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)))#