What is the Cartesian form of $(-16,(-23pi)/4))$ ?

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Answer 1 · 2016-11-15T16:19:38

(- $16/sqrt2, -16/sqrt2$ )

The given point is r= -16 and $theta=(-23pi)/4$

The cartesian coordinates would be x= $rcostheta$ and y = $r sin theta$

Accordingly, x= $-16 cos ((-23pi)/4)$ and y= $-16sin((-23pi)/4)$

Hence x= -16 cos $-(6pi-pi/4)$ and y=-16sin $-(6pi-pi/4)$

x= -16cos $(6pi-pi/4)$ and y= 16sin $(6pi-pi/4)$

x=-16 cos $(-pi/4)$ and y= 16 sin $-(pi/4)$

x= -16 cos $pi/4$ and y= -16 $sin pi/4$

x=- $16/sqrt2$ and y= - $16/sqrt2$

What is the Cartesian form of (-16,(-23pi)/4))(-16,(-23pi)/4))?