What is the Cartesian form of $(-13,(9pi)/8)$ ?

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Answer 1 · 2018-01-10T18:21:44

$<<-13,(9pi)/8>> =((13 sqrt(2 + sqrt(2)))/2,(13 sqrt(2 - sqrt(2)))/2)$

We use the following formulae to convert the polar point $<< r, varphi >>$ to a Cartesian point $x,y$ :

$x=rcosvarphi$

$y=rsinvarphi$

$therefore x= -13cos((9pi)/8)=(13 sqrt(2 + sqrt(2)))/2$

$y=-13sin((9pi)/8)=(13 sqrt(2 - sqrt(2)))/2$

So, $<<-13,(9pi)/8>> =((13 sqrt(2 + sqrt(2)))/2,(13 sqrt(2 - sqrt(2)))/2)$

What is the Cartesian form of (-13,(9pi)/8)(-13,(9pi)/8)?