Unit 5, lesson 64

Double-Angle Formulas

How the sum formulas with two equal angles give sin 2x, the three forms of cos 2x and tan 2x, how to pick the right form, and how to find all of them from one value and a quadrant.

  • Theory part 127
  • Practice part 128
  • Identities and Formulas

Kick a ball from flat ground at speed vv and angle θ\theta above the horizontal, and if air resistance is ignored it lands at the distance

R=2v2sin⁡θcos⁡θg=v2sin⁡2θg,R = \frac{2v^2 \sin \theta \cos \theta}{g} = \frac{v^2 \sin 2\theta}{g},

where gg is the acceleration due to gravity. The second form makes the best angle obvious. The sine of 2θ2\theta is largest when 2θ=90∘2\theta = 90^\circ, so the longest kick leaves at 45∘45^\circ. Rewriting 2sin⁡θcos⁡θ2 \sin \theta \cos \theta as sin⁡2θ\sin 2\theta is one of the double-angle formulas.

Key terms

Double-angle formulas

Definition The double-angle formulas express sin⁡2x\sin 2x, cos⁡2x\cos 2x and tan⁡2x\tan 2x through the trigonometric values of xx. They are the sum formulas with both angles equal to xx,

sin⁡2x=sin⁡(x+x)=sin⁡xcos⁡x+cos⁡xsin⁡x=2sin⁡xcos⁡x,cos⁡2x=cos⁡(x+x)=cos⁡xcos⁡x−sin⁡xsin⁡x=cos⁡2x−sin⁡2x,\begin{aligned} \sin 2x &= \sin(x + x) \\[0.4em] &= \sin x \cos x + \cos x \sin x \\[0.4em] &= 2 \sin x \cos x, \\[0.8em] \cos 2x &= \cos(x + x) \\[0.4em] &= \cos x \cos x - \sin x \sin x \\[0.4em] &= \cos^2 x - \sin^2 x, \end{aligned}

together with a formula for tan⁡2x\tan 2x that comes from the tangent sum formula. The sine and cosine formulas hold for every angle xx, in radians or in degrees. The tangent formula needs tan⁡x\tan x and tan⁡2x\tan 2x to be defined.

In plain words Doubling an angle does not double its sine or cosine. The formulas tell you what happens instead.

Example At x=π6x = \frac{\pi}{6} the doubled angle is π3\frac{\pi}{3}, and sin⁡π3=32≈0.866\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \approx 0.866, while 2sin⁡π6=12 \sin \frac{\pi}{6} = 1. The sum formula gets it right,

sin⁡π6cos⁡π6+cos⁡π6sin⁡π6=2⋅12⋅32=32.\begin{aligned} &\sin \frac{\pi}{6} \cos \frac{\pi}{6} + \cos \frac{\pi}{6} \sin \frac{\pi}{6} \\[0.8em] &\quad = 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}. \end{aligned}

Sine of a double angle

Definition For every real xx,

sin⁡2x=2sin⁡xcos⁡x.\sin 2x = 2 \sin x \cos x.

Read from right to left, it says sin⁡xcos⁡x=12sin⁡2x\sin x \cos x = \frac{1}{2} \sin 2x, which turns a product into a single sine.

In plain words The sine of twice an angle is twice the product of the sine and the cosine of the angle.

Example How large can sin⁡xcos⁡x\sin x \cos x get? It equals 12sin⁡2x\frac{1}{2} \sin 2x, and a sine is at most 11, so the largest value is 12\frac{1}{2}. It is reached when 2x=π2+2kπ2x = \frac{\pi}{2} + 2k\pi, that is, at x=π4+kπx = \frac{\pi}{4} + k\pi for integers kk. At x=π4x = \frac{\pi}{4} the product is 22⋅22=12\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2}.

Cosine of a double angle in three forms

Definition For every real xx,

cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x.\begin{aligned} \cos 2x &= \cos^2 x - \sin^2 x \\[0.4em] &= 2\cos^2 x - 1 \\[0.4em] &= 1 - 2\sin^2 x. \end{aligned}

The first form is the cosine sum formula with both angles equal to xx. The second comes from replacing sin⁡2x\sin^2 x by 1−cos⁡2x1 - \cos^2 x, and the third from replacing cos⁡2x\cos^2 x by 1−sin⁡2x1 - \sin^2 x.

In plain words The three forms are one quantity written three ways. The second uses only the cosine of xx and the third only the sine.

