sum-of-trig-in-ap

Sum of Trig Functions in Arithmetic Progression

Summary: §258–§260. Sines and cosines of angles in arithmetic progression form a recurrent series with scale of relation $2\cos b,-1$. Euler sums the infinite series $\sin a+\sin (a+b)+\sin (a+2b)+\cdots$ in closed form by evaluating the generating rational function at $z=1$, obtaining $(\sin a-\sin (a-b))/(2-2\cos b)$. He then derives the finite sum through $\sin (a+nb)$ by subtracting the tail of the infinite series from the head.

Sources: chapter14 (§258–§260)

Last updated: 2026-05-11

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Setup (§258)

Let the angles $a,a+b,a+2b,a+3b,\dots$ be an arithmetic progression with common difference $b$. Define

$s=\sin a+\sin (a+b)+\sin (a+2b)+\sin (a+3b)+\cdots$

From §129, the sequence of sines is recurrent with scale of relation $2\cos b,-1$, generating rational function denominator $1-2\cos b\cdot Z+Z^2$ (source: chapter14, §258).

The rational generating function

The general theory of recurrent series (Chapter 4 / Chapter 13) says that the generating function of the series $\{a_k\}$ is a rational function $A(z)/B(z)$ where $B(z)$ is read from the scale of relation. Here $B(z)=1-2\cos b\cdot z+z^2$.

For the numerator, Euler uses the §129 initial-value construction. The rational function for ${\sum }_{k=0}^{\infty }\sin (a+kb)z^k$ is:

$F(z)=\frac{\sin a+z(\sin (a+b)-2\cos b\cdot \sin a)}{1-2\cos b\cdot z+z^2}=\frac{\sin a+z(\sin (a+b)-2\sin a\cos b)}{1-2\cos b\cdot z+z^2}$

Using the addition formula $\sin (a+b)-2\sin a\cos b=\sin a\cos b+\cos a\sin b-2\sin a\cos b=-\sin (a-b)+\sin a-\sin a=-\sin (a-b)$… more directly, Euler writes (source: chapter14, §258):

$F(z)=\frac{\sin a+z(\sin (a+b)-2\cos b\cdot \sin a)}{1-2z\cos b+z^2}$

Closed-form sum at $z=1$ (§258)

Setting $z=1$:

$s=F(1)=\frac{\sin a+\sin (a+b)-2\cos b\cdot \sin a}{2-2\cos b}$

Now $\sin (a+b)-2\cos b\cdot \sin a=\sin a\cos b+\cos a\sin b-2\sin a\cos b=-\sin a\cos b+\cos a\sin b=-\sin (a-b)$. Thus:

$\boxed{s=\frac{\sin a-\sin (a-b)}{2(1-\cos b)}}$

(source: chapter14, §258). The denominator $2(1-\cos b)=4{\sin }^2(b/2)$ by the half-angle formula, so equivalently

$s=\frac{\sin a-\sin (a-b)}{4{\sin }^2(b/2)}=\frac{2\cos (a-b/2)\sin (b/2)}{4{\sin }^2(b/2)}=\frac{\cos (a-b/2)}{2\sin (b/2)}$

This is the standard closed form: the infinite sum of sines in arithmetic progression is $\frac{1}{2}\csc (b/2)\cos (a-b/2)$, provided $|z|<1$ (i.e., convergence requires the generating series to converge, which for $z=1$ requires the series to sum in the sense of the rational function’s analytic continuation).

Convergence note

Euler does not address convergence explicitly here. The formula $z=1$ is on the boundary of convergence of the generating function (since the denominator $1-2z\cos b+z^2$ has roots $z=e^{\pm ib}$ on the unit circle). The sum is valid in the sense of the Cesàro or Abel sum; a modern reader would recognize this as the real-part version of the geometric series $\sum e^{i(a+kb)}=e^{ia}/(1-e^{ib})$.

Cosine series

By an identical argument (replacing $\sin$ by $\cos$ throughout):

$\cos a+\cos (a+b)+\cos (a+2b)+\cdots =\frac{\cos a-\cos (a-b)}{2(1-\cos b)}=\frac{\cos (a-b/2)}{2\sin (b/2)}\cdot \frac{\cos (b/2)}{\cos (b/2)}$

More precisely:

$\frac{\cos a-\cos (a-b)}{2(1-\cos b)}=\frac{2\sin (a-b/2)\sin (b/2)}{4{\sin }^2(b/2)}=\frac{\sin (a-b/2)}{2\sin (b/2)}$

Finite sum (§259–§260)

For the sum through $n+1$ terms,

$s=\sin a+\sin (a+b)+\sin (a+2b)+\cdots +\sin (a+nb),$

Euler subtracts the tail of the infinite series from the head (source: chapter14, §259). The infinite head sums to $\cos (a-b/2)/(2\sin (b/2))$ and the tail starting at $\sin (a+(n+1)b)$ sums to $\cos (a+(n+1/2)b)/(2\sin (b/2))$. Their difference is

$s=\frac{\cos (a-\frac{1}{2}b)-\cos (a+(n+\frac{1}{2})b)}{2\sin (\frac{1}{2}b)}=\frac{\sin (a+\frac{1}{2}nb)\sin (\frac{1}{2}(n+1)b)}{\sin (\frac{1}{2}b)}$

after applying the sum-to-product identity. The cosine version (§260) is the same calculation with $\cos$ replacing $\sin$ in the head and a sign flip from $\cos f-\cos g=-2\sin \left(\frac{f+g}{2}\right)\sin \left(\frac{f-g}{2}\right)$, giving

$\cos a+\cos (a+b)+\cdots +\cos (a+nb)=\frac{\cos (a+\frac{1}{2}nb)\sin (\frac{1}{2}(n+1)b)}{\sin (\frac{1}{2}b)}.$

Both finite formulas are convergent in the ordinary sense for any real $b$ with $\sin (b/2)\ne 0$ — the infinite-sum step is just a bookkeeping device.

Connection to Chapter 13 and Chapter 9

This §258 application is the same recurrent-series machinery from §231–§233 applied to a trigonometric sequence. Euler is closing a circle: sines and cosines were first shown to be recurrent in §129, the recurrent-series sum formula was established in §231, and now in §258 the two threads meet to give trig sums in closed form. The same formula also relates to §171–§172, where Newton’s identities were applied to products of shifted sines — the §258 result is the additive rather than multiplicative side of the same structure.

Related pages

• recurrent-series
• trigonometric-recurrent-progression
• sum-of-recurrent-series
• circular-arc-series
• sine-and-cosine
• trigonometric-addition-formulas
• chapter-14-on-the-multiplication-and-division-of-angles
• chapter-13-on-recurrent-series
• chapter-8-on-transcendental-quantities-which-arise-from-the-circle