Question

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

Answer 1

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

Explanation:

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)`