Can You Mathematically Transform Sin(x) to Y=X?

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Homework Help Overview

The discussion revolves around the mathematical transformation of the sine function, specifically how to manipulate the equation of a sine curve (y = sin(x)) to align with a 45-degree line (y = x). Participants are exploring the implications of coordinate transformations and the effects of rotation on the function's representation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of rotation matrices to transform the sine function and question the correct application of angles in the transformation process. There are attempts to derive new equations based on these transformations.

Discussion Status

The conversation includes differing interpretations of the rotation process, with some participants suggesting corrections to the initial approach. There is an ongoing exploration of how to rearrange the resulting equations to express y' in terms of x'.

Contextual Notes

Participants are navigating potential confusion regarding the rotation of coordinate axes versus the function itself, which may impact the transformation results. The discussion reflects a need for clarity on these mathematical concepts.

Tido611
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is there a way to turn a sin curve (sin(x)) from a horizontal graph to a 45 degree, y=x graph mathematically
 
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Yes.

Your initial equation is y=sin x

Put

[tex]x = x' \cos \theta - y' \sin \theta[/tex]
[tex]y = x' \sin \theta + y' \cos \theta[/tex]

where, in your case [itex]\theta = 45[/itex] degrees.

Then, your new equation is:

[tex]\frac{1}{\sqrt{2}}(x' + y') = \sin \left[ \frac{1}{\sqrt{2}}(x' - y') \right][/tex]

Now, the only problem you have is to rearrange this so as to write y' in terms of x'.
 
You rotated over an angle of -45° because your minus-sign was placed wrongly. You have to be careful whether it's the coordinate axis which are being rotated or the function itself. In this case, the rotation matrix is:

[tex]\left( {\begin{array}{*{20}c}<br /> {\cos \theta } & {\sin \theta } \\<br /> { - \sin \theta } & {\cos \theta } \\<br /> \end{array}} \right)[/tex]

giving the new equation:

[tex]\frac{1}{\sqrt{2}}(y' - x') = \sin \left[ \frac{1}{\sqrt{2}}(x' + y') \right][/tex]
 
Fair enough.
 

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