theta} - \sin^2{\theta}
4
Using the Pythagorean identity
cos⁡2θ+sin⁡2θ=1\cos^2{\theta} + \sin^2{\theta} = 1
, we can derive the alternative forms
Derivation of
tan⁡2θ\tan{2\theta}
1
Start with the definition of tangent:
tan⁡θ=sin⁡θcos⁡θ\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}
2
Use the double angle identities for sine and cosine:
sin⁡2θ=2sin⁡θcos⁡θ\sin{2\theta} = 2\sin{\theta}\cos{\theta}
and
cos⁡2θ=cos⁡2θ−sin⁡2θ\cos{2\theta} = \cos^2{\theta} - \sin^2{\theta}
3
Substitute into the tangent formula:
tan⁡2θ=sin⁡2θcos⁡2θ=2sin⁡θcos⁡θcos⁡2θ−sin⁡2θ\tan{2\theta} = \frac{\sin{2\theta}}{\cos{2\theta}} = \frac{2\sin{\theta}\cos{\theta}}{\cos^2{\theta} - \sin^2{\theta}}
4
Divide numerator and denominator by
cos⁡2θ\cos^2{\theta}
to get:
tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan{2\theta} = \frac{2\tan{\theta}}{1-\tan^2{\theta}}

Power Reduction and Multiple Angle Formulas

Power reduction formulas allow us to rewrite powers of trigonometric functions in terms of cosines of multiple angles. This is particularly useful in integration and when simplifying complex expressions.

Power Reduction Formulas

sin⁡2θ=1−cos⁡2θ2\sin^2{\theta} = \frac{1-\cos{2\theta}}{2}
cos⁡2θ=1+cos⁡2θ2\cos^2{\theta} = \frac{1+\cos{2\theta}}{2}
sin⁡3θ=3sin⁡θ−sin⁡3θ4\sin^3{\theta} = \frac{3\sin{\theta} - \sin{3\theta}}{4}
cos⁡3θ=3cos⁡θ+cos⁡3θ4\cos^3{\theta} = \frac{3\cos{\theta} + \cos{3\theta}}{4}

Triple Angle Formulas

sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin{3\theta} = 3\sin{\theta} - 4\sin^3{\theta}
cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}
tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan{3\theta} = \frac{3\tan{\theta} - \tan^3{\theta}}{1 - 3\tan^2{\theta}}
Derivation of
sin⁡3θ\sin{3\theta}
and
cos⁡3θ\cos{3\theta}
1
We can express
3θ3\theta
as
2θ+θ2\theta + \theta
and use the addition formula for sine:
sin⁡3θ=sin⁡(2θ+θ)=sin⁡2θcos⁡θ+cos⁡2θsin⁡θ\sin{3\theta} = \sin{(2\theta + \theta)} = \sin{2\theta}\cos{\theta} + \cos{2\theta}\sin{\theta}
2
Substitute
sin⁡2θ=2sin⁡θcos⁡θ\sin{2\theta} = 2\sin{\theta}\cos{\theta}
and
cos⁡2θ=cos⁡2θ−sin⁡2θ\cos{2\theta} = \cos^2{\theta} - \sin^2{\theta}
:
sin⁡3θ=2sin⁡θcos⁡θ⋅cos⁡θ+(cos⁡2θ−sin⁡2θ)⋅sin⁡θ\sin{3\theta} = 2\sin{\theta}\cos{\theta} \cdot \cos{\theta} + (\cos^2{\theta} - \sin^2{\theta}) \cdot \sin{\theta}
3
Simplify:
sin⁡3θ=2sin⁡θcos⁡2θ+cos⁡2θsin⁡θ−sin⁡3θ\sin{3\theta} = 2\sin{\theta}\cos^2{\theta} + \cos^2{\theta}\sin{\theta} - \sin^3{\theta}
4
Factor out
sin⁡θ\sin{\theta}
:
sin⁡3θ=sin⁡θ(2cos⁡2θ+cos⁡2θ−sin⁡2θ)\sin{3\theta} = \sin{\theta}(2\cos^2{\theta} + \cos^2{\theta} - \sin^2{\theta})
5
Using
cos⁡2θ+sin⁡2θ=1\cos^2{\theta} + \sin^2{\theta} = 1
:
sin⁡3θ=sin⁡θ(3cos⁡2θ−sin⁡2θ)\sin{3\theta} = \sin{\theta}(3\cos^2{\theta} - \sin^2{\theta})
6
Using
sin⁡2θ=1−cos⁡2θ\sin^2{\theta} = 1 - \cos^2{\theta}
:
sin⁡3θ=sin⁡θ(3cos⁡2θ−(1−cos⁡2θ))\sin{3\theta} = \sin{\theta}(3\cos^2{\theta} - (1-\cos^2{\theta}))
7
Simplify:
sin⁡3θ=sin⁡θ(4cos⁡2θ−1)=3sin⁡θ−4sin⁡3θ\sin{3\theta} = \sin{\theta}(4\cos^2{\theta} - 1) = 3\sin{\theta} - 4\sin^3{\theta}
The formula for
cos⁡3θ\cos{3\theta}
can be derived similarly using cosine addition formulas and simplification.
Problem: Simplify
sin⁡4θ\sin^4{\theta}
using power reduction formulas
1
We know that
sin⁡2θ=1−cos⁡2θ2\sin^2{\theta} = \frac{1-\cos{2\theta}}{2}
2
Therefore,
sin⁡4θ=(sin⁡2θ)2=(1−cos⁡2θ2)2\sin^4{\theta} = (\sin^2{\theta})^2 = \left(\frac{1-\cos{2\theta}}{2}\right)^2
3
sin⁡4θ=1−2cos⁡2θ+cos⁡22θ4\sin^4{\theta} = \frac{1 - 2\cos{2\theta} + \cos^2{2\theta}}{4}
4
Now, using
cos⁡22θ=1+cos⁡4θ2\cos^2{2\theta} = \frac{1+\cos{4\theta}}{2}
:
sin⁡4θ=1−2cos⁡2θ+1+cos⁡4θ24\sin^4{\theta} = \frac{1 - 2\cos{2\theta} + \frac{1+\cos{4\theta}}{2}}{4}
5
sin⁡4θ=1−2cos⁡2θ+12+cos⁡4θ24\sin^4{\theta} = \frac{1 - 2\cos{2\theta} + \frac{1}{2} + \frac{\cos{4\theta}}{2}}{4}
6
Simplify:
sin⁡4θ=3−4cos⁡2θ+cos⁡4θ8\sin^4{\theta} = \frac{3 - 4\cos{2\theta} + \cos{4\theta}}{8}

