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Trigonometry · Trigonometry

Difference Formulas for Sine, Cosine, and Tangent

You are standing in Times Square, looking up at a billboard on top of a building. Your eyes see the bottom of the billboard at an angle of elevation $\beta$ and the top at a larger angle $\alpha$. The angle your eyes sweep across the billboard is $\alpha - \beta$. Can you find it from the two elevation angles?

There is a second puzzle. The unit circle gives you $\sin 30^\circ$, $\sin 45^\circ$ and $\sin 60^\circ$ exactly, but what is $\sin 15^\circ$ without a calculator? The difference formulas answer both questions.

The Difference Formulas

You already know the sum formulas:

$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta \qquad \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$

To get the difference formulas, write $\alpha - \beta$ as $\alpha + (-\beta)$. Then use the fact that sine is odd, $\sin(-\beta)=-\sin\beta$, and cosine is even, $\cos(-\beta)=\cos\beta$.

Unit circle with two points at angles alpha and beta and the angle alpha minus beta between their radii. (cos α, sin α) (cos β, sin β) α − β cos(α − β) = cos α cos β + sin α sin β

The formulas also work backward. If you see a pattern like $\sin 50^\circ\cos 20^\circ-\cos 50^\circ\sin 20^\circ$, it is the right-hand side of $\sin(\alpha-\beta)$ with $\alpha=50^\circ$ and $\beta=20^\circ$. So it equals $\sin 30^\circ=\frac12$.

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Simplify the following trigonometric expression as much as possible: $\cos(x) \cos(y) + \sin(x) \sin(y)$. Then, use the simplified result to evaluate the expression…

Simplify the following trigonometric expression as much as possible: $\cos(x) \cos(y) + \sin(x) \sin(y)$. Then, use the simplified result to evaluate the expression exactly when $x = 75^\circ$ and $y = 15^\circ$.

Solve
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Evaluate the following trigonometric expression exactly, without using a calculator, identifying which expansion it corresponds to:…

Evaluate the following trigonometric expression exactly, without using a calculator, identifying which expansion it corresponds to:

$\sin(75^\circ) \cos(15^\circ) - \cos(75^\circ) \sin(15^\circ)$

Solve

Cofunction and Supplementary Identities

Plug in $90^\circ$ or $180^\circ$ as the first angle. Then $\sin$ and $\cos$ of that angle are just $0$, $1$ or $-1$, and the formulas collapse into very useful identities.

Right triangle with acute angles x and 90 degrees minus x, showing that the sine of one equals the cosine of the other. x 90° − x b a c sin(90° − x) = cos x cos(90° − x) = sin x
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Simplify the expression $\sin(90^\circ - x)$ by applying the formula for the sine of the difference of two angles and verify that it is equivalent to one of the basic…

Simplify the expression $\sin(90^\circ - x)$ by applying the formula for the sine of the difference of two angles and verify that it is equivalent to one of the basic trigonometric ratios of the angle $x$.

Solve
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Prove step by step that $\cos(180^\circ - \alpha) = -\cos \alpha$ using the cosine formula for the difference of angles.

Prove step by step that $\cos(180^\circ - \alpha) = -\cos \alpha$ using the cosine formula for the difference of angles.

Solve
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Knowing that two angles $\alpha$ and $\beta$ are complementary ($\alpha + \beta = 90^\circ$), prove step by step that the identity $\sin(\alpha - \beta) = \cos(2\beta)$…

Knowing that two angles $\alpha$ and $\beta$ are complementary ($\alpha + \beta = 90^\circ$), prove step by step that the identity $\sin(\alpha - \beta) = \cos(2\beta)$ holds.

Solve

Exact Values of Non-Special Angles

Angles like $15^\circ$ and $75^\circ$ are not on the standard unit-circle list. But each can be written as a sum or difference of angles that are.

