Answer
a)
$y= cos(sin^{-1}x)$
or
$x^{2}+y^{2}=1$,
$y \geq 0$
b)
(Graph)
As θ increases, x increases and y increases to 1, then decreases to 0.
Work Step by Step
a)
$x= sin(\frac{1}{2}θ)$
$y= cos(\frac{1}{2}θ)$
$-π \leq θ \leq π$
$x= sin(\frac{1}{2}θ)$
$2sin^{-1}x=θ$
$y= cos(\frac{1}{2}θ)$
$y= cos(\frac{1}{2}2sin^{-1}x)$
$y= cos(sin^{-1}x)$
which is the same as,
$x^{2}+y^{2}=1$,
$y \geq 0$
b)
When,
θ=-π x=-1 y=0
θ=-π/2 $x=-(\sqrt 2)/2$ $y=(\sqrt 2)/2$
θ=0 x=0 y=1
θ=π/2 $x=(\sqrt 2)/2$ $y=(\sqrt 2)/2$
θ=π x=1 y=0
From this we can infer that as θ increases, x increases and y increases to 1, then decreases to 0.
