Double-Angle Formulae

Set A in degrees, or as an exam sets it, by one function's value and a quadrant: the triangle in the semicircle shows sin 2A, 1 + cos 2A and 1 − cos 2A as lengths, and at whole multiples of 7.5° the table works them exactly in surds. Equations solves cos 2x or sin 2x against sin x or cos x in an interval, by the form an exam expects.

Sine	$\sin2A=2\sin A\cos A$
Cosine, three forms	$\begin{aligned}\cos2A&=\cos^{2}A-\sin^{2}A\\&=2\cos^{2}A-1\\&=1-2\sin^{2}A\end{aligned}$
Tangent	$\tan2A=\frac{2\tan A}{1-\tan^{2}A}$

Worked example — Double-Angle Formulae

Double angle	60°
Sine of A	0.5
Cosine of A	0.866
Sine of 2A	0.866
Cosine of 2A, three ways	0.5
Tangent of 2A	1.7321
Tangent of A, from 2A	0.5774
Cosine of 2A, exactly	1/2

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Double-Angle Formulae

The double-angle formulae give sin 2A, cos 2A and tan 2A from the functions of A; read backwards, they are the half-angle formulae.

Where the formula comes from

Put B = A in the compound-angle formulae. The sine's two products become equal, and the cosine becomes a difference of squares.
$\sin2A=2\sin A\cos A,\qquad\cos2A=\cos^{2}A-\sin^{2}A$
Trade one square for the other with sin²A + cos²A = 1. That gives two more forms of cos 2A, one with only a cosine in it and one with only a sine.
$\cos2A=2\cos^{2}A-1=1-2\sin^{2}A$
Divide for the tangent. Dividing top and bottom by cos²A needs cos A ≠ 0, and the result has no value when tan²A = 1, which is where cos 2A = 0.
$\tan2A=\frac{2\sin A\cos A}{\cos^{2}A-\sin^{2}A}=\frac{2\tan A}{1-\tan^{2}A}$
Read the cosine forms backwards for the half angle. They give cos²A and sin²A; the sign of the square root is the sign of cos A or sin A in A's quadrant, which no formula decides for you.
$\cos^{2}A=\frac{1+\cos2A}{2},\qquad\sin^{2}A=\frac{1-\cos2A}{2}$
In an equation, leave one function. Beside a sin x, write cos 2x as 1 − 2sin²x; beside a cos x, as 2cos²x − 1. Beside sin 2x, take the common factor out and never divide by it.
$\cos2x=\sin x\ \Rightarrow\ 2\sin^{2}x+\sin x-1=0,\qquad\sin2x=\cos x\ \Rightarrow\ \cos x\,(2\sin x-1)=0$

Problems solved in full

Double angles from a tangent and a quadrant 4 steps
tan A = 3/4 and 180° < A < 270°. Find sin 2A, cos 2A and tan 2A, and the quadrant of 2A.

Open this problem in the tool
1. A 3-4-5 triangle gives the sizes; in quadrant III sine and cosine are both negative:
  
  $\sin A=-\tfrac{3}{5},\qquad\cos A=-\tfrac{4}{5}$
2. The double-angle formulae. The two minus signs cancel in the sine:
  
  $\begin{aligned}\sin2A&=2\left(-\tfrac{3}{5}\right)\left(-\tfrac{4}{5}\right)=\tfrac{24}{25}\\ \cos2A&=\tfrac{16}{25}-\tfrac{9}{25}=\tfrac{7}{25}\end{aligned}$
3. The tangent formula needs only tan A:
  
  $\tan2A=\frac{2\cdot\tfrac{3}{4}}{1-\tfrac{9}{16}}=\frac{3/2}{7/16}=\frac{24}{7}$
4. 2A lies between 360° and 540°, and with a positive sine and cosine it is in quadrant I, one full turn on:
  
  $360^{\circ}<2A<540^{\circ},\qquad\sin2A>0,\quad\cos2A>0$
Answer

sin 2A = 24/25 = 0.96, cos 2A = 7/25 = 0.28, tan 2A = 24/7 ≈ 3.4286, and 2A ≈ 433.7398° is in quadrant I.

$\sin2A=\tfrac{24}{25},\quad\cos2A=\tfrac{7}{25},\quad\tan2A=\tfrac{24}{7}$
An exact value from the half-angle form 3 steps
Use cos 30° to find cos 15° exactly, and show that it equals (√6 + √2)/4.

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Show solution
1. Read cos 2A = 2cos²A − 1 backwards with A = 15°:
  
  $\cos^{2}15^{\circ}=\frac{1+\cos30^{\circ}}{2}=\frac{1+\tfrac{\sqrt{3}}{2}}{2}=\frac{2+\sqrt{3}}{4}$
2. 15° is acute, so take the positive root:
  
  $\cos15^{\circ}=\frac{\sqrt{2+\sqrt{3}}}{2}$
3. Square the compound-angle answer to compare. Both numbers are positive, so equal squares mean equal numbers:
  
  $\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)^{2}=\frac{6+2+2\sqrt{12}}{16}=\frac{8+4\sqrt{3}}{16}=\frac{2+\sqrt{3}}{4}$
Answer

cos 15° = √(2 + √3)/2 = (√6 + √2)/4 ≈ 0.9659.

