Answer:
First, plug-in x and y as 3sinθ-2 and 3cosθ+4 into the equation, respectively:
[tex]((3\sin (\theta)-2)+2)^2 + ((3\cos (\theta)+4)-4)^2 = 9[/tex]
Then, +2 and -2 cancel out and +4 and -4 cancel out as well, leaving you with:
[tex](3\sin(\theta))^2+(3\cos(\theta))^2=9[/tex]
We can factor out 3^2 = 9 from both equations:
[tex]9(\sin(\theta)^2+cos(\theta)^2) = 9[/tex]
We know from a trigonometric identity that [tex]\sin(\theta)^2+cos(\theta)^2 = 1[/tex], meaning we can reduce the equation to:
[tex]9(1) = 9[/tex]
[tex]9=9[/tex]
And therefore, we have shown that (x+2)^2 + (y-4)^2 = 9, if x=3sinθ-2 and y=3cosθ+4.
Hope this helped you.
