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5. Multiple Angle Formulae0%

Complex Numbers · Topic 5 of 5

5. Multiple Angle Formulae

Video coming soon3 worked examples

Theory

De Moivre's Theorem gives two powerful trigonometric tools, working in opposite directions.

Direction 1 — multiple angles into powers. Expand (cos⁡θ+isin⁡θ)n(\cos\theta + i\sin\theta)^n in two different ways. De Moivre says it equals cos⁡nθ+isin⁡nθ\cos n\theta + i\sin n\theta; the Binomial Theorem gives a sum of terms. Since the two must be equal, equate the real parts to get cos⁡nθ\cos n\theta, and the imaginary parts to get sin⁡nθ\sin n\theta.

Direction 2 — powers into multiple angles. Let z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta. Then 1z=z−1=cos⁡θ−isin⁡θ\frac{1}{z} = z^{-1} = \cos\theta - i\sin\theta, so adding and subtracting gives:

z+1z=2cos⁡θ,z−1z=2isin⁡θz + \frac{1}{z} = 2\cos\theta, \qquad z - \frac{1}{z} = 2i\sin\theta

and more generally — these are results you are expected to know:

zk+1zk=2cos⁡kθ,zk−1zk=2isin⁡kθz^k + \frac{1}{z^k} = 2\cos k\theta, \qquad z^k - \frac{1}{z^k} = 2i\sin k\theta

So to convert something like cos⁡4θ\cos^4\theta into multiple angles, expand (z+1z)4\left(z + \frac{1}{z}\right)^4 by the Binomial Theorem and pair the outer terms inwards — each pair collapses to a 2cos⁡kθ2\cos k\theta.

The Golden Rule: read the question to decide the direction. Going from cos⁡nθ\cos n\theta to powers of cos⁡θ\cos\theta uses De Moivre with the Binomial Theorem; going from a power like cos⁡nθ\cos^n\theta to multiple angles uses z+1zz + \frac{1}{z}.

⚠️ Common Examiner Traps

  • Powers of ii: the cycle is i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1. Getting one wrong flips a sign in the middle of the expansion.
  • Dropping the ii when equating: the imaginary part of 4ic3s4ic^3s is 4c3s4c^3s, not 4ic3s4ic^3s — compare coefficients of ii.
  • Not finishing the conversion: “in terms of cos⁡θ\cos\theta” means only cos⁡θ\cos\theta, so use sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta to remove every sine.
  • Forgetting the 2n2^n: (z+1z)n=(2cos⁡θ)n=2ncos⁡nθ\left(z + \frac{1}{z}\right)^n = (2\cos\theta)^n = 2^n\cos^n\theta, so you must divide by 2n2^n at the end.
  • Odd powers bring an ii: (2isin⁡θ)3=−8isin⁡3θ(2i\sin\theta)^3 = -8i\sin^3\theta, since i3=−ii^3 = -i.

Worked examples

Example 1

By considering (cos⁡θ+isin⁡θ)4(\cos\theta + i\sin\theta)^4, show that cos⁡4θ=8cos⁡4θ−8cos⁡2θ+1\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1, and find an expression for sin⁡4θ\sin 4\theta.

Step 1: Write c=cos⁡θc = \cos\theta and s=sin⁡θs = \sin\theta. By De Moivre's Theorem:

(c+is)4=cos⁡4θ+isin⁡4θ(c + is)^4 = \cos 4\theta + i\sin 4\theta

Step 2: Expand the same expression by the Binomial Theorem, with coefficients 1,4,6,4,11,4,6,4,1:

(c+is)4=c4+4c3(is)+6c2(is)2+4c(is)3+(is)4(c+is)^4 = c^4 + 4c^3(is) + 6c^2(is)^2 + 4c(is)^3 + (is)^4

Step 3: Simplify the powers of ii, using i2=−1i^2=-1, i3=−ii^3=-i, i4=1i^4=1:

=c4+4ic3s−6c2s2−4ics3+s4= c^4 + 4ic^3s - 6c^2s^2 - 4ics^3 + s^4

Step 4: Equate real parts to obtain cos⁡4θ\cos 4\theta:

cos⁡4θ=c4−6c2s2+s4\cos 4\theta = c^4 - 6c^2s^2 + s^4

Step 5: The answer must be in terms of cos⁡θ\cos\theta only, so replace s2=1−c2s^2 = 1 - c^2:

cos⁡4θ=c4−6c2(1−c2)+(1−c2)2=c4−6c2+6c4+1−2c2+c4=8c4−8c2+1\begin{aligned} \cos 4\theta &= c^4 - 6c^2(1-c^2) + (1-c^2)^2 \\ &= c^4 - 6c^2 + 6c^4 + 1 - 2c^2 + c^4 \\ &= 8c^4 - 8c^2 + 1 \end{aligned}

