Proving a formula for $\sum ^{n}_{k=1}\sin kt$

Question

Let $t$ be a real number such that $\sin \dfrac{t}{2}\neq 0$. Show that

$$\sum ^{n}_{k=1}\sin kt=\dfrac{\cos\dfrac{t}{2}-\cos \left( n+\dfrac{1}{2}\right) t}{2\sin \dfrac{t}{2}}$$ for every positive $n$. What method should I use?

Any explanation is much appreciated.

Hint: Multiply and divide by $\sin t/2$ and then try to make a telescopic sum. — Koro, Mar 27 '21 at 08:53
See the question "How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?". The limits of summation are different, and the answer there differs in form from the answer you seek, but the approaches are valid. — Blue, Mar 27 '21 at 09:24

score 3 · Answer 1 · edited Mar 27 '21 at 09:31

Just an application of the formula Product-to-Sum

$$ 2 \sin \theta \sin \varphi=\cos (\theta-\varphi)-\cos (\theta+\varphi) $$

$$ \begin{aligned} & \sum_{k=1}^{n} \sin k t \\ =& \sum_{k=1}^{n} \frac{2 \sin \frac{t}{2} \cdot \sin k t}{2 \sin \frac{t}{2}} \\ =& \frac{\left.\sum_{k=1}^{n} [\cos \left(k t-\frac{t}{2}\right)-\cos \left(k t+\frac{t}{2}\right)\right]}{2 \sin \frac{t}{2}} \\ =& \frac{\cos \frac{t}{2}-\cos \left(n+\frac{1}{2}\right) t}{2 \sin \frac{t}{2}} \end{aligned} $$

score 3 · Accepted Answer · answered Mar 27 '21 at 09:05

3

Hint: Plug $z=\cos t+i \sin t$ in $$1+z+z^2+...+z^n=\frac{z^{n+1}-1}{z-1}$$

and equate real and imaginary parts!

Note: In this method, we also get the formula for $\sum_k \cos kt$

answered Mar 27 '21 at 09:05

Chinnapparaj R

11,589

Proving a formula for $\sum ^{n}_{k=1}\sin kt$

2 Answers2

$$ 2 \sin \theta \sin \varphi=\cos (\theta-\varphi)-\cos (\theta+\varphi) $$