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Trigonometry · Trigonometry

Double-angle ratios

A cannon fires a projectile: its range depends on $\sin(2\theta)$, not on $\theta$ alone. An industrial roof has twice the slope of another. In engineering, the double angle turns up all the time, and it is handy to have its trigonometric ratios already "packaged" in a formula, without having to add the angle to itself every time.

Before you start, make sure you know: Reduction to the first quadrant and the fundamental identity (4th ESO)
Before you start, make sure you know: Oriented angles and the unit circle (4th ESO)

Where the formulae come from

You already know the formulae 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 replace $\beta$ with $\alpha$.

Unit circle showing a double angle as the sum of two equal angles alpha. α α 2α = α + α

The cosine of the double angle has three equivalent forms: they are all equal, so use whichever fits best with the data you have (only $\cos\alpha$, only $\sin\alpha$, or both).

Practice★☆☆Not started
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 $\sin\alpha$ or $\cos\alpha$ (and the quadrant), you can find the other ratio using the fundamental identity $\sin^2\alpha+\cos^2\alpha=1$, and then apply the double-angle formula.

Diagram of the procedure for calculating the double-angle ratios from the quadrant and one known ratio. 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

Simplifying expressions

Many expressions involving a product of sine and cosine, or differences of squares, are really a double-angle formula in disguise. Recognising the pattern saves you using a calculator.

Step-by-step diagram of simplifying four times sine of fifteen degrees times cosine of fifteen degrees to obtain one. 4·sin(15°)·cos(15°) = 2·sin(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

Equations with the double angle

When an equation mixes $\sin(2x)$ (or $\cos(2x)$) with ratios of $x$ that are not doubled, the first step is to substitute the double-angle formula so that everything is in terms of the same angle.

Diagram of the steps for solving a trigonometric equation by substituting the double angle and factorising. 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

Technical applications

The double angle appears in the range of projectiles, in roof slopes that double a reference inclination, and in articulated mechanisms.

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

Summary

See all the exercises in this section