Double-Angle Identities
When a quarterback throws a football or a golfer hits a drive, the distance depends on $\sin(2\theta)$, not on the launch angle $\theta$ alone. Why does $45^\circ$ send the ball farthest? And why do two different angles, like $30^\circ$ and $60^\circ$, land at the same spot? The double-angle formulas answer both questions, and they show up all over precalculus and calculus.
Where the Formulas Come From
You already know the sum formulas for sine and cosine. A double angle is just the special case $2\alpha = \alpha + \alpha$: plug $\beta = \alpha$ into the sum formulas and simplify.
Cosine has three equivalent forms. Pick the one that matches what you know: only $\cos\alpha$, only $\sin\alpha$, or both. The tangent formula only works when $\tan\alpha$ exists and $\tan\alpha \ne \pm 1$.
The two graphs sit exactly on top of each other. That is what "identity" means.
Knowing that $\cos(x) = 0.6$, determine the value of $\cos(2x)$ using exclusively the identity that relates the cosine of the double angle to the cosine of the single…
Knowing that $\cos(x) = 0.6$, determine the value of $\cos(2x)$ using exclusively the identity that relates the cosine of the double angle to the cosine of the single angle.
Find the exact value of the following operation using the double angle identities: $\frac{2\tan(22.5^\circ)}{1 - \tan^2(22.5^\circ)}$
Find the exact value of the following operation using the double angle identities:
$\frac{2\tan(22.5^\circ)}{1 - \tan^2(22.5^\circ)}$
Starting from One Known Ratio
Often you know just one ratio of $\alpha$ plus its quadrant. Find the missing ratio with $\sin^2\alpha + \cos^2\alpha = 1$ (or a right triangle), fix its sign from the quadrant, then use a double-angle formula.
If you know $\tan\alpha$, you do not need the other ratios at all: plug straight into $\tan(2\alpha) = \dfrac{2\tan\alpha}{1-\tan^2\alpha}$. If a problem asks for the result as a fraction in simplest form, just reduce it to lowest terms by dividing out common factors.
Knowing that $\sin(\alpha) = \frac{4}{5}$ and that $\alpha$ is an angle in the first quadrant, calculate the exact value of $\sin(2\alpha)$.
Knowing that $\sin(\alpha) = \frac{4}{5}$ and that $\alpha$ is an angle in the first quadrant, calculate the exact value of $\sin(2\alpha)$.
Given that $\cos(\alpha) = -\frac{1}{3}$ and that $\alpha$ is in the second quadrant, determine the exact value of $\cos(2\alpha)$.
Given that $\cos(\alpha) = -\frac{1}{3}$ and that $\alpha$ is in the second quadrant, determine the exact value of $\cos(2\alpha)$.
If $\tan(\alpha) = \frac{3}{2}$, calculate the value of $\tan(2\alpha)$, expressing the result as an irreducible fraction.
If $\tan(\alpha) = \frac{3}{2}$, calculate the value of $\tan(2\alpha)$, expressing the result as an irreducible fraction.
Simplifying Expressions
Many expressions with a product of sine and cosine, or a difference of squares, are a double-angle formula in disguise. Spotting the pattern gives exact values with no calculator.
Simplify the following trigonometric expression to a single exact numerical value: $4\sin(15^\circ)\cos(15^\circ)$
Simplify the following trigonometric expression to a single exact numerical value:
$4\sin(15^\circ)\cos(15^\circ)$
Calculate the exact value of the following expression without using a calculator: $\cos^2\left(\frac{\pi}{8}\right) - \sin^2\left(\frac{\pi}{8}\right)$
Calculate the exact value of the following expression without using a calculator:
$\cos^2\left(\frac{\pi}{8}\right) - \sin^2\left(\frac{\pi}{8}\right)$
Write the following expression in terms of a single trigonometric ratio of the angle $x$, simplifying fully: $\frac{\sin(2x)}{2\sin(x)}$
Write the following expression in terms of a single trigonometric ratio of the angle $x$, simplifying fully:
$\frac{\sin(2x)}{2\sin(x)}$
Going Backward: From $\cos 2\alpha$ to $\alpha$
What if you know $\cos(2\alpha)$ and want the ratios of $\alpha$? Rearrange the two single-ratio forms of the cosine formula:
Then take a square root. The sign is decided by the quadrant of $\alpha$, not of $2\alpha$.
All three curves are the same graph, which is why you can choose whichever form is convenient.
A ray of light is incident on a system of mirrors such that its final angle of deviation is $2\theta$. If laboratory measurements indicate that $\cos(2\theta) = 0.28$…
A ray of light is incident on a system of mirrors such that its final angle of deviation is $2\theta$. If laboratory measurements indicate that $\cos(2\theta) = 0.28$ and we know that the angle of incidence $\theta$ is acute, determine the exact value of $\sin(\theta)$ and $\cos(\theta)$.