Angle $\sin$ $\cos$ $\tan$
$30^\circ$ $\frac12$ $\frac{\sqrt3}{2}$ $\frac{\sqrt3}{3}$
$45^\circ$ $\frac{\sqrt2}{2}$ $\frac{\sqrt2}{2}$ $1$
$60^\circ$ $\frac{\sqrt3}{2}$ $\frac12$ $\sqrt3$
$90^\circ$ $1$ $0$ undefined
$120^\circ$ $\frac{\sqrt3}{2}$ $-\frac12$ $-\sqrt3$
$135^\circ$ $\frac{\sqrt2}{2}$ $-\frac{\sqrt2}{2}$ $-1$
$180^\circ$ $0$ $-1$ $0$
Diagram showing that the angle 15 degrees equals the special angle 45 degrees minus the special angle 30 degrees. 45° − 30° 15° not a special angle
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Calculate the exact value of $\cos(15^\circ)$ without using a calculator, applying the formula for the cosine of the difference of two angles and knowing that…

Calculate the exact value of $\cos(15^\circ)$ without using a calculator, applying the formula for the cosine of the difference of two angles and knowing that $15^\circ = 45^\circ - 30^\circ$.

Solve
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Find the exact value of $\sin(15^\circ)$ by expressing the angle as the difference $60^\circ - 45^\circ$ and applying the corresponding identity.

Find the exact value of $\sin(15^\circ)$ by expressing the angle as the difference $60^\circ - 45^\circ$ and applying the corresponding identity.

Solve

The Tangent Difference Formula

Because $\tan=\frac{\sin}{\cos}$, divide the sine difference by the cosine difference. Then divide the numerator and denominator by $\cos\alpha\cos\beta$:

$$\tan(\alpha-\beta)=\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$$

This is useful when you know tangents (slopes) instead of sines and cosines. The picture shows two lines through the origin with slopes $4$ and $\frac12$. The angle between them is $\alpha-\beta$.

Two rays from the origin with slopes 4 and one half and the angle alpha minus beta between them. tan α = 4 tan β = 1/2 α − β
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Determine the exact value of $\tan(15^\circ)$ from the tangents of the notable angles $45^\circ$ and $30^\circ$. Rationalize the denominator if necessary.

Determine the exact value of $\tan(15^\circ)$ from the tangents of the notable angles $45^\circ$ and $30^\circ$. Rationalize the denominator if necessary.

Solve
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Calculate the exact value of $\tan(105^\circ)$ using the formula for the difference of angles, considering that $105^\circ = 135^\circ - 30^\circ$.

Calculate the exact value of $\tan(105^\circ)$ using the formula for the difference of angles, considering that $105^\circ = 135^\circ - 30^\circ$.

Solve
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In a maritime navigation simulation, a ship must adjust its current course. The necessary correction angle is given by the expression $\alpha - \beta$. Knowing that…

In a maritime navigation simulation, a ship must adjust its current course. The necessary correction angle is given by the expression $\alpha - \beta$. Knowing that $\alpha$ and $\beta$ are acute angles with $\tan \alpha = 3$ and $\tan \beta = \frac{1}{2}$, calculate the exact value of $\tan(\alpha - \beta)$ and deduce the size of the correction angle.

Solve

Starting from Known Ratios

Many problems give you one ratio for each angle (like $\sin\alpha$ and $\cos\beta$) and the quadrant. The formulas need both $\sin$ and $\cos$ of each angle, so you first find the missing ones.

  1. Use $\sin^2\theta+\cos^2\theta=1$ to find the missing ratio.

  2. Choose the sign of the square root from the quadrant.

  3. Substitute into the difference formula.

Coordinate plane divided into four quadrants showing the signs of sine and cosine in each. I sin +, cos + II sin +, cos − III sin −, cos − IV sin −, cos +
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Knowing that $\sin \alpha = \frac{4}{5}$ and $\cos \beta = \frac{12}{13}$, and that both $\alpha$ and $\beta$ are angles in the first quadrant, calculate the exact value…

Knowing that $\sin \alpha = \frac{4}{5}$ and $\cos \beta = \frac{12}{13}$, and that both $\alpha$ and $\beta$ are angles in the first quadrant, calculate the exact value of $\sin(\alpha - \beta)$.

Solve
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A ray of light passes through two different media with angles of refraction $\alpha$ and $\beta$. If, due to the properties of the materials, we know that…

A ray of light passes through two different media with angles of refraction $\alpha$ and $\beta$. If, due to the properties of the materials, we know that $\cos \alpha = \frac{3}{5}$ (where $\alpha$ is an angle in the fourth quadrant) and $\sin \beta = \frac{5}{13}$ (where $\beta$ is an angle in the second quadrant), determine the exact value of $\cos(\alpha - \beta)$.

