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.
![](https://gradesaver.s3.amazonaws.com/uploads/solution/ca857b42-81e6-4818-8f62-da757e9ff2d4/result_image/1513376153.png?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20250215%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20250215T144814Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=3a2cefe33e3c82d6dce509d1fff34dc1c78323d21aa7fa74a218d955fc9c91fe)
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.