Double-angle ratios
A sliotar is struck at Croke Park: how far it travels depends on $\sin(2\theta)$, not on $\theta$ alone. A roof has twice the slope of another. 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. By the end of this page you will be able to say whether a ramp at $30^\circ$ has double the slope of one at $15^\circ$.
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 $2\alpha = \alpha+\alpha$: replace $\beta$ with $\alpha$ in the sum formulae.
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).
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.
Cian has to calculate the value of $\sin(2\alpha)$ knowing that $\alpha \in \left(0, \frac{\pi}{2}\right)$ and that $\sin\alpha = \frac{3}{5}$. On his exam, he writes…
Cian has to calculate the value of $\sin(2\alpha)$ knowing that $\alpha \in \left(0, \frac{\pi}{2}\right)$ and that $\sin\alpha = \frac{3}{5}$. On his exam, he writes the following solution:
"Since this is the double angle, the value of the sine simply doubles: $\sin(2\alpha) = 2 \times \sin\alpha = 2 \times \frac{3}{5} = \frac{6}{5} = 1.2$. However, since the sine of any real angle can never be greater than $1$, this problem has no solution and the proposed angle is impossible."
- a) Identify the two mistakes made by Cian (a conceptual one regarding the supposed linearity of trigonometric ratios and another in his final conclusion).
- b) Correct the exercise by calculating the exact value of $\sin(2\alpha)$ using the double-angle identity and determine with reasoning in which quadrant $2\alpha$ lies by finding the sign of $\cos(2\alpha)$.
Consider an acute angle $\alpha \in \left(0, \frac{\pi}{2}\right)$ for which it is known that $\cos\alpha = 0.8$. a) Before performing any algebraic calculation,…
Consider an acute angle $\alpha \in \left(0, \frac{\pi}{2}\right)$ for which it is known that $\cos\alpha = 0.8$.
- a) Before performing any algebraic calculation, estimate with reasoning the sign that $\cos(2\alpha)$ will have. To do this, compare $\cos\alpha = 0.8$ with the reference value $\cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707$, deduce whether $\alpha$ is greater or smaller than $45^\circ$, and reason in which quadrant the double angle $2\alpha$ must fall.
- b) Select the variant of the formula for $\cos(2\alpha)$ that depends solely on the cosine to calculate its exact numerical value in decimal form. Check whether the sign obtained matches your previous estimate.
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.
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.
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 $\widehat{A}$.
The goal is to design and reconstruct trigonometric equations from their geometric conditions and their solutions. a) It is known that a certain angle $\alpha$ belonging…
The goal is to design and reconstruct trigonometric equations from their geometric conditions and their solutions.
- a) It is known that a certain angle $\alpha$ belonging to the second quadrant $\left(\frac{\pi}{2} < \alpha < \pi\right)$ satisfies $\cos(2\alpha) = -\frac{7}{25}$. Uniquely deduce the exact values of $\cos\alpha$, $\sin\alpha$, and $\tan\alpha$, justifying the choice of sign at each step based on the quadrant of origin.
- b) Construct a trigonometric equation of the form $\cos(2x) + b\cos(x) + c = 0$ (with $b, c \in \mathbb{R}$) whose only solutions in the closed interval $[0, \pi]$ are exactly $x = \frac{\pi}{3}$ and $x = \pi$. Show step by step how you determine the coefficients $b$ and $c$ by first setting up a quadratic equation in the auxiliary variable $u = \cos(x)$, and check that no extraneous solution is introduced in that interval.
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.
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:
$\frac{2\tan(22.5^\circ)}{1 - \tan^2(22.5^\circ)}$
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:
$4\sin(15^\circ)\cos(15^\circ)$
Proving identities and half-angle ratios
In the exam, "Prove that…" or "Show that…" means you start from one side (usually the more complicated one) and transform it, line by line, into the other. Write everything in terms of $\sin x$ and $\cos x$ and use the double-angle formulae to remove the $2x$.
For systems of trigonometric equations, recognise the sum formula and turn each equation into a simple one: e.g. $\sin x\cos y+\cos x\sin y=\sin(x+y)$, and $\cos(2x)-\cos(2y)=0$ gives $2x=2y$ or $2x+2y=2\pi$ in the allowed range, then combine with the first equation.