Inverse Trigonometric Identities

Inverse trigonometric functions (also called arcfunctions) allow us to find angles from trigonometric function values. These identities help to simplify expressions involving inverse trigonometric functions.

Basic Inverse Trigonometric Properties

sin⁡(arcsin⁡x)=xfor−1≤x≤1\sin(\arcsin x) = x \quad \text{for} \quad -1 \leq x \leq 1
cos⁡(arccos⁡x)=xfor−1≤x≤1\cos(\arccos x) = x \quad \text{for} \quad -1 \leq x \leq 1
tan⁡(arctan⁡x)=xfor all x in the domain\tan(\arctan x) = x \quad \text{for all } x \text{ in the domain}
Relations between inverse trigonometric functions:
arcsin⁡x+arccos⁡x=π2for−1≤x≤1\arcsin x + \arccos x = \frac{\pi}{2} \quad \text{for} \quad -1 \leq x \leq 1
arctan⁡x+arctan⁡1x={π2if x>0−π2if x<0\arctan x + \arctan \frac{1}{x} = \begin{cases} \frac{\pi}{2} & \text{if } x > 0 \\ -\frac{\pi}{2} & \text{if } x < 0 \end{cases}
Problem: Evaluate
arctan⁡1+arctan⁡2\arctan 1 + \arctan 2
exactly
1
Using the formula
arctan⁡x+arctan⁡y=arctan⁡(x+y1−xy)\arctan x + \arctan y = \arctan\left(\frac{x + y}{1 - xy}\right)
2
Substitute
x=1x = 1
and
y=2y = 2
:
arctan⁡1+arctan⁡2=arctan⁡(1+21−1⋅2)=arctan⁡(31−2)=arctan⁡(3−1)=arctan⁡(−3)\arctan 1 + \arctan 2 = \arctan\left(\frac{1 + 2}{1 - 1 \cdot 2}\right) = \arctan\left(\frac{3}{1 - 2}\right) = \arctan\left(\frac{3}{-1}\right) = \arctan(-3)
3
Since
arctan⁡(−3)=−arctan⁡(3)\arctan(-3) = -\arctan(3)
, and
arctan⁡(3)\arctan(3)
is in the first quadrant, we must adjust:
arctan⁡1+arctan⁡2=arctan⁡(−3)=−arctan⁡(3)+π\arctan 1 + \arctan 2 = \arctan(-3) = -\arctan(3) + \pi
4
We know
arctan⁡(1)=π4\arctan(1) = \frac{\pi}{4}
and
arctan⁡(2)≈1.107\arctan(2) \approx 1.107
, and their sum is approximately
1.89 radians≈3π51.89 \text{ radians} \approx \frac{3\pi}{5}
5
The exact value is
arctan⁡1+arctan⁡2=π−arctan⁡(3)\arctan 1 + \arctan 2 = \pi - \arctan(3)

Strategies for Verifying Trigonometric Identities

Proving trigonometric identities is a common task in mathematics courses. Here are some strategies to approach these problems systematically.
Key Strategies:
1. Work with one side at a time
Usually, it's best to start with the more complex side and transform it into the other side.
2. Convert to sines and cosines
If the expression involves other trigonometric functions (tan, sec, etc.), convert everything to sines and cosines.
3. Look for Pythagorean identities
The fundamental identity
sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
and its variants are often useful.
4. Find common denominators
When dealing with fractions, combining terms under a common denominator can help.
5. Factor expressions
Look for opportunities to factor expressions to simplify them.