Proving Identities
To prove an identity, work on one side only until it turns into the other side. Double-angle formulas are great for this. A good plan:
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Replace every double angle by single-angle ratios of $x$.
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Choose the form of $\cos(2x)$ that cancels something. For $1 - \cos 2x$ use $1 - 2\sin^2 x$; for $1 + \cos 2x$ use $2\cos^2 x - 1$.
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Factor and cancel.
A scissor mechanism for a hydraulic lift has a central pivot that forms an opening angle $2\alpha$. The shear force on the central pin is given by an expression…
A scissor mechanism for a hydraulic lift has a central pivot that forms an opening angle $2\alpha$. The shear force on the central pin is given by an expression proportional to $F(\alpha) = \frac{1 - \cos(2\alpha)}{\sin(2\alpha)}$. Show that this expression can be simplified to a single trigonometric ratio of the angle $\alpha$.
Prove the following trigonometric identity, transforming the left-hand side until the right-hand side is obtained: $\frac{\sin(2x)}{1 + \cos(2x)} = \tan(x)$
Prove the following trigonometric identity, transforming the left-hand side until the right-hand side is obtained:
$\frac{\sin(2x)}{1 + \cos(2x)} = \tan(x)$
Prove the following trigonometric identity, valid for any angle $x$ where the expressions are defined: $\frac{1 + \sin(2x) - \cos(2x)}{1 + \sin(2x) + \cos(2x)} = \tan(x)$
Prove the following trigonometric identity, valid for any angle $x$ where the expressions are defined:
$\frac{1 + \sin(2x) - \cos(2x)}{1 + \sin(2x) + \cos(2x)} = \tan(x)$
Solving Equations with Double Angles
When an equation mixes $\sin(2x)$ or $\cos(2x)$ with ratios of $x$, substitute a double-angle formula first so everything is in terms of the same angle. Then factor, or treat it as a quadratic in one ratio.
For an equation with $\tan 2x$, write $\tan 2x = \dfrac{2t}{1 - t^2}$ with $t = \tan x$ and solve for $t$. Check at the end that $\tan x$ and $\tan 2x$ are actually defined at each answer.
Solve the following trigonometric equation, finding all solutions in the interval $[0, 2\pi)$: $\sin(2x) + \sqrt{3}\cos(x) = 0$
Solve the following trigonometric equation, finding all solutions in the interval $[0, 2\pi)$:
$\sin(2x) + \sqrt{3}\cos(x) = 0$
Solve the following trigonometric equation and give the solutions in degrees for the interval $[0^\circ, 360^\circ)$: $\cos(2x) - 3\sin(x) = 2$
Solve the following trigonometric equation and give the solutions in degrees for the interval $[0^\circ, 360^\circ)$:
$\cos(2x) - 3\sin(x) = 2$
Solve the following trigonometric equation, determining all solutions in the interval $[0, 2\pi)$: $\tan(2x) \cdot \tan(x) = 1$
Solve the following trigonometric equation, determining all solutions in the interval $[0, 2\pi)$:
$\tan(2x) \cdot \tan(x) = 1$
Harder Equations: Intervals, Parameters, Systems
Equations where the double angle stays. If you solve for $2x$, remember that $x$ lives in $[0, 2\pi)$, so $2x$ lives in $[0, 4\pi)$. You need to go around the circle twice.
Equations with a parameter. Substitute $s = \sin x$ and count the solutions in $s$ that land in the allowed range. On $\left[0, \frac{\pi}{2}\right]$, $s = \sin x$ runs through $[0, 1]$ and each value of $s$ matches exactly one $x$. (The same trick works for $\cos x$ on $[0, \pi]$.)
Systems. Use the sum and difference formulas to collapse an expression like $\sin x\cos y + \cos x\sin y$ into $\sin(x + y)$. Then use the interval to decide which angles are possible.
Find the values of the parameter $k \in \mathbb{R}$ for which the equation $\cos(2x) - 4\sin(x) + k = 0$ has exactly one solution in the interval…
Find the values of the parameter $k \in \mathbb{R}$ for which the equation $\cos(2x) - 4\sin(x) + k = 0$ has exactly one solution in the interval $\left[0, \frac{\pi}{2}\right]$.
First prove the identity $\cos^4(x) - \sin^4(x) = \cos(2x)$. Then, use this result to solve the equation $\cos^4(x) - \sin^4(x) = \frac{1}{2}$ in the interval…
First prove the identity $\cos^4(x) - \sin^4(x) = \cos(2x)$. Then, use this result to solve the equation $\cos^4(x) - \sin^4(x) = \frac{1}{2}$ in the interval $[0^\circ, 360^\circ)$.
Solve the following system of trigonometric equations for $x, y \in [0, \pi/2]$:…
Solve the following system of trigonometric equations for $x, y \in [0, \pi/2]$:
$\begin{cases} \sin(x)\cos(y) + \cos(x)\sin(y) = 1 \\ \cos(2x) - \cos(2y) = 0 \end{cases}$
Applications: Range and Slope
The double angle appears wherever something is launched, or where one slope is built on a doubled angle.
