Question

Given that x=3sin​​θ-2 and y=3cos​​​θ+4.Show that (x+2)²+ (y-4)²=9​​​​​​

Answer 1

Answer:

First, plug-in x and y as 3sin​​θ-2 and 3cos​​​θ+4 into the equation, respectively:

$((3\sin (\theta)-2)+2)^2 + ((3\cos (\theta)+4)-4)^2 = 9$

Then, +2 and -2 cancel out and +4 and -4 cancel out as well, leaving you with:

$(3\sin(\theta))^2+(3\cos(\theta))^2=9$

We can factor out 3^2 = 9 from both equations:

$9(\sin(\theta)^2+cos(\theta)^2) = 9$

We know from a trigonometric identity that $\sin(\theta)^2+cos(\theta)^2 = 1$, meaning we can reduce the equation to:

$9(1) = 9$

$9=9$

And therefore, we have shown that (x+2)^2 + (y-4)^2 = 9, if x=3sin​​θ-2 and y=3cos​​​θ+4.

Hope this helped you.