Example The table compares the three forms at two angles with the value of cos⁡2x\cos 2x itself.

xxcos⁡2x−sin⁡2x\cos^2 x - \sin^2 x2cos⁡2x−12\cos^2 x - 11−2sin⁡2x1 - 2\sin^2 xcos⁡2x\cos 2x
π6\frac{\pi}{6}34−14=12\frac{3}{4} - \frac{1}{4} = \frac{1}{2}32−1=12\frac{3}{2} - 1 = \frac{1}{2}1−12=121 - \frac{1}{2} = \frac{1}{2}cos⁡π3=12\cos \frac{\pi}{3} = \frac{1}{2}
2π3\frac{2\pi}{3}14−34=−12\frac{1}{4} - \frac{3}{4} = -\frac{1}{2}12−1=−12\frac{1}{2} - 1 = -\frac{1}{2}1−32=−121 - \frac{3}{2} = -\frac{1}{2}cos⁡4π3=−12\cos \frac{4\pi}{3} = -\frac{1}{2}

Choosing a form of cos 2x

Definition Use 1−2sin⁡2x1 - 2\sin^2 x when only sin⁡x\sin x is known, 2cos⁡2x−12\cos^2 x - 1 when only cos⁡x\cos x is known, and cos⁡2x−sin⁡2x\cos^2 x - \sin^2 x when both are known or when it factors as (cos⁡x−sin⁡x)(cos⁡x+sin⁡x)(\cos x - \sin x)(\cos x + \sin x). The second and third forms use the square of a single function, so they give cos⁡2x\cos 2x even when the quadrant of xx, and with it the sign of the other function, is unknown. Solving them for the squares gives the power-reducing formulas.

In plain words Pick the form that uses what you already have. If you only know the sine, you never need the cosine.

Example Suppose sin⁡x=13\sin x = \frac{1}{3} and nothing is said about the quadrant. The value of cos⁡x\cos x could be 223\frac{2\sqrt{2}}{3} or −223-\frac{2\sqrt{2}}{3}, but the third form does not care,

cos⁡2x=1−2(13)2=1−29=79.\begin{aligned} \cos 2x &= 1 - 2\left(\frac{1}{3}\right)^2 \\[0.8em] &= 1 - \frac{2}{9} = \frac{7}{9}. \end{aligned}

Tangent of a double angle

Definition For every real xx at which both tan⁡x\tan x and tan⁡2x\tan 2x are defined, that is, x≠π2+kπx \neq \frac{\pi}{2} + k\pi and x≠π4+kπ2x \neq \frac{\pi}{4} + \frac{k\pi}{2} for every integer kk,

tan⁡2x=2tan⁡x1−tan⁡2x.\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}.

It is the tangent sum formula with both angles equal to xx. The denominator 1−tan⁡2x1 - \tan^2 x is zero exactly at x=π4+kπ2x = \frac{\pi}{4} + \frac{k\pi}{2}, where tan⁡2x\tan 2x does not exist.

In plain words The tangent of a double angle needs only the tangent of the angle. Watch for tan⁡x=±1\tan x = \pm 1, where the doubled angle points straight up or down and has no tangent.

Example If tan⁡x=12\tan x = \frac{1}{2}, then

tan⁡2x=2⋅121−14=134=43.\tan 2x = \frac{2 \cdot \frac{1}{2}}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}.

As a check, take xx in Quadrant I with sin⁡x=15\sin x = \frac{1}{\sqrt{5}} and cos⁡x=25\cos x = \frac{2}{\sqrt{5}}. Then sin⁡2x=45\sin 2x = \frac{4}{5} and cos⁡2x=45−15=35\cos 2x = \frac{4}{5} - \frac{1}{5} = \frac{3}{5}, whose quotient is 43\frac{4}{3} again.

All double-angle values from one ratio

Definition To find sin⁡2x\sin 2x, cos⁡2x\cos 2x and tan⁡2x\tan 2x from one trigonometric value of xx and the quadrant of xx, first find sin⁡x\sin x and cos⁡x\cos x with the Pythagorean identities, choosing signs from the quadrant. Then apply the formulas for sin⁡2x\sin 2x and cos⁡2x\cos 2x, and get tan⁡2x\tan 2x as their quotient whenever cos⁡2x≠0\cos 2x \neq 0. Doubling the interval that contains xx gives an interval for 2x2x, and the signs of sin⁡2x\sin 2x and cos⁡2x\cos 2x then fix its quadrant.

In plain words Get the sine and the cosine of xx with the right signs, then plug in. The signs of the results tell you where 2x2x ends up.

