Trigonometric Sums

Graph for reference


Like most of my notes, this one starts with curiosity. I saw this problem and went to Desmos to solve it. Then, I got a bit curious about trigonometric functions and what that pattern would result in \((y=\sin(x)+\sin(2x)+\sin(3x)+\sin(4x))\); the resulting graph looked a bit like the graph of a heart rate monitor in hospitals. I used \(\sum\) to make it easier, giving me the function \[f(x)= \displaystyle\sum_{n=1}^a \sin(n\times x)\] Since Desmos can't handle infinity, I used the next best thing: \(a=99\). This gave me a graph that was a bit strange: it was so squished together that it looked like a solid shape instead of a continuous line (imagine the graph of the derivative...); the shape looked suspiciously like the graph of the cotangent function. After messing around a bit, I found the the two graphs that contained this strange function/shape: \[y=\frac{\cot\left(\frac{x}{4}\right)}{2}\]

\[y=\frac{\cot\left(\frac{x}{4}+\frac{\pi}{2}\right)}{2}\] I was baffled by the connection between the two functions, sine and cotangent. However, this is only half of the story.


I went through the same process with the cosine function, giving me \[g(x)= \displaystyle\sum_{n=1}^b \cos(n\times x)\] \(b=99\) This time, the graph/shape looked suspiciously like the secant/cosecant function. Once again, after some messing and mapping, I managed to get the two defining functions of this new cosine sum: \[y=\frac{\csc\left(\frac{x}{2}\right)-1}{2}\]

\[y=\frac{\csc\left(\frac{x}{2}\right)+1}{-2}\]


My question to the brilliant community of Brilliant: How on Earth are these functions related? The seemed to pop up out of nowhere.

Note by Blan Morrison
3 months ago

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Graph for reference

First,

\(\sin {x}+\sin {2x}+\sin {3x}+\cdots +\sin {nx}\\ =\dfrac { \sin { \frac { x }{ 2 } } (\sin {x}+\sin {2x}+\sin {3x}+\cdots +\sin {nx}) }{ \sin { \frac { x }{ 2 } } } \\ =\frac { \sin { \frac { x }{ 2 } } \sin { x } +\sin { \frac { x }{ 2 } } \sin { 2x }+\sin { \frac { x }{ 2 } } \sin { 3x }+\cdots +\sin { \frac { x }{ 2 } } \sin { nx } } { \sin { \frac { x }{ 2 } } }\)

Using the Product-to-Sum Formula:

\(=\frac { \cos { (\frac { x }{ 2 } -x) } -\cos { (\frac { x }{ 2 } +x) } +\cos { (\frac { x }{ 2 } -2x) } -\cos { (\frac { x }{ 2 } +2x) } \cdots +\cos { (\frac { x }{ 2 } -nx) } -\cos { (\frac { x }{ 2 } +nx) } }{ 2\sin { \frac { x }{ 2 } } } \\ =\frac { \cos { (-\frac { x }{ 2 } ) } -\cos { (\frac { 3x }{ 2 } ) } +\cos { (-\frac { 3x }{ 2 } ) } -\cos { (\frac { 5x }{ 2 } ) } \cdots +\cos { (\frac { x }{ 2 } -nx) } -\cos { (\frac { x }{ 2 } +nx) } }{ 2\sin { \frac { x }{ 2 } } } \)

Since \(cos{(a)}=cos{(-a)}\), we can cancel the terms:

\(=\frac { \cos { (\frac { x }{ 2 } ) } -\cos { (\frac { x }{ 2 } +nx) } }{ 2\sin { \frac { x }{ 2 } } } \)

Now, as \(n\) approaches infinity, \(\cos { (\frac { x }{ 2 } +nx) } \) fluctuates continuously and infinitely frequently between \(-1\) and \(1\), and the whole expression fluctuates continuously and infinitely frequently between

\(\frac { \cos { (\frac { x }{ 2 } ) } +1 }{ 2\sin { \frac { x }{ 2 } } } \) and \(\frac { \cos { (\frac { x }{ 2 } ) } -1 }{ 2\sin { \frac { x }{ 2 } } } \)

near a point.

Using the Double-Angle Formula:

\(\frac { \cos { (\frac { x }{ 2 } ) } +1 }{ 2\sin { \frac { x }{ 2 } } } \\ =\frac { 2\cos ^{ 2 }{ \frac { x }{ 4 } } -1+1 }{ 2\cdot 2\sin { \frac { x }{ 4 } } \cos { \frac { x }{ 4 } } } \\ =\frac { \cos ^{ 2 }{ \frac { x }{ 4 } } }{ 2\sin { \frac { x }{ 4 } } \cos { \frac { x }{ 4 } } } \\ =\frac { \cot { \frac { x }{ 4 } } }{ 2 } \)

Similarly:

\(\frac { \cos { (\frac { x }{ 2 } ) } -1 }{ 2\sin { \frac { x }{ 2 } } } \\ =\frac { 1-2\sin ^{ 2 }{ \frac { x }{ 4 } } -1 }{ 2\cdot 2\sin { \frac { x }{ 4 } } \cos { \frac { x }{ 4 } } } \\ =\frac { -\sin ^{ 2 }{ \frac { x }{ 4 } } }{ 2\sin { \frac { x }{ 4 } } \cos { \frac { x }{ 4 } } } \\ =\frac { -\tan { \frac { x }{ 4 } } }{ 2 } =\frac { \cot { (\frac { x }{ 4 } +\frac { \pi } { 2 } ) } }{ 2 } \)

So the expression fluctuates frequently between \(\frac { \cot { \frac { x }{ 4 } } }{ 2 }\) and \(\frac { \cot { (\frac { x }{ 4 } +\frac { \pi } { 2 } ) } }{ 2 } \).

In other words, \(\displaystyle\sum_{k=1}^n \sin(k\times x)\) looks "sandwiched" between \(\frac { \cot { \frac { x }{ 4 } } }{ 2 }\) and \(\frac { \cot { (\frac { x }{ 4 } +\frac { \pi } { 2 } ) } }{ 2 } \).

In the same way, we can also find the functions that "sandwich" \(\displaystyle\sum_{k=1}^n \cos(k\times x)\).

Icreiuhe Zhu - 2 months, 3 weeks ago

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Looks tough- glad I’m in Geometry

Ella Collard - 2 months, 3 weeks ago

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