Unit 5, lesson 65

Power-Reducing and Half-Angle Formulas

How to trade a square of sine or cosine for a cosine of twice the angle, and how to find exact values at half an angle with the correct sign.

  • Theory part 129
  • Practice part 130
  • Identities and Formulas

A calculator says cos⁡22.5∘≈0.9239\cos 22.5^\circ \approx 0.9239, yet no special triangle has a 22.5∘22.5^\circ angle. It is half of 45∘45^\circ, though, and the cosine of 45∘45^\circ is known exactly. The formulas here turn a known angle into exact values at half that angle. Read the other way, they turn squares like sin⁡2x\sin^2 x into first powers, which is exactly what calculus needs to integrate them.

Key terms

Power-reducing formulas

Definition For every real number xx, in radians or in degrees,

sin⁡2x=1−cos⁡2x2,cos⁡2x=1+cos⁡2x2.\begin{gathered} \sin^2 x = \frac{1 - \cos 2x}{2}, \\[0.8em] \cos^2 x = \frac{1 + \cos 2x}{2}. \end{gathered}

They come from solving the double-angle forms cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x and cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1 for the square. Some books, including the calculus course, also call them half-angle identities.

In plain words A square of sine or cosine equals a constant plus a multiple of the cosine of twice the angle. The power drops from 2 to 1, and the angle doubles in exchange.

Example At x=π8x = \frac{\pi}{8} the doubled angle π4\frac{\pi}{4} has a known cosine, so

sin⁡2π8=1−cos⁡π42=1−222=2−24.\begin{aligned} \sin^2 \frac{\pi}{8} &= \frac{1 - \cos \frac{\pi}{4}}{2} \\[0.8em] &= \frac{1 - \frac{\sqrt{2}}{2}}{2} = \frac{2 - \sqrt{2}}{4}. \end{aligned}

As a check, sin⁡π8≈0.3827\sin \frac{\pi}{8} \approx 0.3827, and its square 0.14640.1464 matches 2−24≈0.1464\frac{2 - \sqrt{2}}{4} \approx 0.1464.

Power reduction for tangent

Definition For every real xx with cos⁡x≠0\cos x \neq 0,

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

The condition cos⁡x≠0\cos x \neq 0 is the same as cos⁡2x≠−1\cos 2x \neq -1, so the right side is defined exactly where tan⁡x\tan x is.

In plain words Divide the formula for sin⁡2x\sin^2 x by the formula for cos⁡2x\cos^2 x. The factors of 2 cancel.

Example Take x=π12x = \frac{\pi}{12}, so that 2x=π62x = \frac{\pi}{6} and cos⁡2x=32\cos 2x = \frac{\sqrt{3}}{2}. Then

tan⁡2π12=1−321+32=2−32+3=(2−3)2=7−43.\begin{aligned} \tan^2 \frac{\pi}{12} &= \frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{\sqrt{3}}{2}} = \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \\[0.8em] &= (2 - \sqrt{3})^2 = 7 - 4\sqrt{3}. \end{aligned}

The third step multiplies top and bottom by 2−32 - \sqrt{3} and uses (2+3)(2−3)=1(2 + \sqrt{3})(2 - \sqrt{3}) = 1.

Reducing higher powers

Definition A fourth power sin⁡4x\sin^4 x or cos⁡4x\cos^4 x is reduced in two rounds. Write it as the square of sin⁡2x\sin^2 x or cos⁡2x\cos^2 x, replace that square by its power-reducing form, expand, and reduce the new term cos⁡22x\cos^2 2x with

cos⁡22x=1+cos⁡4x2.\cos^2 2x = \frac{1 + \cos 4x}{2}.

The result, valid for every real xx, is a constant plus multiples of cos⁡2x\cos 2x and cos⁡4x\cos 4x. In the same way, for every real xx,

sin⁡2xcos⁡2x=14sin⁡22x=1−cos⁡4x8.\begin{aligned} \sin^2 x \cos^2 x &= \frac{1}{4} \sin^2 2x \\[0.8em] &= \frac{1 - \cos 4x}{8}. \end{aligned}

Sixth and higher even powers also produce odd powers such as cos⁡32x\cos^3 2x, and those are reduced with the help of the product-to-sum formulas.

