Difference Formulas for Sine, Cosine, and Tangent
Safety guidelines say to lean a ladder against a wall at about $75^\circ$. For a 12-foot ladder, the top reaches $12\sin 75^\circ$ feet up the wall, and the base sits $12\cos 75^\circ$ feet from it. A calculator gives decimals, but $75^\circ$ is not one of the angles you memorized ($30^\circ$, $45^\circ$, $60^\circ$). Can we get exact values anyway?
The Difference Formulas
You already know the exact values of sine and cosine at the special angles. The difference formulas turn those into exact values for angles like $15^\circ$ and $75^\circ$. They need almost no new memorizing, because they come from the sum formulas by changing one sign.
Start with the sum formulas:
$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta \qquad \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$A difference is a sum with a negative angle: $\alpha-\beta=\alpha+(-\beta)$. Cosine is even, so $\cos(-\beta)=\cos\beta$. Sine is odd, so $\sin(-\beta)=-\sin\beta$. Substituting $-\beta$ in the sum formulas gives the difference formulas.
The formulas also work in reverse. If you see an expression shaped like the right side, you can collapse it into one function of a single angle.
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 $y = 15^\circ$.
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:
$\sin(75^\circ) \cos(15^\circ) - \cos(75^\circ) \sin(15^\circ)$
Cofunction and Supplementary Angles
When one of the two angles is $90^\circ$ or $180^\circ$, the formulas collapse into short identities that show up everywhere. The reason is that $\sin$ and $\cos$ at $90^\circ$ and $180^\circ$ are just $0$, $1$, or $-1$.
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 $x$.
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.
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.
Exact Values for Angles Like $15^\circ$ and $75^\circ$
To find an exact value, write the angle as a difference of two special angles. Then plug their known values into the formula.
Keep these values handy:
| Angle | $\sin$ | $\cos$ | $\tan$ |
|---|---|---|---|
| $30^\circ$ | $\frac12$ | $\frac{\sqrt3}{2}$ | $\frac{\sqrt3}{3}$ |
| $45^\circ$ | $\frac{\sqrt2}{2}$ | $\frac{\sqrt2}{2}$ | $1$ |
| $60^\circ$ | $\frac{\sqrt3}{2}$ | $\frac12$ | $\sqrt3$ |
Angles bigger than $90^\circ$ use a reference angle. For $120^\circ$, the reference angle is $60^\circ$, so $\sin 120^\circ=\frac{\sqrt3}{2}$ and $\cos 120^\circ=-\frac12$ (cosine is negative in Quadrant II).
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 $15^\circ = 45^\circ - 30^\circ$.
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.
The Tangent Difference Formula
The tangent formula comes from dividing sine by cosine:
$$\tan(\alpha-\beta)=\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos\alpha\cos\beta+\sin\alpha\sin\beta}$$Divide the top and bottom by $\cos\alpha\cos\beta$. Each term becomes a tangent, and you get
$$\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}.$$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.
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 $105^\circ = 135^\circ - 30^\circ$.
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 $\tan(30^\circ) = \frac{\sqrt{3}}{3}$.
Using Given Ratios and Quadrants
Many problems give you $\sin\alpha$ and $\cos\beta$, for example, and ask for $\sin(\alpha-\beta)$. The formula needs all four values: $\sin\alpha$, $\cos\alpha$, $\sin\beta$, $\cos\beta$. So the plan has three steps.
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Find the missing ratios with $\sin^2x+\cos^2x=1$ (or a reference triangle).
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Choose the signs from the quadrant.
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Substitute into the formula and simplify.
The sign chart is: Quadrant I, all positive. Quadrant II, only sine positive. Quadrant III, only tangent positive. Quadrant IV, only cosine positive. Drag the point and watch the signs of $x$ and $y$ (cosine and sine are $x$ and $y$ divided by the distance from the origin).
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 $\sin(\alpha - \beta)$.
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 $\cos(\alpha - \beta)$.
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 $\cos(x - y)$.
Applications: Angles of Elevation, Forces, and Slopes
A difference of two angles shows up whenever you look at something between two directions. This happens with two lines of sight, two forces measured from the horizontal, or two lines with different slopes. The angle between the two directions is $\alpha-\beta$.
For two forces, the situation is the same. If force 1 makes angle $\alpha$ with the horizontal and force 2 makes angle $\beta$, the angle between the forces is $\alpha-\beta$. You usually know the sine and cosine of each angle from the force components.
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.
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, $\sin(\alpha - \beta)$.
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 $d$.
Simplifying and Proving Identities
To prove an identity, work on one side only until it turns into the other side. Start with the side that has the angle sum or difference, then expand it. The three most common moves are:
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expand with the sum and difference formulas,
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divide the top and bottom by a product like $\cos a\cos b$ to create tangents,
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replace $\sin^2$ with $1-\cos^2$ (or the reverse) when squares appear.
Adding or subtracting a sum formula and a difference formula makes terms cancel. This is very useful.
The sine version of that last identity is just as useful. It turns a product of two sines into a difference of squares.
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 $b$:
$\frac{\sin(a - b)}{\cos a \cos b}$
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:
$\frac{\cos(a - b) - \cos(a + b)}{\sin(a + b) + \sin(a - b)} = \tan b$
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 $\sin(75^\circ) \cdot \sin(15^\circ)$.
Solving Trigonometric Equations
Difference formulas help in three kinds of equations: an angle shifted by a constant, systems with products like $\sin x\cos y$, and equations where a ratio of tangents leads to a quadratic.
A shifted angle
Don't expand $\sin(x-20^\circ)$. Treat the whole angle as one unknown, $u=x-20^\circ$, solve for $u$, then shift back. Watch the interval: if $0^\circ\le x<360^\circ$, then $u$ runs from $-20^\circ$ up to (but not including) $340^\circ$.
A system of products
When you see $\sin x\cos y$ and $\cos x\sin y$ together, add and subtract the equations. The sum is $\sin(x+y)$ and the difference is $\sin(x-y)$. Then list every angle that fits, and check each candidate in the original system.
A quadratic in the tangent
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 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.
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 $\tan \beta$.
Combining Waves: $A\sin(\theta+\varphi)$
The current from a US wall outlet is a sine wave that repeats 60 times per second. Subtracting an angle inside a sine just slides the graph sideways, so the difference of two same-frequency waves is another wave. The difference formulas let you write it as a single $A\sin(\theta+\varphi)$.
The same add-and-subtract idea solves some equations. Combining $\cos(x-45^\circ)+\cos(x+45^\circ)$ makes the $\sin x$ terms cancel.
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:
$\cos(x - 45^\circ) + \cos(x + 45^\circ) = 1$
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 $\phi$.
Back to the start: For the ladder, $\sin 75^\circ=\sin(120^\circ-45^\circ)=\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt2}{2}-\left(-\dfrac12\right)\dfrac{\sqrt2}{2}=\dfrac{\sqrt6+\sqrt2}{4}$. The top of the 12-foot ladder reaches $12\cdot\dfrac{\sqrt6+\sqrt2}{4}=3(\sqrt6+\sqrt2)\approx 11.6$ ft up the wall. The base is $3(\sqrt6-\sqrt2)\approx 3.1$ ft from the wall.