Cosine and sine function midpoints

Mar 2017
356
3
Massachusetts
Hi,

I hope someone can help. I'm trying to understand why a function must be cosine if x = 1 for the midline point and is a sine function when x = 0 for the midline point? I'm currently taking advanced functions, so please give a response at that knowledge-level. Thank you!

- Olivia
 

romsek

MHF Helper
Nov 2013
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California
That's just how they are defined.

$\cos(0)=1$

$\sin(0)=0$
 
Mar 2017
356
3
Massachusetts
but to my understanding of functions whatever is inside the function is the x-value and the output is the y-value. So in both cases it appears that x = 0, just cosine outputs 1 and sine outputs 0. This is where my confusion resides.
 

skeeter

MHF Helper
Jun 2008
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North Texas
I'm trying to understand why a function must be cosine if x = 1 for the midline point and is a sine function when x = 0 for the midline point?
What is a midline point anyway? Can you be more context specific about how cosine & sine relate to this midline point?
 

romsek

MHF Helper
Nov 2013
6,665
3,002
California
What is a midline point anyway? Can you be more context specific about how cosine & sine relate to this midline point?
I just assumed it to be $x=0$ (Tongueout)
 
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Plato

MHF Helper
Aug 2006
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What is a midline point anyway? Can you be more context specific about how cosine & sine relate to this midline point?
Thank you for that question. The OP is about midpoints. Midlines are not defined is metric spaces, example $\mathbb{R}^2$. qie
Midpoints are defined in terms of two other points. So a question about midpoints on a sine curve is meaningless.

I am prepared to argue that $M_s: \left(\frac{\pi}{2},1\right)$ is the midpoint of the graph $\left\{(x,\sin(x): 0\le x\le\pi\right\}$
Clearly, $M_s$ is both equally distant from $(0,0)~\&~(\pi, 0)$ in an ordinary sense as well as the arc-lengths.
However, as this question is posted in basic algebra arc-length is not applicable.

Likewise, $M_c: \left(\frac{\pi}{2},0\right)$ is the midpoint of the graph $\left\{(x,\cos(x): 0\le x\le\pi\right\}$.