Question

What is the Cartesian form of `(36,(-pi)/16))`?

Answer 1

`(x, y) = (rcostheta, rsintheta)`

`= (18sqrt(2+sqrt(2+sqrt(2))), -18sqrt(2-sqrt(2+sqrt(2))))`

Explanation:

To convert from polar to rectangular coordinates use:

`{ (x = rcos theta), (y = rsin theta) :}`

First let's find algebraic expressions for `sin(pi/16)` and `cos(pi/16)`

Use:

`sin(pi/4) = cos(pi/4) = sqrt(2)/2`

`sin^2 theta + cos^2 theta = 1`

`cos(2theta) = cos^2 theta - sin^2 theta = 2cos^2 theta - 1 = 1 - 2sin^2 theta`

`sin(-theta) = -sin(theta)`

`cos(-theta) = cos(theta)`

From `cos2theta = 2cos^2 theta - 1` we can deduce:

`cos theta = +-sqrt((cos 2theta + 1)/2) = +-1/2sqrt(2+2cos2theta)`

From `cos2theta = 1 - 2sin^2 theta` we can deduce:

`sin theta = +-sqrt((1-cos 2theta)/2) = +-1/2sqrt(2-2cos2theta)`

Since `pi/8` and `pi/16` are both in Q1, we know that `sin` and `cos` are both positive, so we can use the positive square root each time to find:

`sin (pi/16) = 1/2sqrt(2-2cos(pi/8))`

`color(white)(sin (pi/16)) = 1/2sqrt(2-sqrt(2+2cos(pi/4)))`

`color(white)(sin (pi/16)) = 1/2sqrt(2-sqrt(2+sqrt(2)))`

`cos (pi/16) = 1/2sqrt(2+2cos(pi/8))`

`color(white)(cos(pi/16)) = 1/2sqrt(2+sqrt(2+sqrt(2)))`

So:

`sin(-pi/16) = -sin(pi/16) = -1/2sqrt(2-sqrt(2+sqrt(2)))`

`cos(-pi/16) = cos(pi/16) = 1/2sqrt(2+sqrt(2+sqrt(2)))`

So with `r=36` and `theta = -pi/16` we find:

`(x, y) = (rcostheta, rsintheta)`

`color(white)((x, y)) = (36(1/2sqrt(2+sqrt(2+sqrt(2)))), 36(-1/2sqrt(2-sqrt(2+sqrt(2)))))`

`color(white)((x, y)) = (18sqrt(2+sqrt(2+sqrt(2))), -18sqrt(2-sqrt(2+sqrt(2))))`