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Trigonometry · Trigonometry

Double-Angle Ratios

Before you start, make sure you know: Reduction to the first quadrant and the fundamental identity (Secondary Year 3)

A Liga MX player takes a free kick: the range of the ball depends on $\operatorname{sen}(2\theta)$, not just on $\theta$. In an industrial warehouse, one roof has twice the slope of the one next to it. The double angle shows up all the time, and it helps to have its ratios already "packaged" in a formula, without adding the angle to itself every time.

Where the formulas come from

You already know the formulas for the sine and cosine of a sum. The double angle is just the special case $2\alpha = \alpha+\alpha$: all you need to do is substitute $\beta$ with $\alpha$.

Unit circle showing a double angle as the sum of two equal alpha angles, with alpha equal to 45 degrees and 2 alpha equal to 90 degrees. α α 2α = α + α

The cosine of the double angle has three equivalent forms: use the one that best fits your data (only $\cos\alpha$, only $\operatorname{sen}\alpha$, or both).

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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:

$4\sin(15^\circ)\cos(15^\circ)$

Solve
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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.

Solve

With only one known ratio

If you only know $\operatorname{sen}\alpha$ or $\cos\alpha$ (and the quadrant), you find the other ratio with $\operatorname{sen}^2\alpha+\cos^2\alpha=1$ and then apply the double-angle formula. For the tangent, you only need $\tan\alpha$.

Right triangle with sides 3, 4, and 5 with angle alpha at the left vertex. α 3 4 5
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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 $\sin(2\alpha)$.

Solve
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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 $\cos(2\alpha)$.

Solve
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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.

Solve
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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 $\cos(\theta)$.

Solve
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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 $A$.…
α α A B C D

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 $\widehat{A}$.

Solve

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.

Diagram of the simplification of four times sine of 15 degrees times cosine of 15 degrees down to 1. 4·sen(15°)·cos(15°) = 2·sen(30°) = 1
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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:

$\cos^2\left(\frac{\pi}{8}\right) - \sin^2\left(\frac{\pi}{8}\right)$

Solve
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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:

$\frac{\sin(2x)}{2\sin(x)}$

Solve
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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:

$\frac{2\tan(22.5^\circ)}{1 - \tan^2(22.5^\circ)}$

Solve

Proving identities

Proving an identity means transforming one side, with justified steps, until you get the other. With the double angle, the plan is almost always: rewrite everything in terms of $x$ (not $2x$), factor, and simplify.

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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:

$\frac{\sin(2x)}{1 + \cos(2x)} = \tan(x)$

Solve
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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 $\alpha$.

Solve
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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:

$\frac{1 + \sin(2x) - \cos(2x)}{1 + \sin(2x) + \cos(2x)} = \tan(x)$

Solve
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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

Equations with the double angle

When an equation mixes $\operatorname{sen}(2x)$ or $\cos(2x)$ with ratios of $x$, the first step is to substitute the double-angle formula to put everything in terms of the same angle and then factor.

Unit circle with the solutions pi over 2, 3 pi over 2, 4 pi over 3, and 5 pi over 3. π/2 3π/2 4π/3 5π/3
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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
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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
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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 $[0, 2\pi)$:

$\tan(2x) \times \tan(x) = 1$

Solve
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Solve the following system of trigonometric equations for $x, y \in [0, \pi/2]$:…

Solve the following system of trigonometric equations for $x, y \in [0, \pi/2]$:

$\begin{cases} \sin(x)\cos(y) + \cos(x)\sin(y) = 1 \\ \cos(2x) - \cos(2y) = 0 \end{cases}$

Solve

Equations with a parameter

Sometimes the equation has a letter $k$ and the question is how many solutions it has depending on its value. The method is: reduce everything to a single ratio $u$ (for example $u=\operatorname{sen}x$), isolate $k$ on one side and study the graph of the other side, but only for the values that $u$ can take in the given interval.

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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 $\left[0, \frac{\pi}{2}\right]$.

Solve

Maximums and minimums with the double angle

There are expressions like $5\operatorname{sen}x\cos x - 2\operatorname{sen}^2x + 1$ whose maximum value can't be seen at a glance. With the double angle they turn into a combination of $\operatorname{sen}2x$ and $\cos2x$, and there you can bound them.

For exercises of this type, always follow the same steps: bring the squares and products down to $2x$ with the double-angle formulas, gather the terms into $a\operatorname{sen}2x+b\cos2x+c$, and apply $c+\sqrt{a^2+b^2}$.

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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.

Solve

When you mix $\cos\alpha$ and $\cos2\alpha$

If the expression contains both $\alpha$ and $2\alpha$ (not just $2x$), the trick is different: substitute $\cos2\alpha=2\cos^2\alpha-1$ and everything becomes a second-degree polynomial in $c=\cos\alpha$. Then you can factor it, find its vertex, or evaluate at the endpoints of $c$.

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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…
X Y P α 2α

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

Applications

The double angle appears in the range of projectiles (a ball, a cannon), in roof slopes that double an inclination, and in articulated mechanisms like a robotic arm.

Profile of two industrial warehouse roofs where the second has twice the inclination of the first. α roof 1 roof 2

Central angle, inscribed angle, and area with sine

Some geometry problems with circles also use the double angle. You need two support tools.

For example, a triangle $OAB$ with $OA=OB=R$ and angle $2\alpha$ at $O$ has area $\tfrac12R^2\operatorname{sen}(2\alpha)=R^2\operatorname{sen}\alpha\cos\alpha$. There the double angle turns everything into ratios of $\alpha$. For the other triangle, the height is obtained with the radius and $\cos\alpha$.

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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.

Solve
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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.

Solve
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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…
O A B P 2α α

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 $\triangle PAB$.

Solve

Back to the beginning: the range of the kick depends on $\operatorname{sen}(2\theta)$, and that's why two different angles ($\theta$ and $90^\circ-\theta$) reach the same point.

Summary

See all the exercises in this section