Sin curve

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

James R

Science Advisor
Homework Helper
Gold Member
601
15
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'.
 

TD

Homework Helper
1,020
0
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}
{\cos \theta } & {\sin \theta } \\
{ - \sin \theta } & {\cos \theta } \\
\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]
 

James R

Science Advisor
Homework Helper
Gold Member
601
15
Fair enough.
 

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