Example Let sin⁡x=35\sin x = \frac{3}{5} with π2<x<π\frac{\pi}{2} < x < \pi. Cosine is negative there, so cos⁡x=−45\cos x = -\frac{4}{5}. Then

sin⁡2x=2⋅35⋅(−45)=−2425,cos⁡2x=1−2⋅925=725,\begin{gathered} \sin 2x = 2 \cdot \frac{3}{5} \cdot \left(-\frac{4}{5}\right) = -\frac{24}{25}, \\[0.8em] \cos 2x = 1 - 2 \cdot \frac{9}{25} = \frac{7}{25}, \end{gathered}

and tan⁡2x=−247\tan 2x = -\frac{24}{7}. Doubling gives π<2x<2π\pi < 2x < 2\pi, and with a negative sine and a positive cosine the angle 2x2x is in Quadrant IV.

Triple angles from double angles

Definition Writing 3x=2x+x3x = 2x + x and using a sum formula together with the double-angle formulas gives, for every real xx,

cos⁡3x=4cos⁡3x−3cos⁡x,sin⁡3x=3sin⁡x−4sin⁡3x.\begin{gathered} \cos 3x = 4\cos^3 x - 3\cos x, \\[0.4em] \sin 3x = 3\sin x - 4\sin^3 x. \end{gathered}

For the cosine,

cos⁡3x=cos⁡2xcos⁡x−sin⁡2xsin⁡x=(2cos⁡2x−1)cos⁡x−2sin⁡2xcos⁡x=2cos⁡3x−cos⁡x−2(1−cos⁡2x)cos⁡x=4cos⁡3x−3cos⁡x.\begin{aligned} \cos 3x &= \cos 2x \cos x - \sin 2x \sin x \\[0.4em] &= (2\cos^2 x - 1)\cos x \\[0.4em] &\quad - 2\sin^2 x \cos x \\[0.4em] &= 2\cos^3 x - \cos x \\[0.4em] &\quad - 2(1 - \cos^2 x)\cos x \\[0.4em] &= 4\cos^3 x - 3\cos x. \end{aligned}

In plain words Double-angle formulas are building blocks. Combined with the sum formulas they reach any whole multiple of an angle.

Example At x=π4x = \frac{\pi}{4} the left side is cos⁡3π4=−22\cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}, and with cos⁡π4=22\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} the right side is

4(22)3−3⋅22=2−322=−22.\begin{aligned} 4 \left(\frac{\sqrt{2}}{2}\right)^3 - 3 \cdot \frac{\sqrt{2}}{2} &= \sqrt{2} - \frac{3\sqrt{2}}{2} \\[0.8em] &= -\frac{\sqrt{2}}{2}. \end{aligned}

Check yourself

  1. Let cos⁡x=−513\cos x = -\frac{5}{13} with π<x<3π2\pi < x < \frac{3\pi}{2}. Find sin⁡2x\sin 2x, cos⁡2x\cos 2x and tan⁡2x\tan 2x.
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In Quadrant III sine is negative, so sin⁡x=−1213\sin x = -\frac{12}{13}. Then

sin⁡2x=2(−1213)(−513)=120169,cos⁡2x=2⋅25169−1=−119169,\begin{gathered} \sin 2x = 2 \left(-\frac{12}{13}\right)\left(-\frac{5}{13}\right) = \frac{120}{169}, \\[0.8em] \cos 2x = 2 \cdot \frac{25}{169} - 1 = -\frac{119}{169}, \end{gathered}

and tan⁡2x=−120119\tan 2x = -\frac{120}{119}. The angle 2x2x lies between 2π2\pi and 3π3\pi, with a positive sine and a negative cosine, so it points into Quadrant II.

  1. Write cos⁡4x−sin⁡4x\cos^4 x - \sin^4 x as a single cosine.
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Factor as a difference of squares and use the first form of cos⁡2x\cos 2x.

cos⁡4x−sin⁡4x=(cos⁡2x−sin⁡2x)⋅(cos⁡2x+sin⁡2x)=cos⁡2x.\begin{aligned} \cos^4 x - \sin^4 x &= (\cos^2 x - \sin^2 x) \\[0.4em] &\quad \cdot (\cos^2 x + \sin^2 x) \\[0.4em] &= \cos 2x. \end{aligned}
  1. Without a calculator, find 2sin⁡15∘cos⁡15∘2 \sin 15^\circ \cos 15^\circ and cos⁡215∘−sin⁡215∘\cos^2 15^\circ - \sin^2 15^\circ.
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These are the double-angle formulas at x=15∘x = 15^\circ, so they equal sin⁡30∘=12\sin 30^\circ = \frac{1}{2} and cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2}.

How this lesson connects

Where this leads in Calculus