What is the Cartesian form of $(4, \frac{23 π}{12})$ $(4,(23pi )/12)$ ?

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What is the Cartesian form of $(4, \frac{23 π}{12})$ $(4,(23pi )/12)$ ?

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Answer 1 · 2018-03-31T10:12:19

$(x, y) \to (\sqrt{6} + \sqrt{2}, \sqrt{2} - \sqrt{6})$ $(x,y) -> (sqrt(6) +sqrt(2) , sqrt(2) - sqrt(6) )$

Explanation:

![https://www.google.co.uk/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&ved=2ahUKEwjZh4aOqJbaAhVE7BQKHSYGCTgQjRx6BAgAEAU&url=https%3A%2F%2Fmathinsight.org%2Fimage%2Fpolar_coordinates_cartesian&psig=AOvVaw1BLPuvER19bm1YM21HxgXQ&ust=1522577163343344]( useruploads.socratic.org )

As seen in the image, we see that the $x$ $x$ coordinate is $r cos θ$ $rcostheta$ , and the $y$ $y$ is $r sin θ$ $rsintheta$ obtained by simple trig

$r = 4, θ = \frac{23 π}{12}$ $r = 4 , theta = (23pi)/12$

$\Rightarrow x = 4 cos (\frac{23 π}{12}) = \sqrt{6} + \sqrt{2}$ $=> x = 4cos( (23pi)/12 ) = sqrt(6) + sqrt(2)$

$\Rightarrow y = 4 sin (\frac{23 π}{12}) = \sqrt{2} - \sqrt{6}$ $=> y = 4sin( (23pi) / 12 ) = sqrt(2) - sqrt(6)$

$(r, θ) \to (4, \frac{23 π}{12})$ $(r, theta ) -> (4, (23pi)/12 )$

$(x, y) \to (\sqrt{6} + \sqrt{2}, \sqrt{2} - \sqrt{6})$ $color(blue)((x,y) -> (sqrt(6) +sqrt(2) , sqrt(2) - sqrt(6) )$

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What is the Cartesian form of $(4, \frac{23 π}{12})$ $(4,(23pi )/12)$ ?

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