Step 6: Now equate the imaginary parts, dropping the factor of ii:

sin⁡4θ=4c3s−4cs3=4sin⁡θcos⁡θ(cos⁡2θ−sin⁡2θ)\sin 4\theta = 4c^3s - 4cs^3 = 4\sin\theta\cos\theta\left(\cos^2\theta - \sin^2\theta\right)

Check: that last form is 2(2sin⁡θcos⁡θ)(cos⁡2θ−sin⁡2θ)=2sin⁡2θcos⁡2θ=sin⁡4θ2(2\sin\theta\cos\theta)(\cos^2\theta - \sin^2\theta) = 2\sin 2\theta\cos 2\theta = \sin 4\theta ✓

Example 2

By considering z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta, show that cos⁡4θ=18[cos⁡4θ+4cos⁡2θ+3]\cos^4\theta = \dfrac{1}{8}\bigl[\cos 4\theta + 4\cos 2\theta + 3\bigr].

Step 1: Since z+1z=2cos⁡θz + \dfrac{1}{z} = 2\cos\theta, raising both sides to the fourth power gives:

(z+1z)4=(2cos⁡θ)4=16cos⁡4θ\left(z + \frac{1}{z}\right)^4 = (2\cos\theta)^4 = 16\cos^4\theta

Step 2: Expand the left-hand side by the Binomial Theorem. Each term is a power of zz times a power of 1z\frac{1}{z}, so the powers partly cancel:

(z+1z)4=z4+4z2+6+4z2+1z4\left(z + \frac{1}{z}\right)^4 = z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4}

Step 3: Pair the outer terms inwards, so each pair has the form zk+1zkz^k + \frac{1}{z^k}:

=(z4+1z4)+4(z2+1z2)+6= \left(z^4 + \frac{1}{z^4}\right) + 4\left(z^2 + \frac{1}{z^2}\right) + 6

Step 4: Apply zk+1zk=2cos⁡kθz^k + \frac{1}{z^k} = 2\cos k\theta to each pair:

=2cos⁡4θ+4(2cos⁡2θ)+6=2cos⁡4θ+8cos⁡2θ+6= 2\cos 4\theta + 4(2\cos 2\theta) + 6 = 2\cos 4\theta + 8\cos 2\theta + 6

Step 5: Equate the two expressions and divide by 1616:

16cos⁡4θ=2cos⁡4θ+8cos⁡2θ+6  ⟹  cos⁡4θ=18[cos⁡4θ+4cos⁡2θ+3]16\cos^4\theta = 2\cos 4\theta + 8\cos 2\theta + 6 \implies \cos^4\theta = \frac{1}{8}\bigl[\cos 4\theta + 4\cos 2\theta + 3\bigr]

Example 3

Show that sin⁡3θ=14[3sin⁡θ−sin⁡3θ]\sin^3\theta = \dfrac{1}{4}\bigl[3\sin\theta - \sin 3\theta\bigr].

Step 1: This time the function is sine, so use the subtraction result z−1z=2isin⁡θz - \dfrac{1}{z} = 2i\sin\theta. Cube both sides, remembering i3=−ii^3 = -i:

(z−1z)3=(2isin⁡θ)3=8i3sin⁡3θ=−8isin⁡3θ\left(z - \frac{1}{z}\right)^3 = (2i\sin\theta)^3 = 8i^3\sin^3\theta = -8i\sin^3\theta

Step 2: Expand the left-hand side, with alternating signs and coefficients 1,3,3,11,3,3,1:

(z−1z)3=z3−3z+3z−1z3\left(z - \frac{1}{z}\right)^3 = z^3 - 3z + \frac{3}{z} - \frac{1}{z^3}

Step 3: Pair the outer terms inwards. Note the middle pair carries a minus sign:

=(z3−1z3)−3(z−1z)= \left(z^3 - \frac{1}{z^3}\right) - 3\left(z - \frac{1}{z}\right)

Step 4: Apply zk−1zk=2isin⁡kθz^k - \frac{1}{z^k} = 2i\sin k\theta to each pair:

=2isin⁡3θ−3(2isin⁡θ)=2isin⁡3θ−6isin⁡θ= 2i\sin 3\theta - 3(2i\sin\theta) = 2i\sin 3\theta - 6i\sin\theta

Step 5: Equate with Step 1 and divide through by −2i-2i:

−8isin⁡3θ=2isin⁡3θ−6isin⁡θ  ⟹  4sin⁡3θ=3sin⁡θ−sin⁡3θ-8i\sin^3\theta = 2i\sin 3\theta - 6i\sin\theta \implies 4\sin^3\theta = 3\sin\theta - \sin 3\theta

Step 6: Divide by 44:

sin⁡3θ=14[3sin⁡θ−sin⁡3θ]\sin^3\theta = \frac{1}{4}\bigl[3\sin\theta - \sin 3\theta\bigr]