Chapter 4: Addition Theorem
Overview
The addition theorem is a set of formulas that express the trigonometric functions of the sum or difference of two angles in terms of products of the trigonometric functions of the individual angles. It is one of the most fundamental and important of all trigonometric identities, and many other formulas are derived from it.
The addition theorem formulas
Addition theorem
$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$ $$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$Below we prove these formulas in four different ways.
Proof 1: Distance between two points on the unit circle
This proof can be understood within the scope of high-school mathematics and is the standard method adopted in many textbooks.
Proof of the cos addition theorem (subtraction form)
We take four points on the unit circle and use the fact that the distances between two pairs of points are equal.
Fig 1: Four points A, B, P, Q on the unit circle (chords AB and PQ have equal length)
Step 1: Setting up the four points
Take the following four points on the unit circle:
- $A = (1, 0)$ (angle $0$)
- $B = (\cos(\alpha - \beta), \sin(\alpha - \beta))$ (angle $\alpha - \beta$)
- $P = (\cos\alpha, \sin\alpha)$ (angle $\alpha$)
- $Q = (\cos\beta, \sin\beta)$ (angle $\beta$)
Step 2: Central angles and arcs
Fig 2: Arc AB and arc QP subtend the same central angle (= α−β)
The central angle from $A$ to $B$ is $\alpha - \beta$.
The central angle from $Q$ to $P$ is also $\alpha - \beta$ (since $\alpha - \beta = \alpha - \beta$).
Therefore the arcs $AB$ and $QP$ have equal length, and the chords $AB$ and $PQ$ also have equal length.
$$|AB| = |PQ|$$Step 3: Computing |AB|²
Fig 3: |AB|² = (horizontal distance)² + (vertical distance)²
By the distance formula between two points:
$$|AB|^2 = (\cos(\alpha - \beta) - 1)^2 + (\sin(\alpha - \beta) - 0)^2$$ $$= \cos^2(\alpha - \beta) - 2\cos(\alpha - \beta) + 1 + \sin^2(\alpha - \beta)$$Since $\sin^2 + \cos^2 = 1$:
$$= 2 - 2\cos(\alpha - \beta)$$Step 4: Computing |PQ|²
Fig 4: |PQ|² = (cos α − cos β)² + (sin α − sin β)²
Step 5: Deriving from |AB|² = |PQ|²
Fig 5: Equal chord lengths → equate the distance expressions
cos subtraction formula
$$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta \quad \blacksquare$$Deriving the cos addition formula and the sin formulas
Step 6: Deriving cos(α+β)
Fig 6: Replacing β → −β (symmetry about the x-axis)
Replace $\beta$ with $-\beta$. Since $\cos(-\beta) = \cos\beta$ and $\sin(-\beta) = -\sin\beta$:
$$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$Step 7: Deriving sin(α+β)
Fig 7: Complementary-angle relation: sin θ = cos(π/2 − θ)
Use the complementary-angle relation $\sin\theta = \cos\left(\dfrac{\pi}{2} - \theta\right)$:
$$\sin(\alpha + \beta) = \cos\left(\dfrac{\pi}{2} - (\alpha + \beta)\right) = \cos\left(\left(\dfrac{\pi}{2} - \alpha\right) - \beta\right)$$Apply the cos subtraction formula:
$$= \cos\left(\dfrac{\pi}{2} - \alpha\right)\cos\beta + \sin\left(\dfrac{\pi}{2} - \alpha\right)\sin\beta$$ $$= \sin\alpha\cos\beta + \cos\alpha\sin\beta$$Addition theorem (from Proof 1)
$$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$ $$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \quad \blacksquare$$Proof 2: Geometric proof using right triangles
We combine two right triangles to prove the addition theorem geometrically. This method is intuitive and easy to follow when $0 < \alpha, \beta, \alpha + \beta < \dfrac{\pi}{2}$.
Fig 8: Construction decomposing the segment $|OP| = 1$ along the direction of angle $\alpha$ (e.g. $\alpha = 35°,\ \beta = 25°$)
Step 1: Construction
From the origin $O$, draw a ray making angle $\alpha$ with the $x$-axis and another ray making angle $\alpha + \beta$.
