A trigonometry question relating to parametric curves?

Question

Eliminate the parameter to find a Cartesian equation of the curve:

x=sine of theta, y=cosine of theta, 0 (>or=) theta (>or=) pi

The solution I found was:

y=cos(sin^-1(x))

however, the solution in the book is:

x^2 + y^2 = 1

I don't understand why the answer isn't in "y=" form like every other problem in this section until now. Also, I don't know how to break down the equation past the inverse trig functions. I appreciate any help.

Answer 1

Note: there is a typo in you problem statement though: 0 >= theta >= pi has no solutions.  I assume this means 0 <= theta <= pi

x^2 + y^2 = 1 is correct, and is equivalent to your solution y = cos(sin^-1(x)).

However, between 0 and pi, this is a function:

y = sqrt(1 - x^2)

and is the correct solution to the original problem, expressed as y = f(x)

Both of these give the equation of a circle of radius 1 in the plane, centered at (0,0).  However, given the limits 0 <= theta <= pi, one has the upper-half circle.

The reason the circle is not _properly_ expressed in the form y=f(x), is that the circle is not the graph of a function (it does not pass the "vertical line test" -- for each value of x there are 2 values of y).

y=cos(sin^-1(x)) is ok, but it is important to note that sin^-1 (inverse sine) is not a single-valued function.  There are 2 different values of sin^-1(x) (between 0 and 2pi), for and x such that -1 < x < 1, but the fact is that cos() of these 2 values is the same.

The trig identity sin^2+cos^2=1 is the key to understanding why these various expression are the same.