In plain words Each round halves the highest power and doubles the angle. Keep going until only first powers of cosines are left.

Example The first round for sin⁡4x\sin^4 x gives

sin⁡4x=(1−cos⁡2x2)2=1−2cos⁡2x+cos⁡22x4.\begin{aligned} \sin^4 x &= \left( \frac{1 - \cos 2x}{2} \right)^2 \\[0.8em] &= \frac{1 - 2\cos 2x + \cos^2 2x}{4}. \end{aligned}

Replacing cos⁡22x\cos^2 2x by 1+cos⁡4x2\frac{1 + \cos 4x}{2} and collecting terms gives

sin⁡4x=3−4cos⁡2x+cos⁡4x8.\sin^4 x = \frac{3 - 4\cos 2x + \cos 4x}{8}.

At x=π4x = \frac{\pi}{4} both sides equal 14\frac{1}{4}, a quick check worth doing.

Half-angle formulas for sine and cosine

Definition For every real number θ\theta,

sin⁡θ2=±1−cos⁡θ2,cos⁡θ2=±1+cos⁡θ2.\begin{gathered} \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}}, \\[0.8em] \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}}. \end{gathered}

The sign in front of each root is not a free choice. It is the sign of the left side, which depends on where the angle θ2\frac{\theta}{2} lies. The radicands are never negative, because −1≤cos⁡θ≤1-1 \le \cos \theta \le 1. These are the power-reducing formulas with x=θ2x = \frac{\theta}{2}, solved for the first power.

In plain words To get the sine or cosine of half an angle, start from the cosine of the whole angle. Subtract it from 1 or add it to 1, halve, take the square root, and then attach the right sign.

Example The angle 15∘15^\circ is half of 30∘30^\circ and lies in the first quadrant, where cosine is positive. So

cos⁡15∘=1+cos⁡30∘2=1+322=2+32≈0.9659.\begin{aligned} \cos 15^\circ &= \sqrt{\frac{1 + \cos 30^\circ}{2}} \\[0.8em] &= \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} \\[0.8em] &= \frac{\sqrt{2 + \sqrt{3}}}{2} \approx 0.9659. \end{aligned}

The same number equals 6+24\frac{\sqrt{6} + \sqrt{2}}{4}, the value the difference formula gives for cos⁡(45∘−30∘)\cos(45^\circ - 30^\circ).

Choosing the sign of the square root

Definition Without a sign, the root gives the absolute value. For every real θ\theta,

1−cos⁡θ2=∣sin⁡θ2∣,1+cos⁡θ2=∣cos⁡θ2∣.\begin{gathered} \sqrt{\frac{1 - \cos \theta}{2}} = \left\lvert \sin \frac{\theta}{2} \right\rvert, \\[0.8em] \sqrt{\frac{1 + \cos \theta}{2}} = \left\lvert \cos \frac{\theta}{2} \right\rvert. \end{gathered}

The sign is the sign of sin⁡θ2\sin \frac{\theta}{2} or cos⁡θ2\cos \frac{\theta}{2} itself, so it is decided by the quadrant of the half angle θ2\frac{\theta}{2}, not of θ\theta.

Quadrant of θ2\frac{\theta}{2}Sign of sin⁡θ2\sin \frac{\theta}{2}Sign of cos⁡θ2\cos \frac{\theta}{2}
I++++
II++−-
III−-−-
IV−-++

Angles θ\theta and θ+2π\theta + 2\pi have the same cosine, but their halves differ by π\pi, so the sine and the cosine of the half angle change sign. So the sign depends on the actual value of θ\theta, not only on its terminal side.

In plain words The square root always comes out positive. Find where half the angle points and give the answer the sign that sine or cosine has there.

