Double-Angle Ratios
At Estadio Quisqueya, a player throws the ball from the outfield toward home plate. How far it goes depends on $\operatorname{sen}(2\theta)$, not on $\theta$ alone. Why does double the angle appear? In engineering and physics the double angle comes up all the time, and it helps to have its ratios already "packaged" in a formula, without adding the angle to itself every time. At the end of the page we will come back to this throw.
Where the formulas come from
The double angle is a sum of two equal angles: $2\alpha = \alpha+\alpha$. To derive its ratios we start from the angle-sum formulas, which we remind you of here.
It is enough to substitute $\alpha$ for $\beta$ in each one.
The cosine of the double angle has three equivalent forms: they all have the same value. Use the one that best fits your data (only $\cos\alpha$, only $\operatorname{sen}\alpha$, or both).
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:
$4\sin(15^\circ)\cos(15^\circ)$
Given that $\cos(x) = 0.6$, determine the value of $\cos(2x)$ using exclusively the identity relating the cosine of a double angle to the cosine of a single angle.
Given that $\cos(x) = 0.6$, determine the value of $\cos(2x)$ using exclusively the identity relating the cosine of a double angle to the cosine of a single angle.
With only one known ratio
If you only know $\operatorname{sen}\alpha$ or $\cos\alpha$ (and the quadrant), find the other ratio with $\operatorname{sen}^2\alpha+\cos^2\alpha=1$ and then apply the double-angle formula. If you know $\tan\alpha$, the tangent formula is used directly.
Given 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 $\sin(\alpha) = \frac{4}{5}$ 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 $\alpha$ belongs to the second quadrant, determine the exact value of $\cos(2\alpha)$.
Given that $\cos(\alpha) = -\frac{1}{3}$ and that $\alpha$ belongs to 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 with differences of squares, are a double-angle formula in disguise. Recognizing the pattern saves you the calculator.
Verifying identities
The curriculum asks you to verify identities: you transform one side until you get the other. With the double angle, the trick is to substitute and simplify.
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 as much as possible: $\frac{\sin(2x)}{2\sin(x)}$
Write the following expression in terms of a single trigonometric ratio of the angle $x$, simplifying as much as possible:
$\frac{\sin(2x)}{2\sin(x)}$
Find the exact value of the following operation using double-angle identities: $\frac{2\tan(22.5^\circ)}{1 - \tan^2(22.5^\circ)}$
Find the exact value of the following operation using double-angle identities:
$\frac{2\tan(22.5^\circ)}{1 - \tan^2(22.5^\circ)}$
A scissor mechanism for a hydraulic lift has a central pivot forming 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 forming an opening angle $2\alpha$. The shear force on the central pin is given by an expression proportional to $F(\alpha) = \frac{1 - \cos(2\alpha)}{\sin(2\alpha)}$. Prove 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 to obtain the right-hand side: $\frac{\sin(2x)}{1 + \cos(2x)} = \tan(x)$
Prove the following trigonometric identity, transforming the left-hand side to obtain the right-hand side:
$\frac{\sin(2x)}{1 + \cos(2x)} = \tan(x)$
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:
$\frac{1 + \sin(2x) - \cos(2x)}{1 + \sin(2x) + \cos(2x)} = \tan(x)$
Equations with the double angle
When an equation mixes $\operatorname{sen}(2x)$ (or $\cos(2x)$) with ratios of $x$ that are not doubled, the first step is to substitute the double-angle formula so everything is in terms of the same angle. Then you factor or solve it as a quadratic in one ratio.
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 $[0, 2\pi)$: $\tan(2x) \times \tan(x) = 1$
Solve the following trigonometric equation, determining all solutions in the interval $[0, 2\pi)$:
$\tan(2x) \times \tan(x) = 1$
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 $[0^\circ, 360^\circ)$.
Equations with a parameter
Sometimes the equation has a letter $k$ and you are asked how many solutions it has depending on the value of $k$. The method is: substitute the double angle, call the remaining ratio $u$, solve for $k$, and see how many times the expression in $u$ takes each value of $k$.
There is one key detail: on $\left[0,\frac{\pi}{2}\right]$, the sine increases from $0$ to $1$, so each $u=\operatorname{sen}x\in[0,1]$ corresponds to exactly one $x$. Counting solutions in $x$ is the same as counting solutions in $u\in[0,1]$.
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 $\left[0, \frac{\pi}{2}\right]$.
A sine and a cosine together: maximum value
Sometimes an expression mixes $\operatorname{sen}x\cos x$ and $\operatorname{sen}^2x$, and you are asked for its maximum value. The idea is to bring it to the form $a\operatorname{sen}(2x)+b\cos(2x)$ and read the maximum from there.
To go from squares to the double angle we use the two forms of the cosine that you already know, solved for the squares:
$$\operatorname{sen}^2x=\dfrac{1-\cos(2x)}{2}\qquad\cos^2x=\dfrac{1+\cos(2x)}{2}$$And to combine sine and cosine of the same angle $u$ we use this property.
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 elapsed time 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 elapsed time 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.
Area of a triangle with a double angle
If you know two sides of a triangle and the angle between them, the area does not need the height: it is calculated directly with the sine of that angle.
When the angle between the sides is double another angle, $2\alpha$, the area is $\frac12ab\operatorname{sen}(2\alpha)=ab\operatorname{sen}\alpha\cos\alpha$, and the double angle lets you calculate it with a single ratio of $\alpha$.
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 $\widehat{APB}$ measuring $\alpha$. Given that $\cos(\alpha) = \frac{4}{5}$, calculate the ratio (quotient) between the area of the central triangle $\triangle OAB$ and the area of the inscribed triangle $\triangle PAB$.
Applications and problems
The double angle appears in projectile range, in factory roofs that double a reference slope, and in articulated mechanisms. It is also useful when you know $\cos(2\theta)$ and want the angle $\theta$: the forms $\cos(2\theta)=1-2\operatorname{sen}^2\theta$ and $\cos(2\theta)=2\cos^2\theta-1$ solve for $\operatorname{sen}\theta$ and $\cos\theta$.
In the construction of an industrial warehouse, two adjacent roofs have been designed. The first roof has an inclination $\alpha$ to the horizontal such that…
In the construction of an industrial warehouse, two adjacent roofs have been designed. The first roof has an inclination $\alpha$ 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 twice 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 modelled using 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 modelled using 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 situated $45 \text{ m}$ away, find the possible launch angles $\theta$ in degrees.
A ray of light is incident on a mirror system such that its final deviation angle is $2\theta$. If laboratory measurements indicate that $\cos(2\theta) = 0.28$ and we…
A ray of light is incident on a mirror system such that its final deviation angle 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 $\cos(\theta)$.
In rhombus $ABCD$ shown in the figure, the diagonals measure $AC = 16 \text{ cm}$ and $BD = 12 \text{ cm}$. Let $\alpha$ be half of the interior angle at vertex $A$.…
In rhombus $ABCD$ shown in the figure, the diagonals measure $AC = 16 \text{ cm}$ and $BD = 12 \text{ cm}$. Let $\alpha$ be half of 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 $2\alpha$ 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 situated exactly on the y-axis ($x_P = 0$).
Solve the following system of trigonometric equations for $x, y \in [0, \pi/2]$:…
Solve the following system of trigonometric equations for $x, y \in [0, \pi/2]$:
$\begin{cases} \sin(x)\cos(y) + \cos(x)\sin(y) = 1 \\ \cos(2x) - \cos(2y) = 0 \end{cases}$