2.5 - Multiple-angle identities

2.5 - Multiple-angle identities

De Moivre's theorem turns one binomial expansion into two trigonometric identities. You will compare real and imaginary parts, derive tangent multiple-angle formulae with their restrictions, and use the identities to solve exact equations without losing solutions.

Comparing real and imaginary parts

How can a cube tell us the value of cos⁡3θ\cos3\theta? Set c=cos⁡θc=\cos\theta and s=sin⁡θs=\sin\theta. De Moivre's theorem says

(c+is)3=cos⁡3θ+isin⁡3θ.(c+is)^3=\cos3\theta+i\sin3\theta.

The binomial expansion gives another expression for exactly the same number:

(c+is)3=c3+3ic2s−3cs2−is3.(c+is)^3=c^3+3ic^2s-3cs^2-is^3.

The powers i2=−1i^2=-1 and i3=−ii^3=-i explain the last two signs. Real parts must agree, and imaginary parts must agree. This is stronger than comparing moduli, which would discard the signs and angles.

Comparing complex components

If a+ib=c+ida+ib=c+id, where a,b,c,da,b,c,d are real, then a=ca=c and b=db=d.

The imaginary part of a+iba+ib is the real number bb, not ibib.

First extract the real part, then use c2+s2=1c^2+s^2=1 to leave only powers of cosine.

Worked example 1: a cosine triple-angle identity

Question. Use de Moivre's theorem to express cos⁡3θ\cos3\theta in terms of cos⁡θ\cos\theta.

Plan. Take the real part of the cubic expansion and eliminate sin⁡2θ\sin^2\theta.

cos⁡3θ=c3−3cs2compare real parts=c3−3c(1−c2)use s2=1−c2=4c3−3c.\begin{aligned} \cos3\theta&=c^3-3cs^2 &&\text{compare real parts}\\ &=c^3-3c(1-c^2) &&\text{use }s^2=1-c^2\\ &=4c^3-3c. \end{aligned}

Answer. cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta, for every real θ\theta.

Check. At θ=π/3\theta=\pi/3, the right-hand side is 4(1/2)3−3(1/2)=−14(1/2)^3-3(1/2)=-1, agreeing with cos⁡π\cos\pi.

Keep the same expansion, but change the component being extracted. Eliminating c2c^2, rather than s2s^2, now leaves powers of sine.

Worked example 2: a sine triple-angle identity

Question. Use de Moivre's theorem to express sin⁡3θ\sin3\theta in terms of sin⁡θ\sin\theta.

Plan. Take the imaginary part of the cubic expansion and replace cos⁡2θ\cos^2\theta.

sin⁡3θ=3c2s−s3compare imaginary parts=3(1−s2)s−s3=3s−4s3.\begin{aligned} \sin3\theta&=3c^2s-s^3 &&\text{compare imaginary parts}\\ &=3(1-s^2)s-s^3\\ &=3s-4s^3. \end{aligned}

Answer. sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta=3\sin\theta-4\sin^3\theta, for every real θ\theta.

Check. At θ=π/6\theta=\pi/6, the right-hand side is 3/2−4(1/8)=13/2-4(1/8)=1, agreeing with sin⁡(π/2)\sin(\pi/2).

The expansion has already proved each identity for all real angles. A numerical check can catch an algebraic sign error; it cannot replace that proof.

Watch out

The real term from 3c(is)23c(is)^2 is −3cs2-3cs^2. Replacing i2i^2 by 11 produces a plausible-looking but false identity. Cycle the powers 1,i,−1,−i1,i,-1,-i before collecting components.

Use one higher power with the same comparison method.

Tangent multiple angles

Dividing the imaginary part by the real part produces a tangent identity wherever that division is defined.

The cubic expansion gives

tan⁡3θ=3c2s−s3c3−3cs2.\tan3\theta=\frac{3c^2s-s^3}{c^3-3cs^2}.

If c≠0c\ne0, divide every term in numerator and denominator by c3c^3. Each s/cs/c becomes t=tan⁡θt=\tan\theta. There is a second restriction: the original real part cos⁡3θ\cos3\theta must also be nonzero.

Tangent triple angle

For t=tan⁡θt=\tan\theta,

tan⁡3θ=3t−t31−3t2,\tan3\theta=\frac{3t-t^3}{1-3t^2},

provided cos⁡θ≠0\cos\theta\ne0 and 1−3t2≠01-3t^2\ne0. These conditions ensure that tan⁡θ\tan\theta and tan⁡3θ\tan3\theta are both defined.

Not in the formula booklet: learn how to derive it. The angle-addition formulae are supplied.

The restrictions come from the original functions, not from a preference about how to write the final fraction.

Worked example 3: deriving the tangent ratio

Question. Derive the tangent triple-angle formula using de Moivre's theorem, stating its restrictions.

Plan. Divide the cubic's imaginary part by its real part, then divide both by cos⁡3θ\cos^3\theta.

With c=cos⁡θc=\cos\theta and s=sin⁡θs=\sin\theta, de Moivre and the binomial expansion give

cos⁡3θ=c3−3cs2,sin⁡3θ=3c2s−s3.\cos3\theta=c^3-3cs^2,\qquad \sin3\theta=3c^2s-s^3.

Where cos⁡3θ≠0\cos3\theta\ne0,

tan⁡3θ=3c2s−s3c3−3cs2=3(s/c)−(s/c)31−3(s/c)2divide by c3, c≠0=3t−t31−3t2.\begin{aligned} \tan3\theta&=\frac{3c^2s-s^3}{c^3-3cs^2}\\ &=\frac{3(s/c)-(s/c)^3}{1-3(s/c)^2} &&\text{divide by }c^3,\ c\ne0\\ &=\frac{3t-t^3}{1-3t^2}. \end{aligned}

Answer. The formula holds when cos⁡θ≠0\cos\theta\ne0 and 1−3t2≠01-3t^2\ne0, with t=tan⁡θt=\tan\theta.

Check. At θ=π/4\theta=\pi/4, t=1t=1 gives (3−1)/(1−3)=−1(3-1)/(1-3)=-1, agreeing with tan⁡(3π/4)\tan(3\pi/4). At θ=π/6\theta=\pi/6, the denominator vanishes, correctly signalling that tan⁡(π/2)\tan(\pi/2) is undefined.

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