Example The angles 300∘300^\circ and −60∘-60^\circ share a terminal side, and both have cosine 12\frac{1}{2}. The formula gives

±1+122=±32\pm \sqrt{\frac{1 + \frac{1}{2}}{2}} = \pm \frac{\sqrt{3}}{2}

for the cosine of the half angle. Half of 300∘300^\circ is 150∘150^\circ, in Quadrant II, so cos⁡150∘=−32\cos 150^\circ = -\frac{\sqrt{3}}{2}. Half of −60∘-60^\circ is −30∘-30^\circ, in Quadrant IV, so cos⁡(−30∘)=32\cos(-30^\circ) = \frac{\sqrt{3}}{2}. Same cosine of θ\theta, opposite signs.

Half-angle formulas for tangent

Definition For every real θ\theta with cos⁡θ≠−1\cos \theta \neq -1, that is, θ\theta not an odd multiple of π\pi,

tan⁡θ2=sin⁡θ1+cos⁡θ.\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}.

For every real θ\theta with sin⁡θ≠0\sin \theta \neq 0, that is, θ\theta not an integer multiple of π\pi,

tan⁡θ2=1−cos⁡θsin⁡θ.\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}.

Neither form needs a sign choice. The quantities 1+cos⁡θ1 + \cos \theta and 1−cos⁡θ1 - \cos \theta are never negative, and wherever tan⁡θ2\tan \frac{\theta}{2} is defined, the number sin⁡θ=2sin⁡θ2cos⁡θ2\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} has the same sign as tan⁡θ2\tan \frac{\theta}{2}.

In plain words The tangent of half an angle needs no square root at all. Both forms are fractions built from the sine and cosine of the whole angle, and the sign comes out right by itself.

Example With θ=π4\theta = \frac{\pi}{4} the second form gives

tan⁡π8=1−2222=2−22=2−1≈0.4142.\begin{aligned} \tan \frac{\pi}{8} &= \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}} \\[0.8em] &= \sqrt{2} - 1 \approx 0.4142. \end{aligned}

The first form, 2/21+2/2\frac{\sqrt{2}/2}{1 + \sqrt{2}/2}, simplifies to the same 2−1\sqrt{2} - 1.

Sine and cosine from the tangent of the half angle

Definition Let θ\theta be a real number that is not an odd multiple of π\pi, and let t=tan⁡θ2t = \tan \frac{\theta}{2}. Then

sin⁡θ=2t1+t2,cos⁡θ=1−t21+t2.\sin \theta = \frac{2t}{1 + t^2}, \qquad \cos \theta = \frac{1 - t^2}{1 + t^2}.

Both follow from the double-angle formulas and cos⁡2θ2=11+t2\cos^2 \frac{\theta}{2} = \frac{1}{1 + t^2}.

In plain words One number, the tangent of the half angle, fixes both the sine and the cosine of the whole angle, using fractions without any roots. Calculus uses this to turn fractions of sines and cosines into ordinary fractions.

Example Suppose tan⁡θ2=12\tan \frac{\theta}{2} = \frac{1}{2}. Then

sin⁡θ=2⋅121+14=45,cos⁡θ=1−141+14=35.\begin{gathered} \sin \theta = \frac{2 \cdot \frac{1}{2}}{1 + \frac{1}{4}} = \frac{4}{5}, \\[0.8em] \cos \theta = \frac{1 - \frac{1}{4}}{1 + \frac{1}{4}} = \frac{3}{5}. \end{gathered}

As expected, (45)2+(35)2=1\left( \frac{4}{5} \right)^2 + \left( \frac{3}{5} \right)^2 = 1.

