Double-Angle Identities
A water balloon launcher at the school fair, a skate ramp built at twice the angle of another, a revenue model that swings with the seasons. In each case the angle gets doubled, and the answer depends on $\sin(2\theta)$ or $\tan(2\alpha)$ instead of on $\theta$ alone. Instead of adding an angle to itself every time, we can pack the result into one identity. By the end of this page you will know why a launcher fires farthest at exactly $45^\circ$.
Where the Double-Angle Identities Come From
You already know the formulas for the sine and cosine of a sum. A double angle is just the special case $2\alpha = \alpha + \alpha$, so we plug $\beta = \alpha$ into those formulas.
All three forms of $\cos(2\alpha)$ have the same value. Pick the one that matches what you know: only $\cos\alpha$, only $\sin\alpha$, or both.
Here the two graphs sit on top of each other, so $\sin(2x)$ and $2\sin x\cos x$ are the same function.
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.
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:
$\cos^2\left(\frac{\pi}{8}\right) - \sin^2\left(\frac{\pi}{8}\right)$
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 $\cos(2\alpha)$.
Starting From One Known Ratio
Often you know only one ratio, say $\sin\alpha$, plus the quadrant. Find the missing ratio with $\sin^2\alpha + \cos^2\alpha = 1$, then apply a double-angle identity. Here is the picture for $\sin\alpha = \dfrac35$ with $\alpha$ in Quadrant II.
The hypotenuse is $5$, the vertical side is $3$ (positive), and the horizontal side is $-4$ because the point is left of the $y$-axis. So $\cos\alpha = -\dfrac45$.
Working backward: from $\cos(2\theta)$ to $\sin\theta$ and $\cos\theta$
Sometimes you know the double angle's cosine and need the single angle's ratios. Solve the second and third forms of the identity for the squares:
$$\sin^2\theta = \dfrac{1 - \cos(2\theta)}{2}, \qquad \cos^2\theta = \dfrac{1 + \cos(2\theta)}{2}.$$These are called the power-reducing identities. They come directly from $\cos 2\theta = 1 - 2\sin^2\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$.
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 $\sin(2\alpha)$.
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.
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 $\cos(\theta)$.
Recognizing the Patterns: Exact Values and Simplifying
Many expressions are a double-angle identity in disguise. Spotting the pattern lets you find exact values without a 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:
$4\sin(15^\circ)\cos(15^\circ)$
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:
$\frac{\sin(2x)}{2\sin(x)}$
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:
$\frac{2\tan(22.5^\circ)}{1 - \tan^2(22.5^\circ)}$
Proving Identities
To prove an identity, work on one side only (usually the more complicated one) until it turns into the other side. For double angles, the plan is almost always the same:
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Replace every double angle by the form that matches the rest of the expression.
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Factor and cancel.
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Recognize what is left.
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:
$\frac{\sin(2x)}{1 + \cos(2x)} = \tan(x)$
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, 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 $x$, first substitute a double-angle identity so everything uses the same angle. Then either factor or solve a quadratic in one ratio.
The blue points are where $\cos x = 0$. The orange points are where $\sin x = \tfrac12$. Together they solve $\sin(2x) = \cos x$.
When you solve for the ratio, always check that the value is possible: $-1 \le \sin x \le 1$ and $-1 \le \cos x \le 1$. In the second example, $\sin x = 2$ has no solution.
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: Parameters, Identities, and Systems
Three strategies cover the toughest problems.
1. Counting solutions with a parameter. Rewrite the equation using one ratio, let $u = \sin x$ (or $\cos x$), and track which values $u$ can take on the given interval. Then count how many $x$ each $u$ produces.
2. Prove first, then solve. If a problem asks you to prove an identity and then use it, the identity usually turns a hard equation into an easy one. For example, an expression like $\cos^4 x - \sin^4 x$ factors as a difference of squares and collapses to a single double-angle ratio.
3. Systems. Compress an equation with the sum formula $\sin(x + y) = \sin x\cos y + \cos x\sin y$. Remember that cosine is one-to-one on $[0, \pi]$: if $\cos A = \cos B$ with both $A, B \in [0, \pi]$, then $A = B$. Watch how the interval for $x$ and $y$ changes when you double them.
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: Ramps, Projectiles, and Models
Doubling an angle shows up in physics, building design, and models with periodic behavior.
The slope of ramp B is $\dfrac34$, which is more than twice $\dfrac13$. Tangent is not linear, so doubling the angle does not double the slope.
Now close the hook. Since $R = 50\sin(2\theta)$ is largest when $\sin(2\theta) = 1$, we need $2\theta = 90^\circ$. So the launcher reaches its farthest distance at $\theta = 45^\circ$, whatever the launch speed. The two angles that hit the same target always add up to $90^\circ$.
Maximum value of $a\sin u + b\cos u$
Double-angle identities also turn products like $\sin x\cos x$ and squares like $\sin^2 x$ into $\sin(2x)$ and $\cos(2x)$. That makes a periodic model easy to analyze, because a combination $a\sin u + b\cos u$ is just a shifted sine wave. Its amplitude is $R = \sqrt{a^2 + b^2}$, so its maximum is $R$ and its minimum is $-R$.
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.
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.