Solve
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Two coplanar forces are applied to an object, forming angles $\alpha$ and $\beta$ with the horizontal, respectively. If $\sin \alpha = \frac{8}{17}$ and…

Two coplanar forces are applied to an object, forming angles $\alpha$ and $\beta$ with the horizontal, respectively. If $\sin \alpha = \frac{8}{17}$ and $\cos \beta = \frac{4}{5}$ (both angles being in the first quadrant), calculate the sine of the angle formed between the two forces, that is, $\sin(\alpha - \beta)$.

F₂ F₁ β α α - β
Solve

Proving Identities with the Difference Formulas

To prove an identity, work on one side only (usually the more complicated one) until it becomes the other side. Expand the difference formula, then split fractions or rewrite in terms of $\sin$ and $\cos$.

Graph the two sides and move $b$: the curves land on top of each other for every $b$. (Here angles are in radians.)

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Simplify the following trigonometric algebraic expression and express it exclusively in terms of the tangents of the angles $a$ and $b$:…

Simplify the following trigonometric algebraic expression and express it exclusively in terms of the tangents of the angles $a$ and $b$:

$\frac{\sin(a - b)}{\cos a \cos b}$

Solve
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Prove analytically that $\tan(45^\circ - x) = \frac{1 - \tan x}{1 + \tan x}$. Use this result to find the exact value of $\tan(15^\circ)$ knowing that…

Prove analytically that $\tan(45^\circ - x) = \frac{1 - \tan x}{1 + \tan x}$. Use this result to find the exact value of $\tan(15^\circ)$ knowing that $\tan(30^\circ) = \frac{\sqrt{3}}{3}$.

Solve

Using Sum and Difference Together

The sum and difference formulas differ only in a few signs. So if you add or subtract them, half of the terms cancel.

Adding the sine sum formula and the sine difference formula cancels the cos a sin b terms and leaves 2 sin a cos b. sin(a+b) = sin a cos b + cos a sin b sin(a−b) = sin a cos b − cos a sin b sin(a+b) + sin(a−b) = 2 sin a cos b +

The second tool is a product of a sum and a difference, which is a difference of squares. For $\sin(a-b)\sin(a+b)$, write $u=\sin a\cos b$ and $v=\cos a\sin b$. The product is $(u-v)(u+v)=u^2-v^2$.

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Prove the following trigonometric identity by appropriately combining the angle addition and subtraction formulas for sine and cosine:…

Prove the following trigonometric identity by appropriately combining the angle addition and subtraction formulas for sine and cosine:

$\frac{\cos(a - b) - \cos(a + b)}{\sin(a + b) + \sin(a - b)} = \tan b$

Solve
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Prove analytically that $\sin(\alpha - \beta) \cdot \sin(\alpha + \beta) = \sin^2 \alpha - \sin^2 \beta$. Then, use this algebraic result to mentally calculate the exact…

Prove analytically that $\sin(\alpha - \beta) \cdot \sin(\alpha + \beta) = \sin^2 \alpha - \sin^2 \beta$. Then, use this algebraic result to mentally calculate the exact value of the expression $\sin(75^\circ) \cdot \sin(15^\circ)$.

Solve

Solving Trigonometric Equations

When an equation looks like $\sin(x-30^\circ)=k$, do not expand it. Treat $x-30^\circ$ as one angle $u$, solve for $u$, then add $30^\circ$ back. The one catch is that the interval for $x$ shifts, so the interval for $u$ shifts too.

Unit circle with a horizontal line at height square root of 3 over 2 crossing the circle at angles 60 and 120 degrees. u = 60° u = 120° y = √3/2
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Solve the following trigonometric equation, finding all possible solutions for the interval $0^\circ \leq x < 360^\circ$: $\sin(x - 30^\circ) = \frac{\sqrt{3}}{2}$

Solve the following trigonometric equation, finding all possible solutions for the interval $0^\circ \leq x < 360^\circ$:

$\sin(x - 30^\circ) = \frac{\sqrt{3}}{2}$

Solve
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Solve the following trigonometric equation, determining all possible solutions in the interval $[0, 2\pi)$ radians: $\cos(x - 45^\circ) + \cos(x + 45^\circ) = 1$

Solve the following trigonometric equation, determining all possible solutions in the interval $[0, 2\pi)$ radians:

$\cos(x - 45^\circ) + \cos(x + 45^\circ) = 1$

Solve

Systems and Relations Between Angles

Some problems link two angles by a system. The key idea is to add and subtract the equations so that sum and difference formulas appear.

