What is the Cartesian form of $( 7 , (23pi)/3 )$ ?

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Answer 1 · 2016-02-05T04:56:45

$(7/2, -(7sqrt(3))/2)$

To find cartesian form of $(r,theta)$ we use the formula

$x=rcos(theta)$ and $y=rsin(theta)$

We are given the $(7, (23pi)/3)$

$r=7$ and $theta=(23pi)/3$

Let us simplify this $(23pi)/3$ into something which is easier to handle.
$(23pi)/3 + pi/3 = (24pi)/3$
$(23pi)/3 + pi/3 = 8pi$
$(23pi)/3 = 8pi-pi/3$

Also note $cos(2npi-theta) = cos(theta)$
and $sin(2npi-theta) =-sin(theta)$

$x=rcos(theta)$
$x=7cos(8pi-pi/3)$
$x=7cos(pi/3)$
$x=7(1/2)$
$x=7/2$

$y=rsin(theta)$
$y=7sin(8pi-pi/3)$
$y=7(-sin(pi/3))$
$y=-7(sqrt(3)/2)$
$y=-(7sqrt(3))/2$

$(7/2, -(7sqrt(3))/2)$

What is the Cartesian form of ( 7 , (23pi)/3 ) ( 7 , (23pi)/3 ) ?