What is the Cartesian form of (45,(-13pi)/8)?

1 Answer
Sep 22, 2017

(45 cos(-(13pi)/8), 45 sin(-(13pi)/8))=(45/2sqrt(2-sqrt(2)), 45/2sqrt(2+sqrt(2)))

Explanation:

The cartesian form of (r, theta) is (rcos theta, rsin theta)

Let's find cos(-(13pi)/8) and sin(-(13pi)/8) first.

Note that -(13pi)/4 is coterminal with (3pi)/4 in Q2.

Hence:

{ (cos(-(13pi)/4) = -sqrt(2)/2), (sin(-(13pi)/4) = sqrt(2)/2) :}

Then note that:

cos 2theta = 2cos^2 theta - 1 = 1 - 2sin^2 theta

Hence:

cos(-(13pi)/8) = +-sqrt((1+cos(-(13pi)/4))/2)

color(white)(cos(-(13pi)/8)) = +-1/2sqrt(2+2cos(-(13pi)/4))

color(white)(cos(-(13pi)/8)) = +-1/2sqrt(2-sqrt(2))

sin(-(13pi)/8) = +-sqrt((1-cos(-(13pi)/4))/2)

color(white)(sin(-(13pi)/8)) = +-1/2sqrt(2-2cos(-(13pi)/4))

color(white)(sin(-(13pi)/8)) = +-1/2sqrt(2+sqrt(2))

Which signs are correct?

Note that -(13pi)/8 is coterminal with (3pi)/8 which is in Q1.

So cos and sin are both positive and:

{ (cos(-(13pi)/8) = 1/2sqrt(2-sqrt(2))), (sin(-(13pi)/8) = 1/2sqrt(2+sqrt(2))) :}

So polar coordinates (45, -(13pi)/8) in rectangular coordinates is:

(45 cos(-(13pi)/8), 45 sin(-(13pi)/8))=(45/2sqrt(2-sqrt(2)), 45/2sqrt(2+sqrt(2)))