Domain conditions

Definition An identity holds only where both of its sides are defined. For the formulas above, with kk any integer,

FormulaValid for
power reduction for sin⁡2x\sin^2 x and cos⁡2x\cos^2 xall real xx
half-angle formulas for sine and cosineall real θ\theta
tan⁡2x=1−cos⁡2x1+cos⁡2x\tan^2 x = \frac{1 - \cos 2x}{1 + \cos 2x}x≠π2+kπx \neq \frac{\pi}{2} + k\pi
tan⁡θ2=sin⁡θ1+cos⁡θ\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}θ≠π+2kπ\theta \neq \pi + 2k\pi
tan⁡θ2=1−cos⁡θsin⁡θ\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta}θ≠kπ\theta \neq k\pi

The form with 1+cos⁡θ1 + \cos \theta is valid exactly where tan⁡θ2\tan \frac{\theta}{2} exists. The form with sin⁡θ\sin \theta in the denominator also fails at the multiples of 2π2\pi, where tan⁡θ2=0\tan \frac{\theta}{2} = 0 does exist.

In plain words A formula is only true where both sides make sense. Before you use one, make sure your angle avoids every zero denominator.

Example At θ=0\theta = 0 the half-angle tangent is tan⁡0=0\tan 0 = 0, and the first form agrees,

sin⁡01+cos⁡0=02=0.\frac{\sin 0}{1 + \cos 0} = \frac{0}{2} = 0.

The second form gives 1−10\frac{1 - 1}{0}, which is undefined, so it cannot be used at θ=0\theta = 0. At θ=π\theta = \pi neither form works, and tan⁡π2\tan \frac{\pi}{2} itself is undefined.

Check yourself

  1. Find the exact value of cos⁡112.5∘\cos 112.5^\circ.
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The angle 112.5∘112.5^\circ is half of 225∘225^\circ, and cos⁡225∘=−22\cos 225^\circ = -\frac{\sqrt{2}}{2}. The half angle is in Quadrant II, where cosine is negative, so

cos⁡112.5∘=−1−222=−2−22≈−0.3827.\begin{aligned} \cos 112.5^\circ &= -\sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} \\[0.8em] &= -\frac{\sqrt{2 - \sqrt{2}}}{2} \\[0.8em] &\approx -0.3827. \end{aligned}
  1. Suppose cos⁡θ=513\cos \theta = \frac{5}{13} and 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi. Find sin⁡θ2\sin \frac{\theta}{2}, cos⁡θ2\cos \frac{\theta}{2} and tan⁡θ2\tan \frac{\theta}{2}.
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Halving the inequalities gives 3π4<θ2<π\frac{3\pi}{4} < \frac{\theta}{2} < \pi, which is in Quadrant II. So the sine of the half angle is positive and the cosine is negative.

sin⁡θ2=1−5132=21313,cos⁡θ2=−1+5132=−31313.\begin{gathered} \sin \frac{\theta}{2} = \sqrt{\frac{1 - \frac{5}{13}}{2}} = \frac{2\sqrt{13}}{13}, \\[0.8em] \cos \frac{\theta}{2} = -\sqrt{\frac{1 + \frac{5}{13}}{2}} = -\frac{3\sqrt{13}}{13}. \end{gathered}

Their quotient is tan⁡θ2=−23\tan \frac{\theta}{2} = -\frac{2}{3}. As a check, sin⁡θ=−1213\sin \theta = -\frac{12}{13} in Quadrant IV, and the tangent formula gives

sin⁡θ1+cos⁡θ=−12131813=−23.\frac{\sin \theta}{1 + \cos \theta} = \frac{-\frac{12}{13}}{\frac{18}{13}} = -\frac{2}{3}.
  1. Write cos⁡4x\cos^4 x using only constants and first powers of cosines.
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Square the power-reducing formula and reduce cos⁡22x\cos^2 2x once more.

cos⁡4x=1+2cos⁡2x+cos⁡22x4=3+4cos⁡2x+cos⁡4x8.\begin{aligned} \cos^4 x &= \frac{1 + 2\cos 2x + \cos^2 2x}{4} \\[0.8em] &= \frac{3 + 4\cos 2x + \cos 4x}{8}. \end{aligned}

At x=0x = 0 both sides equal 11.

How this lesson connects