On the ray at angle $\alpha + \beta$, take the point $P$ with $|OP| = 1$.
Let $E$ be the foot of the perpendicular dropped from $P$ to the $x$-axis, and let $Q$ be the foot of the perpendicular dropped from $P$ to the ray at angle $\alpha$.
Step 2: Sides of the two right triangles
In the right triangle $OPE$, since $\angle POE = \alpha + \beta$ and $|OP| = 1$,
$$|OE| = \cos(\alpha+\beta), \quad |PE| = \sin(\alpha+\beta)$$In the right triangle $OPQ$, since $\angle POQ = \beta$ and $|OP| = 1$,
$$|OQ| = \cos\beta, \quad |QP| = \sin\beta$$Step 3: Decompose into components at the foot Q
Drop a perpendicular from $Q$ to the $x$-axis and let its foot be $B$. Also let $R$ be the intersection of the horizontal line through $Q$ with $PE$. Then the quadrilateral $QBER$ is a rectangle.
Fig 9: Component decomposition of $\sin(\alpha+\beta)$ (splitting the height $|PE|$ into an upper and a lower segment)
In the right triangle $OQB$, since $\angle QOB = \alpha$ and $|OQ| = \cos\beta$,
$$|QB| = \cos\beta\,\sin\alpha = \sin\alpha\cos\beta, \quad |OB| = \cos\beta\,\cos\alpha = \cos\alpha\cos\beta$$Since $PQ \perp OQ$ and $PE \perp OE$, we have $\angle QPR = \alpha$. In the right triangle $PRQ$, since $|PQ| = \sin\beta$,
$$|PR| = \sin\beta\,\cos\alpha = \cos\alpha\sin\beta, \quad |RQ| = \sin\beta\,\sin\alpha = \sin\alpha\sin\beta$$Step 4: Add the heights and the base
From the rectangle $QBER$ we have $|RE| = |QB| = \sin\alpha\cos\beta$. Therefore the height $|PE|$ of $P$ is
$$\sin(\alpha+\beta) = |PE| = |PR| + |RE| = \cos\alpha\sin\beta + \sin\alpha\cos\beta$$Also $|EB| = |RQ| = \sin\alpha\sin\beta$, so the horizontal position $|OE|$ of $P$ is
$$\cos(\alpha+\beta) = |OE| = |OB| - |EB| = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$Addition theorem (from Proof 2)
$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$ $$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \quad \blacksquare$$Note: This geometric proof is valid when $0 < \alpha, \beta, \alpha + \beta < \dfrac{\pi}{2}$. For general angles, use the method of Proof 1 or the rotation-matrix proof given below.
Proof 3: Proof using the rotation matrix
Note: This proof assumes knowledge of the linear algebra (rotation matrices) taught at university. If you want to understand it within the scope of high-school mathematics, please refer to Proof 1 or Proof 2.
This proof uses linear algebra, but it is the most elegant and general of all.
Consider two points on the unit circle.
Fig 10: Point P (angle α) and point Q (angle α+β) on the unit circle
Step 1: Set the coordinates of the two points
On the unit circle, let $P$ be the point at angle $\alpha$ measured counterclockwise from the $x$-axis, and let $Q$ be the point at angle $\alpha + \beta$.
$$P = (\cos\alpha, \sin\alpha)$$ $$Q = (\cos(\alpha+\beta), \sin(\alpha+\beta))$$Step 2: Think of it as a rotation
Rotating the point $P$ about the origin by angle $\beta$ yields the point $Q$.