A system with products sin x cos y equal to P and cos x sin y equal to Q becomes sin of x plus y when added and sin of x minus y when subtracted. sin x cos y = P cos x sin y = Q add: P + Q = sin(x + y) subtract: P − Q = sin(x − y)

The second type uses a relation such as $\tan\alpha=3\tan\beta$. Let $t=\tan\beta$ and substitute into the tangent formula.

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Solve the following system of trigonometric equations knowing that both $x$ and $y$ are angles in the first quadrant: $\sin x \cos y = \frac{3}{4}$…

Solve the following system of trigonometric equations knowing that both $x$ and $y$ are angles in the first quadrant:

$\sin x \cos y = \frac{3}{4}$

$\cos x \sin y = \frac{1}{4}$

Hint: First calculate the value of $\sin(x - y)$ and $\sin(x + y)$ by adding and subtracting both equations.

Solve
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In the design of a robotic arm, the angle of rotation of the final joint depends on the subtraction of two internal control angles $\alpha$ and $\beta$. If the tangent…

In the design of a robotic arm, the angle of rotation of the final joint depends on the subtraction of two internal control angles $\alpha$ and $\beta$. If the tangent of the resulting angle is $\tan(\alpha - \beta) = 2$ and due to mechanical constraints we know that $\tan \alpha = 3 \tan \beta$, determine the possible exact values of $\tan \beta$.

Solve

Applications: Sight Lines and Signals

Whenever two angles are measured from the same starting line, the angle between them is a difference. That covers sight lines from a surveyor, light rays, and phase shifts of electric signals.

Sight lines. Back to the billboard. Draw a right triangle to the base of the billboard and another to its top. Then use $\tan\beta=\frac{\text{height of the building}}{\text{distance}}$ and $\tan\alpha=\frac{\text{height to the top}}{\text{distance}}$.

A viewer on the ground looks at the base and top of a billboard on a building of height H at horizontal distance d, with elevation angles beta and alpha. β α d H h billboard

Signals. Two out-of-phase alternating signals can be combined into one wave $A\sin(\theta+\varphi)$. Expand it with the sum formula, $A\sin(\theta+\varphi)=(A\cos\varphi)\sin\theta+(A\sin\varphi)\cos\theta$, and match coefficients. This gives $A=\sqrt{p^2+q^2}$ for coefficients $p$ and $q$, and you choose $\varphi$ from the signs of $\cos\varphi$ and $\sin\varphi$.

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In a market study, the seasonal sales cycles of two products are modeled using trigonometric functions. The phase difference that determines direct competition is given…

In a market study, the seasonal sales cycles of two products are modeled using trigonometric functions. The phase difference that determines direct competition is given by the value of $\cos(x - y)$. If $\sin x = \frac{1}{3}$ and $\cos y = \frac{2}{5}$, with $x$ located in the first quadrant and $y$ in the fourth quadrant, find the exact value of $\cos(x - y)$.

Solve
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A surveyor observes a cliff of height $H$ and, situated right on the edge above it, a lighthouse of height $h$. From their position at a horizontal distance $d$, the…

A surveyor observes a cliff of height $H$ and, situated right on the edge above it, a lighthouse of height $h$. From their position at a horizontal distance $d$, the angle of elevation to the base of the lighthouse is $\beta$ and to the dome of the lighthouse is $\alpha$. Express the tangent of the angle subtended exclusively by the lighthouse, that is $\tan(\alpha - \beta)$, solely in terms of the distances $H$, $h$ and $d$.

β α α - β H h d
Solve
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In an alternating current circuit, the potential difference between two nodes is given by the superposition of two out-of-phase signals:…

In an alternating current circuit, the potential difference between two nodes is given by the superposition of two out-of-phase signals: $V(t) = 10 \sin(\omega t) - 10 \sin(\omega t - \frac{\pi}{3})$. Use the trigonometric formulas for angle subtraction to express the potential difference $V(t)$ in the simplified form $A \sin(\omega t + \phi)$, determining the values of $A$ and $\phi$.

Solve

Back to the start: The billboard angle comes from $\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$, with the two tangents read from the distances. And $\sin 15^\circ=\sin(45^\circ-30^\circ)=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6-\sqrt2}{4}\approx0.259$.

Summary

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