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De Moivre: cos nθ and sin nθ in powers

Edexcel IAL Further Mathematics · WFM02 Further Pure Mathematics 2

Expanding (cos θ + i sin θ)n with the binomial theorem and comparing it with cos nθ + i sin nθ gives identities such as cos 3θ = 4cos3 θ - 3cos θ. These are a favourite exam question, often followed by solving an equation.

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The method

Let c = cos θ and s = sin θ.

1. By de Moivre: cos nθ + i sin nθ = (c + is)n.

2. Expand the right-hand side with the binomial theorem; use i2 = -1, i3 = -i, i4 = 1.

3. Equate real parts (for cos nθ) or imaginary parts (for sin nθ).

4. Replace s2 with 1 - c2 (or c2 with 1 - s2) to get a single function.

Results you should be able to derive

If the question says "use de Moivre's theorem", other methods (such as compound angle formulae) may get no credit.

tan nθ

Divide the imaginary part by the real part, then divide top and bottom by the highest power of c:

tan 3θ = (3c2 s - s3)/(c3 - 3cs2) = (3t - t3)/(1 - 3t2), where t = tan θ.

Solving equations

A polynomial equation such as 8x3 - 6x - 1 = 0 can be solved by putting x = cos θ: it becomes 2cos 3θ = 1.

When you set an expansion equal to zero, such as sin 5θ = 0, include every solution, including θ = 0. The June 2025 examiners' report notes some students squaring (1 - s2) incorrectly: (1 - s2)2 = 1 - 2s2 + s4 has three terms.

Worked example

Use de Moivre's theorem to show that cos 4θ = 8cos4 θ - 8cos2 θ + 1.

  1. cos 4θ + i sin 4θ = (c + is)4 = c4 + 4c3(is) + 6c2(is)2 + 4c(is)3 + (is)4.
  2. Real part: cos 4θ = c4 - 6c2 s2 + s4.
  3. Replace s2 = 1 - c2: c4 - 6c2(1 - c2) + (1 - c2)2 = c4 - 6c2 + 6c4 + 1 - 2c2 + c4.

Answer: cos 4θ = 8c4 - 8c2 + 1

Key idea: Expand (c + is)n, equate real or imaginary parts, then use s2 + c2 = 1 to write the result in one function.

Check yourself

  1. Which identity follows from the real part of (cos θ + i sin θ)3?
    • cos 3θ = 4cos3 θ - 3cos θ
    • cos 3θ = 3cos θ - 4cos3 θ
    • cos 3θ = 4cos3 θ + 3cos θ
    • cos 3θ = cos3 θ - 3cos θ
    Show answer

    cos 3θ = 4cos3 θ - 3cos θ. Real part: c3 - 3cs2 = c3 - 3c(1 - c2) = 4c3 - 3c.

  2. sin 5θ = a sin5 θ + b sin3 θ + c sin θ. Find a.
    Show answer

    16. Imaginary part of (c + is)5: 5c4 s - 10c2 s3 + s5. With c2 = 1 - s2: 5(1 - s2)2 s - 10(1 - s2)s3 + s5 = 16s5 - 20s3 + 5s.

  3. cos 6θ = 32cos6 θ + p cos4 θ + q cos2 θ - 1. Find q.
    Show answer

    18. Equating real parts and replacing s2 by 1 - c2 gives cos 6θ = 32c6 - 48c4 + 18c2 - 1, so q = 18.

  4. cos 5θ = 16cos5 θ + k cos3 θ + 5cos θ. Find k.
    Show answer

    -20. Real part: c5 - 10c3 s2 + 5cs4. Replacing s2 gives 16c5 - 20c3 + 5c.

  5. Which expression equals tan 3θ, where t = tan θ?
    • (3t + t3)/(1 - 3t2)
    • (3t - t3)/(1 - 3t2)
    • (3t - t3)/(1 + 3t2)
    • 3t/(1 - t2)
    Show answer

    (3t - t3)/(1 - 3t2). tan 3θ = (3c2 s - s3)/(c3 - 3cs2). Divide top and bottom by c3.

  6. Using x = cos θ and cos 3θ = 4cos3 θ - 3cos θ, the equation 8x3 - 6x - 1 = 0 becomes cos 3θ = 1/2. Find the largest root, to 4 decimal places.
    Show answer

    0.9397. cos 3θ = 1/2 gives 3θ = π/3, 5π/3, 7π/3, so θ = π/9, 5π/9, 7π/9. The roots are cos(π/9), cos(5π/9), cos(7π/9); the largest is cos(π/9) = 0.9397.

  7. A student uses sin 5θ = sin θ(16sin4 θ - 20sin2 θ + 5) to solve sin 5θ = 0 for 0 ≤ θ < π. What must not be missed?
    • Nothing: only the quartic factor gives solutions
    • The solution θ = 0 from the factor sin θ = 0
    • Solutions with θ greater than π
    • The value sin θ = 5
    Show answer

    The solution θ = 0 from the factor sin θ = 0. sin θ = 0 gives θ = 0, and the quartic factor gives the other four solutions π/5, 2π/5, 3π/5, 4π/5. The June 2025 examiners' report notes that many students missed θ = 0.

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Last updated 8 November 2026. Spotted a mistake? Tell us.