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.

 

[blue_raver22]

These are special cases because they represent the sums of cosine and sine functions with a specific number of terms, rather than an infinite number of terms. They are also familiar because they are commonly used in trigonometry and have specific values that are often memorized, such as $\sin\frac{\pi}{2} = 1$ and $\cos\frac{\pi}{2} = 0$.