In the construction of an industrial building, two adjacent roofs have been designed. The first roof has an inclination $\alpha$ with respect to the horizontal such that…
In the construction of an industrial building, two adjacent roofs have been designed. The first roof has an inclination $\alpha$ with respect to the horizontal such that $\tan(\alpha) = 0.4$. For water drainage reasons, the second roof has been designed with an angle of inclination that is exactly double that of the first, that is, $2\alpha$. Calculate the exact slope (tangent) of the second roof.
In physics, the maximum range $R$ of a projectile launched over horizontal ground is modeled by the formula $R = \frac{v_0^2 \sin(2\theta)}{g}$, where $v_0$ is the…
In physics, the maximum range $R$ of a projectile launched over horizontal ground is modeled by the formula $R = \frac{v_0^2 \sin(2\theta)}{g}$, where $v_0$ is the initial velocity, $\theta$ is the launch angle and $g \approx 10 \text{ m/s}^2$ is the acceleration due to gravity. If a test cannon fires a projectile with an initial velocity of $30 \text{ m/s}$ and it is desired to hit a target located at a distance of $45 \text{ m}$, find the possible launch angles $\theta$ in degrees.
Doubled Angles in Geometry
In a rhombus, the diagonals bisect the angles. So an angle of the rhombus is $2\alpha$, where $\alpha$ is the angle between a side and a diagonal. The half-diagonals form a right triangle that gives you $\tan\alpha$, and the double-angle formulas do the rest.
The same idea works on a circle. If $P$ is at angle $\alpha$ on a circle of radius $r$, then the point at angle $2\alpha$ is $(r\cos 2\alpha,\ r\sin 2\alpha)$.
In the rhombus $ABCD$ in the figure, the diagonals measure $AC = 16 \text{ cm}$ and $BD = 12 \text{ cm}$. Let $\alpha$ be half the interior angle at vertex $A$.…
In the rhombus $ABCD$ in the figure, the diagonals measure $AC = 16 \text{ cm}$ and $BD = 12 \text{ cm}$. Let $\alpha$ be half the interior angle at vertex $A$. Calculate the exact value of $\sin(2\alpha)$ and $\cos(2\alpha)$ to determine the sine and cosine of the full angle $\widehat{A}$.
A planar robotic arm consists of two links, both of length $L$. The first link forms an angle $\alpha$ with the positive horizontal axis. The second link forms an angle…
A planar robotic arm consists of two links, both of length $L$. The first link forms an angle $\alpha$ with the positive horizontal axis. The second link forms an angle $2\alpha$ with respect to the horizontal. The $x$-coordinate of the end effector $P$ is given by the expression $x_P = L\cos(\alpha) + L\cos(2\alpha)$. Determine the exact value of the angle $\alpha \in [0, \pi/2]$ for which the end effector is located exactly on the Y-axis ($x_P = 0$).
In a circle with center $O$ and radius $R$, the central angle $\widehat{AOB}$ measures $2\alpha$ and subtends the chord $AB$. A point $P$ located on the major arc…
In a circle with center $O$ and radius $R$, the central angle $\widehat{AOB}$ measures $2\alpha$ and subtends the chord $AB$. A point $P$ located on the major arc subtends an inscribed angle $\widehat{APB}$ which measures $\alpha$. Knowing that $\cos(\alpha) = \frac{4}{5}$, calculate the ratio (the quotient) between the area of the central triangle $\triangle OAB$ and the area of the inscribed triangle $\triangle PAB$.
Combining into One Wave: Maximum and Minimum
Double-angle formulas turn products and squares into $\sin 2x$ and $\cos 2x$. The result is often $A\sin u + B\cos u$. This always has the same shape as a sine wave, with maximum $\sqrt{A^2 + B^2}$ and minimum $-\sqrt{A^2 + B^2}$.
Why? Let $R = \sqrt{A^2 + B^2}$ and choose $\varphi$ with $\cos\varphi = \dfrac{A}{R}$ and $\sin\varphi = \dfrac{B}{R}$. By the sum formula, $A\sin u + B\cos u = R(\sin u\cos\varphi + \cos u\sin\varphi) = R\sin(u + \varphi)$, which ranges from $-R$ to $R$.
A company's revenue function from the sale of a seasonal product is given by the expression $I(x) = 5\sin(x)\cos(x) - 2\sin^2(x) + 1$, where $x$ is the time elapsed in…
A company's revenue function from the sale of a seasonal product is given by the expression $I(x) = 5\sin(x)\cos(x) - 2\sin^2(x) + 1$, where $x$ is the time elapsed in months ($x \ge 0$). Express $I(x)$ in terms of $\sin(2x)$ and $\cos(2x)$, and determine the maximum revenue the company can achieve.