Sine and Cosine Series Sums
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Direct answer
Multiply the whole sum by sin(θ/2) and every term pairs up into cosines that cancel in a chain — one move produces both standard closed forms. For θ not a multiple of 2π, Σ(r=1)^n sin rθ = [sin(nθ/2)·sin((n+1)θ/2)]/sin(θ/2) and Σ(r=1)^n cos rθ = [sin(nθ/2)·cos((n+1)θ/2)]/sin(θ/2). The special case θ = π/n is worth memorising separately: Σ_(r=1)^(n−1) sin(rπ/n) = cot(π/2n), the most quoted numerical-answer identity in this chapter. Derivations beat memorisation here because the exam asks for them.
What you must remember
- The two closed forms: Σ_(r=1)^n sin rθ = sin(nθ/2)sin((n+1)θ/2)/sin(θ/2); Σ cos rθ = sin(nθ/2)cos((n+1)θ/2)/sin(θ/2); both valid when sin(θ/2) ≠ 0.
- Full-range form: 1 + Σ_(r=1)^n cos rθ = [sin((2n+1)θ/2)]/[2 sin(θ/2)] — the sum starting from r = 0 has this slightly different shape.
- Half-angle multiples: Σ_(r=1)^n sin(2r − 1)θ = sin²(nθ)/sin θ; check n = 1 and n = 2 to confirm the formula before quoting it.
- Special case: Σ_(r=1)^(n−1) sin(rπ/n) = cot(π/2n), obtained by substituting θ = π/n and using sin(nθ/2) = 1.
- Complex route: Σ e^(irθ) is a geometric series with ratio e^(iθ); taking real and imaginary parts reproduces both sums — the De Moivre derivation Advanced prefers.
- Weighted sums: differentiating Σ cos rθ with respect to θ gives Σ r sin rθ = −d/dθ(Σ cos rθ), extending the toolkit to r-weighted series.
- Guard cases: θ = 0 gives sum 0 for sines and n for cosines; θ = π gives 0 and the alternating ±1 pattern — handle these before dividing by sin(θ/2).
Deriving, then using, the sum
Derive the sine sum. Let S = Σ_(r=1)^n sin rθ and multiply by 2 sin(θ/2): each product 2 sin(θ/2)sin rθ becomes cos(rθ − θ/2) − cos(rθ + θ/2) by the identity 2 sin A sin B = cos(A − B) − cos(A + B). Written out, the chain is [cos(θ/2) − cos(3θ/2)] + [cos(3θ/2) − cos(5θ/2)] + ... + [cos((2n−1)θ/2) − cos((2n+1)θ/2)] — every interior cosine appears twice with opposite signs, leaving 2 sin(θ/2)·S = cos(θ/2) − cos((2n+1)θ/2). Convert the right side by cos C − cos D = 2 sin((C+D)/2)sin((D−C)/2) and divide: S = sin(nθ/2)sin((n+1)θ/2)/sin(θ/2). The cosine sum follows the same way with 2 cos(θ/2) as the multiplier.
Apply it: evaluate Σ_(r=1)^10 sin(rπ/11). Substituting θ = π/n with n = 11 makes sin(nθ/2) = sin(π/2) = 1 and sin((n+1)θ/2) = sin(π/2 + π/22) = cos(π/22), while the denominator is sin(π/22); the quotient is cot(π/22). One substitution, one clean answer — this is the numerical-value form Main favours.
How the exam uses it
Main asks direct substitution with θ = π/6, π/4 or the π/n family, and the cot(π/2n) special case appears repeatedly in numerical questions. Advanced derives: prove the formula, or sum Σ (cos rθ + i sin rθ) as a geometric progression and extract the real part — the complex route is often shorter than the real one. Traps: dividing by sin(θ/2) without checking that θ is a multiple of 2π (the vanishing denominator is precisely the excluded case); swapping the two sine factors in the cosine formula, where the endpoint angle (n+1)θ/2 multiplies a cosine; and degree–radian confusion — these questions are set in radians, and a 180° left in degrees wrecks the substitution.
Frequently asked questions
What is Σ_(r=1)^n sin rθ when θ is a multiple of 2π?
Each term vanishes, so the sum is 0 — the same case where sin(θ/2) in the denominator of the closed form also vanishes.
How do I remember which formula carries which factor?
Sine series: product of two sines over sin(θ/2); cosine series: sin(nθ/2) times cos((n+1)θ/2) over sin(θ/2) — the endpoint angle (n+1)θ/2 marks the difference.
What does Σ_(r=1)^(n−1) sin(rπ/n) equal?
cot(π/2n) — the standard special case behind most numerical-answer questions in this chapter.
How does the complex-numbers route work?
Σ e^(irθ) is geometric with ratio e^(iθ); its real part is the cosine sum and its imaginary part the sine sum, no telescoping needed.
How do I sum r·sin rθ?
Differentiate the cosine series termwise: d/dθ of Σ cos rθ equals −Σ r sin rθ, so the weighted sum follows from the standard closed form.