Rearranging equation of a circle

[Georgepowell]

[solved] rearranging equation of a circle

the equation of a circle with radius 1 around the origin is normally given by:

x²+y²=1

or parametrically:

x= sin(t)
y= cos(t)

If you rearrange the parametric equations to get rid of t you get:

arcsin(x)=arccos(y)

which should also be the equation of the same circle.

So my question is, how do you rearrange arcsin(x)=arccos(y) to get x²+y²=1 ?

 

[derek e]

x = sin(arccos(y)) = \sqrt{1 - y^2},

because

<br /> \begin{equation*}\begin{split}<br /> arccos(y) &amp;= \theta \\<br /> y &amp;= cos(\theta) \\<br /> \sqrt{1-y^2} &amp;= sin(\theta) \\<br /> \end{split}\end{equation*}<br />

 

[Georgepowell]

Thanks, that was more simple than I thought.

 

[ThirstyDog]

To go from parametric form
x = sin(t) \mbox{ and } y =cos(t),
to the form
x^{2} + y^{2} = 1
you do not use the rearrangement as you have given. What you do is consider
x^{2} + y^{2} = sin^{2}(t) + cos^{2}(t) = 1.
The last step is using a trigonometric identity.

The rearrangement you have given breaks down as sin and cos are not 'one-to-one' functions thus arcsin and arccos are not the inverse functions. e.g
sin(0) = sin(\pi) = 0 \mbox{ but what about } arcsin(0) = 0 \mbox{ or } \pi.
For arcsin to be well defined it can only be one. This is where trouble can occur.

P.S. My example is not as general as it could have been as sin(n\pi) = 0 for all integers n.

 

[Georgepowell]

The rearrangement you have given breaks down as sin and cos are not 'one-to-one' functions thus arcsin and arccos are not the inverse functions. e.g
sin(0) = sin(\pi) = 0 \mbox{ but what about } arcsin(0) = 0 \mbox{ or } \pi.
For arcsin to be well defined it can only be one. This is where trouble can occur.

I don't understand the problem is with using a one-to-many 'function'. (note the inverted commas)

You would NEED to use one to get an equation of a circle because there are two values for anyone x or y value. Why is using sqrt(1-x²) any better because this can also be two values?

I'm not saying your wrong, just that I don't understand.

 

[derek e]

One could be careful and restrict the domain of sin and cos, such that the functions are 1-1 and hence invertible. It sounds like a bit of argument and specification/patchwork, but not unimaginable.

 

[ThirstyDog]

The point I was making was that although through the rearranging you can get the right answer in the end in the middle you made statement which were not entirely true. This is does not matter significantly if you are just looking for a sketch of the proof or aren't required to be completely rigorous.

One technical error that was made was using
x = sin(t) \Rightarrow t = arcsin(x),
this is not actually true. As derek e said this can be fixed by restricting domains etc but this might become tedious.