Answer to Question #255680 in Analytic Geometry for Samson

Question

Given that x=2cosθ and y=2sinθ evaluate x2+y2

Answer 1

Solution:

$x=2\cosθ , y=2\sinθ$

On squaring both sides of each equation:

$x^2=2^2\cos^2θ ,y^2=2^2\sin^2θ \\ \Rightarrow x^2=4\cos^2θ ,y^2=4\sin^2θ$

Now,

$x^2+y^2=4\cos^2θ +4\sin^2θ \\=4(\cos^2θ +\sin^2θ) \\=4(1) \ \ [\because \cos^2A +\sin^2A=1] \ \ \\=4$