The same formulae also work backwards. Rearranging $\cos(2\theta)=2\cos^2\theta-1$ gives $\cos^2\theta=\dfrac{1+\cos(2\theta)}{2}$, and $\cos(2\theta)=1-2\sin^2\theta$ gives $\sin^2\theta=\dfrac{1-\cos(2\theta)}{2}$.
The double-angle ratios can also be written in terms of $t=\tan\alpha$ alone. Divide the top and bottom of $\cos(2\alpha)=\cos^2\alpha-\sin^2\alpha$ by $\cos^2\alpha$, using $1=\cos^2\alpha+\sin^2\alpha$ in the denominator: $$\cos(2\alpha)=\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha}=\dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}$$ In the same way, $$\sin(2\alpha)=\dfrac{2\sin\alpha\cos\alpha}{\cos^2\alpha+\sin^2\alpha}=\dfrac{2\tan\alpha}{1+\tan^2\alpha}$$
To write $a\sin x\cos x+b\cos^2x$ in terms of $2x$, use $\sin x\cos x=\tfrac12\sin(2x)$ and $\cos^2x=\tfrac12(1+\cos(2x))$ (and similarly for $\sin^2x$).
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:
$\frac{\sin(2x)}{1 + \cos(2x)} = \tan(x)$
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 $\cos(\theta)$.
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.
When the equation has $\cos(2x)$ and $\sin x$, choose the form $\cos(2x)=1-2\sin^2x$ so that everything is in terms of $\sin x$. You get a quadratic in $\sin x$.
If an equation has a parameter, for example $\cos(2x)-4\sin x+k=0$, the same substitution gives a quadratic in $s=\sin x$. Then check which values of $s$ are allowed by the interval: on $\left[0,\tfrac{\pi}{2}\right]$, $s$ lies in $[0,1]$ and each such $s$ gives exactly one $x$.
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 $[0^\circ, 360^\circ)$:
$\cos(2x) - 3\sin(x) = 2$
To solve the trigonometric equation $\sin(2x) - \sqrt{3}\cos(x) = 0$ in the interval $[0, 2\pi)$, two classmates propose different strategies: Aoife proposes: "I divide…
To solve the trigonometric equation $\sin(2x) - \sqrt{3}\cos(x) = 0$ in the interval $[0, 2\pi)$, two classmates propose different strategies:
-
Aoife proposes: "I divide both sides by $\cos(x)$ to get $\frac{\sin(2x)}{\cos(x)} - \sqrt{3} = 0$, substitute $\sin(2x) = 2\sin(x)\cos(x)$, simplify the factor $\cos(x)$, and solve directly for $\sin(x) = \frac{\sqrt{3}}{2}$."
-
Seán proposes: "I first substitute $\sin(2x) = 2\sin(x)\cos(x)$ into the equation, factor out $\cos(x)$ without dividing by any unknown, and set each factor equal to zero."
- a) Critically analyze Aoife's method: explain what algebraic violation she commits by dividing by $\cos(x)$ and which specific solutions in the interval $[0, 2\pi)$ are lost if her procedure is applied.
- b) Develop Seán's method fully and rigorously, and write all exact solutions of the equation in the interval $[0, 2\pi)$ expressed in radians.
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 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
The double angle appears in the range of projectiles, in roof slopes that double a reference inclination, and in articulated mechanisms.
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.
Greatest value of $a\sin\theta+b\cos\theta$
To find the greatest value of an expression like $a\sin\theta+b\cos\theta$, write it as a single sine: $R\sin(\theta+\varphi)$. Expanding gives $R\sin\theta\cos\varphi+R\cos\theta\sin\varphi$, so $R\cos\varphi=a$ and $R\sin\varphi=b$. Squaring and adding gives $R=\sqrt{a^2+b^2}$. Since $\sin(\theta+\varphi)$ lies between $-1$ and $1$, the greatest value is $R$ and the least is $-R$.
To use this with the double angle, first rewrite products and squares in terms of $2\theta$ using the formulae above, for example $\sin x\cos x=\tfrac12\sin(2x)$ and $\sin^2x=\tfrac12(1-\cos(2x))$. The result is a constant plus $a\sin(2x)+b\cos(2x)$, whose greatest value is the constant plus $\sqrt{a^2+b^2}$.
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 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.