$\cos15^{\circ}=\frac{\sqrt{2+\sqrt{3}}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0{.}9659$
An exact tangent from a quadratic 3 steps
Use tan 45° = 1 and the formula for tan 2A to find tan 22.5° exactly.

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Show solution
1. Put t = tan 22.5° into the double-angle formula with 2A = 45°:
  
  $\frac{2t}{1-t^{2}}=1\ \Rightarrow\ t^{2}+2t-1=0$
2. Solve the quadratic:
  
  $t=\frac{-2\pm\sqrt{4+4}}{2}=-1\pm\sqrt{2}$
3. 22.5° is acute, so its tangent is positive; the other root, −1 − √2, is tan 112.5°, whose double also has tangent 1. The triangle in the semicircle gives the same number another way:
  
  $\tan22{.}5^{\circ}=\frac{\sin45^{\circ}}{1+\cos45^{\circ}}=\frac{\sqrt{2}/2}{1+\sqrt{2}/2}=\frac{\sqrt{2}}{2+\sqrt{2}}=\sqrt{2}-1$
Answer

tan 22.5° = √2 − 1 ≈ 0.4142.

$\tan22{.}5^{\circ}=\sqrt{2}-1\approx0{.}4142$
Choosing the form of cos 2x 3 steps
Solve cos 2x = sin x for 0° ≤ x < 360°.

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Show solution
1. The other term is a sine, so use the form of cos 2x with only sines in it:
  
  $1-2\sin^{2}x=\sin x\ \Rightarrow\ 2\sin^{2}x+\sin x-1=0$
2. Factorise as a quadratic in sin x:
  
  $(2\sin x-1)(\sin x+1)=0\ \Rightarrow\ \sin x=\tfrac{1}{2},\quad\sin x=-1$
3. sin x = 1/2 at 30° and 180° − 30°; sin x = −1 only at 270°:
  
  $x=30^{\circ},\ 150^{\circ},\ 270^{\circ}$
Answer

x = 30°, 150° or 270°. The form cos²x − sin²x would have left a cosine and a sine in one equation.

$x=30^{\circ},\ 150^{\circ},\ 270^{\circ}$
Factorise, do not divide 3 steps
Solve sin 2x = cos x for 0° ≤ x < 360°.

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Show solution
1. Expand sin 2x and bring everything to one side:
  
  $2\sin x\cos x-\cos x=0$
2. cos x is in both terms. Take it out:
  
  $\cos x\,(2\sin x-1)=0$
3. A product is zero when either factor is:
  
  $\cos x=0:\ x=90^{\circ},\ 270^{\circ},\qquad\sin x=\tfrac{1}{2}:\ x=30^{\circ},\ 150^{\circ}$
Answer

x = 30°, 90°, 150° or 270°. Dividing by cos x at the first line gives only 30° and 150°: it assumes cos x ≠ 0 and loses exactly the solutions where cos x = 0.

$x=30^{\circ},\ 90^{\circ},\ 150^{\circ},\ 270^{\circ}$
A square into a double angle, for an integral 3 steps
Write sin²x in terms of cos 2x, and use it to find the exact value of the integral of sin²x from 0 to π/2.
Show solution
1. Rearrange cos 2x = 1 − 2sin²x:
  
  $\sin^{2}x=\tfrac{1}{2}-\tfrac{1}{2}\cos2x$
2. Each term now has an antiderivative you know:
  
  $\int_{0}^{\pi/2}\sin^{2}x\,dx=\left[\frac{x}{2}-\frac{\sin2x}{4}\right]_{0}^{\pi/2}$
3. sin π = sin 0 = 0, so only x/2 counts:
  
  $=\frac{\pi}{4}-0=\frac{\pi}{4}$
Answer

The integral is π/4 ≈ 0.7854, half the length of the interval, because sin²x averages 1/2 over it.

$\int_{0}^{\pi/2}\sin^{2}x\,dx=\frac{\pi}{4}\approx0{.}7854$
Proving a half-angle identity 3 steps
Prove that sin 2A/(1 + cos 2A) = tan A wherever both sides are defined, and check it at A = 30°.

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Show solution
1. Replace each double angle by the form that factorises with the other:
  
  $\frac{\sin2A}{1+\cos2A}=\frac{2\sin A\cos A}{1+(2\cos^{2}A-1)}=\frac{2\sin A\cos A}{2\cos^{2}A}$
2. Cancel 2 cos A, which is not zero wherever tan A is defined:
  
  $=\frac{\sin A}{\cos A}=\tan A$
3. At A = 30°:
  
  $\frac{\sin60^{\circ}}{1+\cos60^{\circ}}=\frac{\sqrt{3}/2}{3/2}=\frac{\sqrt{3}}{3}=\tan30^{\circ}$
Answer

sin 2A/(1 + cos 2A) = tan A; at 30° both sides are √3/3 ≈ 0.5774. In the semicircle it is P's height over the longer piece of the diameter.

