MHB Special cases for sine and cosine sum

[Dustinsfl]

State the special cases of the above two formulas for $n = 0, 1,$ and $2$.
These should be familiar formulas.

I don't see what is so special and familiar about when n = 2 or for cosine n = 1.When $n = 0$, we have
$$
\sum\limits_{k = 0}^0\cos k\theta = \frac{\sin\left(\frac{\theta}{2}\right)}{\sin\frac{\theta}{2}} = 1
$$
and
$$
\sum\limits_{k = 0}^0\sin k\theta = \frac{\sin\left(\frac{ \theta}{2}\right)}{\sin\frac{ \theta}{2}}\times 0 = 0.
$$
When $n = 1$, we have
$$
\sum\limits_{k = 0}^1\cos k\theta = \frac{\sin\theta}{\sin\frac{ \theta}{2}}\cos\frac{\theta}{2} = \sin\theta\cot\frac{\theta}{2}
$$
and
$$
\sum\limits_{k = 0}^1\sin k\theta = \frac{\sin\theta}{\sin\frac{\theta}{2}}\sin\frac{ \theta}{2} = \sin\theta.
$$
When $n = 2$, we have
$$
\sum\limits_{k = 0}^2\cos k\theta = \frac{\sin\left(\frac{3}{2}\theta\right)}{\sin \frac{ \theta}{2}}\cos\theta
$$
and
$$
\sum\limits_{k = 0}^2\sin k\theta = \frac{\sin\left(\frac{3}{2}\theta\right)}{\sin \frac{ \theta}{2}}\sin\theta.
$$

 

[Sudharaka]

State the special cases of the above two formulas for $n = 0, 1,$ and $2$.
These should be familiar formulas.

I don't see what is so special and familiar about when n = 2 or for cosine n = 1.When $n = 0$, we have
$$
\sum\limits_{k = 0}^0\cos k\theta = \frac{\sin\left(\frac{\theta}{2}\right)}{\sin\frac{\theta}{2}} = 1
$$
and
$$
\sum\limits_{k = 0}^0\sin k\theta = \frac{\sin\left(\frac{ \theta}{2}\right)}{\sin\frac{ \theta}{2}}\times 0 = 0.
$$
When $n = 1$, we have
$$
\sum\limits_{k = 0}^1\cos k\theta = \frac{\sin\theta}{\sin\frac{ \theta}{2}}\cos\frac{\theta}{2} = \sin\theta\cot\frac{\theta}{2}
$$
and
$$
\sum\limits_{k = 0}^1\sin k\theta = \frac{\sin\theta}{\sin\frac{\theta}{2}}\sin\frac{ \theta}{2} = \sin\theta.
$$
When $n = 2$, we have
$$
\sum\limits_{k = 0}^2\cos k\theta = \frac{\sin\left(\frac{3}{2}\theta\right)}{\sin \frac{ \theta}{2}}\cos\theta
$$
and
$$
\sum\limits_{k = 0}^2\sin k\theta = \frac{\sin\left(\frac{3}{2}\theta\right)}{\sin \frac{ \theta}{2}}\sin\theta.
$$

Hi dwsmith, :)

I don't understand what you meant by the "above two formulas". Is there anything missing here? :)

Kind Regards,
Sudharaka.

 

[Dustinsfl]

It was the formulas for cosine and sine.

$\sum\limits_{k = 0}^n\sin k\theta = \frac{\sin\left(\frac{n + 1}{2}\theta\right)}{\sin\frac{\theta}{2}}\sin\frac{n}{2}\theta$

and

$\sum\limits_{k = 0}^n\cos k\theta = \frac{\sin\left(\frac{n + 1}{2}\theta\right)}{\sin\frac{\theta}{2}}\cos\frac{n}{2}\theta$

 

[Sudharaka]

It was the formulas for cosine and sine.

$\sum\limits_{k = 0}^n\sin k\theta = \frac{\sin\left(\frac{n + 1}{2}\theta\right)}{\sin\frac{\theta}{2}}\sin\frac{n}{2}\theta$

and

$\sum\limits_{k = 0}^n\cos k\theta = \frac{\sin\left(\frac{n + 1}{2}\theta\right)}{\sin\frac{\theta}{2}}\cos\frac{n}{2}\theta$

For \(n=1\) in the cosine summation you can express the result using only a cosine function as,

\[\sum\limits_{k = 0}^1\cos k\theta = \frac{\sin\theta}{\sin\frac{ \theta}{2}}\cos\frac{\theta}{2} = 2\cos^{2}\frac{\theta}{2}\]

Similarly for \(n=2\) in the sine summation,

\[\sum\limits_{k = 0}^2\sin k\theta = \frac{\sin\left(\frac{3}{2}\theta\right)}{\sin \frac{ \theta}{2}}\sin\theta=2\sin\left(\frac{3}{2}\theta\right)\cos \frac{ \theta}{2}\]

Apart from these minor simplifications, I don't see anything further that could be done to the results that you have obtained.

Kind Regards,
Sudharaka.