Question

Given a parametric curve x = 2sin(t); y = sin(2t), where t takes all values in R.
(a) Find an implicit equation of this curve by eliminating t.
(b) Sketch this parametric curve in the xy-plane.

For part (a), I rearranged the x function to get t=sin^-1(x/2) and subbed that into the y function to get y = sin(2sin^-1(x/2)). However, when putting both the parametric and implicit equations into wolfram alpha, I got different equations with different graphs...
What did I do wrong, or how am I supposed to approach this question?

Answer 1

When you take the inverse of a function, be careful about the domain and range of the function. The function should be both one-one and onto within that domain that you fix.
When x=2sint, inverse t=sin^-1(x/2), exists only when \[t \in [2n \pi -\frac{ \pi }{ 2 }, 2n \pi+\frac{ \pi }{ 2 }]\].
Here, let say it is [-pi/2,pi/2]. Domain of t is restricted.
When you fix the domain of t, x is also restricted. t=sin^-1(x/2) only when t belongs to [-pi/2,pi/2] and x belongs to [-2,2].
Now, with that domain, you are allowed to plug the value of t in y=sin2t.

When you write the equation
y=sin(2sin^-1(x/2))
in wolfram alpha, it understands to plot the graph of the equation for all real x, which is not the case (since the substitution is restricted to only for x belongs to [-2,2] ).

Answer 2

If you command wolfram alpha to plot the graph of y=sin(sin^-1(x/2)) for \[x \in [-2,2]\]
then it may show you the graph as that of the parametric equation's.