$\frac{\sin2A}{1+\cos2A}=\tan A$

Supplementary problems, answers only

tan A = 2 and A is acute. Find tan 2A.

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−4/3

$\tan2A=-\tfrac{4}{3}$
Solve cos 2x + 3 cos x + 2 = 0 for 0° ≤ x < 360°.

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x = 120°, 180° or 240°

$x=120^{\circ},\ 180^{\circ},\ 240^{\circ}$
Solve sin 2x = sin x for 0° ≤ x < 360°.

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x = 0°, 60°, 180° or 300°

$x=0^{\circ},\ 60^{\circ},\ 180^{\circ},\ 300^{\circ}$
Find sin 22.5° exactly.

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√(2 − √2)/2 ≈ 0.3827

$\sin22{.}5^{\circ}=\frac{\sqrt{2-\sqrt{2}}}{2}$
cos A = 1/3 and A is in quadrant IV. Find sin 2A and cos 2A exactly.

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sin 2A = −4√2/9 ≈ −0.6285, cos 2A = −7/9

$\sin2A=-\tfrac{4\sqrt{2}}{9},\quad\cos2A=-\tfrac{7}{9}$
Show that cos 3A = 4cos³A − 3cos A.

Show answer

Expand cos(2A + A) = cos 2A cos A − sin 2A sin A, use 2cos²A − 1 and 2 sin A cos A, then sin²A = 1 − cos²A.

$\cos3A=4\cos^{3}A-3\cos A$

The double-angle formulae are one triangle in a semicircle

Join both ends of a diameter, W and E, to the point P at 2A on the circle. The chord WP leaves W at angle A, half the angle at the centre: Euclid's theorem that the angle at the circumference is half the angle at the centre on the same arc (Elements III.20). The triangle WPE is right-angled at P (III.31) and its hypotenuse is the diameter, 2, so WP = 2 cos A and EP = 2 sin A. Now measure it. P's height is WP · sin A, which is 2 sin A cos A, and it is also sin 2A. P's foot F splits the diameter into WF = WP · cos A = 2cos²A and FE = EP · sin A = 2sin²A, and those two pieces are 1 + cos 2A and 1 − cos 2A. The height over WF is tan A = sin 2A/(1 + cos 2A), the half-angle formula, without a line of algebra. Past A = 90° the chord WP points the other way and 2 cos A turns negative; each product carries its sign with it, so every one of these measurements holds at every angle.

Dividing by cos x throws half the answers away

Press sin 2x = cos x. Written as 2 sin x cos x = cos x, it invites you to cancel cos x, and 2 sin x = 1 gives 30° and 150°. The page finds four solutions: 90° and 270° as well, where cos x = 0 and both sides are 0. Cancelling cos x was dividing by it, which is allowed only where it is not zero, so the step assumed away exactly the solutions it lost. Bring everything to one side and take the common factor out instead, cos x(2 sin x − 1) = 0: a product is zero when either factor is. The same trap waits in sin 2x = sin x, which loses 0° and 180°.

sin²x averages one half, which is why mains is rated below its peak

Read cos 2x = 1 − 2sin²x backwards: sin²x = 1/2 − (1/2) cos 2x. The square of a sine wave is another wave, twice as fast and half as tall, riding on the level 1/2. Over any whole number of its periods the cosine part averages to zero, so sin²x averages exactly 1/2. A resistor turns voltage into heat in proportion to the square of the voltage, so an alternating voltage V₀ sin ωt heats it exactly as a steady V₀/√2 would. That steady equivalent is the root-mean-square voltage, and it is the figure on the plate: European mains is rated 230 V, and swings to 230√2 ≈ 325 V and back fifty times a second.

Learning path

Trigonometric identities and equations

References (3)

The double-angle formulae with all three forms of cos 2θ, the half-angle formulae with their quadrant sign and tan(α/2) = sin α/(1 + cos α), and the power-reducing formulae: J. Abramson, Precalculus 2e, OpenStax (2021), section 7.3, “Double-Angle, Half-Angle, and Reduction Formulas”.
Insight block 1: the angle at the centre is double the angle at the circumference on the same arc, which puts A at W when 2A is at the centre: Euclid, Elements, Book III, Proposition 20. D. E. Joyce (ed.), Clark University.
Insight block 1: the angle in a semicircle is a right angle, so WPE is right-angled at P: Euclid, Elements, Book III, Proposition 31. D. E. Joyce (ed.), Clark University.

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.

Double-Angle Formulae

Where the formula comes from

Supplementary problems, answers only

The double-angle formulae are one triangle in a semicircle 🖖

Dividing by cos x throws half the answers away 🖖

sin²x averages one half, which is why mains is rated below its peak 🖖

The double-angle formulae are one triangle in a semicircle

Dividing by cos x throws half the answers away

sin²x averages one half, which is why mains is rated below its peak