Using the rotation matrix, the point $(x, y)$ rotated by angle $\beta$ becomes $(x', y')$ where:
$$ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$Step 3: Rotate P by β
Apply the rotation matrix to the point $P = (\cos\alpha, \sin\alpha)$:
$$ \begin{pmatrix} \cos(\alpha+\beta) \\ \sin(\alpha+\beta) \end{pmatrix} = \begin{pmatrix} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{pmatrix} \begin{pmatrix} \cos\alpha \\ \sin\alpha \end{pmatrix} $$Step 4: Compute the matrix product
Computing the first component (the $x$-coordinate):
$$\cos(\alpha+\beta) = \cos\beta \cdot \cos\alpha + (-\sin\beta) \cdot \sin\alpha$$ $$= \cos\alpha\cos\beta - \sin\alpha\sin\beta$$Computing the second component (the $y$-coordinate):
$$\sin(\alpha+\beta) = \sin\beta \cdot \cos\alpha + \cos\beta \cdot \sin\alpha$$ $$= \sin\alpha\cos\beta + \cos\alpha\sin\beta$$Addition theorem (from Proof 3)
$$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$ $$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \quad \blacksquare$$Proof 4: Proof using complex numbers
Note: This proof assumes Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$. Euler's formula is proved rigorously in university mathematics (complex analysis). If you want to understand it within the scope of high-school mathematics, please refer to Proof 1 or Proof 2.
This proof, using Euler's formula, is the most concise and elegant method. The addition theorem follows naturally from the properties of the complex exponential function.
Euler's formula
$$e^{i\theta} = \cos\theta + i\sin\theta$$Here $i$ is the imaginary unit ($i^2 = -1$).
Fig 11: $e^{i\theta}$ on the complex plane (real part $\cos\theta$, imaginary part $\sin\theta$)
Step 1: Apply the law of exponents
The complex exponential function satisfies the usual law of exponents:
$$e^{i(\alpha + \beta)} = e^{i\alpha} \cdot e^{i\beta}$$Step 2: Expand the left side with Euler's formula
$$e^{i(\alpha + \beta)} = \cos(\alpha + \beta) + i\sin(\alpha + \beta)$$Step 3: Expand the right side with Euler's formula
$$e^{i\alpha} \cdot e^{i\beta} = (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta)$$Compute the product of the complex numbers:
$$= \cos\alpha\cos\beta + i\cos\alpha\sin\beta + i\sin\alpha\cos\beta + i^2\sin\alpha\sin\beta$$Since $i^2 = -1$:
$$= (\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta)$$Step 4: Compare the real and imaginary parts
Set the left and right sides equal:
$$\cos(\alpha + \beta) + i\sin(\alpha + \beta) = (\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta)$$When two complex numbers are equal, their real parts and their imaginary parts are equal, so:
- Real part: $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$
- Imaginary part: $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$
Addition theorem (from Proof 4)
$$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$ $$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \quad \blacksquare$$The beauty of this proof: Using complex numbers, the addition theorems for both cos and sin are derived simultaneously from the single identity $e^{i(\alpha+\beta)} = e^{i\alpha} e^{i\beta}$. This reflects the deep connection between the trigonometric functions and the complex exponential function.
Derivation of the subtraction formulas
Replacing $\beta$ with $-\beta$ in the addition theorem gives the subtraction formulas.
Negative-angle formulas
By the even/odd symmetry of the trigonometric functions:
$$\cos(-\beta) = \cos\beta, \quad \sin(-\beta) = -\sin\beta$$Substituting these into the addition theorem:
Subtraction formulas
$$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$$ $$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$$Addition theorem for tan
The addition theorem for $\tan$ can be derived from the addition theorems for $\sin$ and $\cos$.
Derivation
$$\tan(\alpha + \beta) = \dfrac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}$$ $$= \dfrac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta}$$Dividing numerator and denominator by $\cos\alpha\cos\beta$:
$$= \dfrac{\dfrac{\sin\alpha}{\cos\alpha} + \dfrac{\sin\beta}{\cos\beta}}{1 - \dfrac{\sin\alpha}{\cos\alpha} \cdot \dfrac{\sin\beta}{\cos\beta}}$$Addition theorem for tan
$$\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$$Similarly, the subtraction formula is:
$$\tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}$$Worked examples
Example 1: Evaluate sin 75°
Problem
Use the addition theorem to evaluate $\sin 75°$.
Step 1: Decompose the angle
Write $75° = 45° + 30°$.
Step 2: Apply the sin addition theorem
$$\sin 75° = \sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30°$$Step 3: Substitute the known values
$$= \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4}$$Result
$$\sin 75° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$$Example 2: Evaluate cos 15°
Problem
Use the addition theorem to evaluate $\cos 15°$.
Step 1: Decompose the angle
Write $15° = 45° - 30°$.
Step 2: Apply the cos subtraction formula
$$\cos 15° = \cos(45° - 30°) = \cos 45°\cos 30° + \sin 45°\sin 30°$$Step 3: Substitute the known values
$$= \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4}$$Result
$$\cos 15° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$$(This also confirms that $\sin 75° = \cos 15°$.)
Example 3: Evaluate tan 75°
Problem
Use the addition theorem to evaluate $\tan 75°$.
Step 1: Decompose the angle
Write $75° = 45° + 30°$.
Step 2: Apply the tan addition theorem
$$\tan 75° = \tan(45° + 30°) = \dfrac{\tan 45° + \tan 30°}{1 - \tan 45°\tan 30°}$$Step 3: Substitute the known values
$$= \dfrac{1 + \dfrac{1}{\sqrt{3}}}{1 - 1 \cdot \dfrac{1}{\sqrt{3}}} = \dfrac{\dfrac{\sqrt{3} + 1}{\sqrt{3}}}{\dfrac{\sqrt{3} - 1}{\sqrt{3}}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}$$Step 4: Rationalize
$$= \dfrac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \dfrac{3 + 2\sqrt{3} + 1}{3 - 1} = \dfrac{4 + 2\sqrt{3}}{2}$$Result
$$\tan 75° = 2 + \sqrt{3}$$Exercises
Exercise 1
Use the addition theorem to evaluate $\sin 105°$.
Hint
Write $105° = 60° + 45°$ and apply the $\sin$ addition theorem.
Solution
Exercise 2
Use the addition theorem to evaluate $\cos 75°$.
Hint
Write $75° = 45° + 30°$ and use the $\cos$ addition theorem. Note that $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$.
Solution
Exercise 3
Use the addition theorem to evaluate $\tan 15°$.
Hint
Write $15° = 45° - 30°$ and use the $\tan$ subtraction formula. Finally, rationalize the denominator.
Solution
Rationalizing:
$$= \dfrac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \dfrac{3 - 2\sqrt{3} + 1}{2} = 2 - \sqrt{3}$$Exercise 4
If $\alpha$ is a first-quadrant angle with $\sin\alpha = \dfrac{3}{5}$, find the value of $\sin\!\left(\alpha + \dfrac{\pi}{3}\right)$.
Hint
Find $\cos\alpha$ from $\sin^2\alpha + \cos^2\alpha = 1$ (positive, since $\alpha$ is in the first quadrant), then apply the addition theorem.
Solution
$\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$ (positive, since $\alpha$ is in the first quadrant).
$$\sin\!\left(\alpha + \dfrac{\pi}{3}\right) = \sin\alpha\cos\dfrac{\pi}{3} + \cos\alpha\sin\dfrac{\pi}{3}$$ $$= \dfrac{3}{5} \cdot \dfrac{1}{2} + \dfrac{4}{5} \cdot \dfrac{\sqrt{3}}{2} = \dfrac{3 + 4\sqrt{3}}{10}$$Summary
List of addition theorems
Addition formulas:
$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$ $$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$ $$\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$$Subtraction formulas:
$$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$$ $$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$$ $$\tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}$$References
Frequently Asked Questions
Q1: What is the addition theorem?
A: The addition theorem is a set of formulas that express the trigonometric functions of the sum or difference of two angles in terms of products of the trigonometric functions of the individual angles. For example, sin(α+β) = sin α cos β + cos α sin β.
Q2: What is the formula for sin(α+β)?
A: sin(α+β) = sin α cos β + cos α sin β. This is one of the basic addition-theorem formulas: it expresses the sine of a sum of two angles as a sum of products of the sines and cosines of each angle.
Q3: What methods are there for proving the addition theorem?
A: The main methods are: (1) using the distance between two points on the unit circle, (2) a geometric proof with right triangles, (3) a proof using the rotation matrix, and (4) a proof using complex numbers (Euler's formula).
Q4: Where is the addition theorem used?
A: The addition theorem is used to evaluate the trigonometric functions of special angles (for example, sin 75° = sin(45°+30°)), to derive the double-angle and half-angle formulas, to combine trigonometric functions, and in calculus computations — in many areas of mathematics.