What is the Cartesian form of #(9,(1pi )/12)#?

1 Answer
Apr 27, 2018

#P(9,{1π}/12)# in rectangular form is:

#( 9 cos(pi/12), 9 sin(pi/12) ) = ( 9/4 (sqrt{6} + sqrt{2}), 9/4 (sqrt{6} - sqrt{2}) ) #

Explanation:

#(x,y) = ( 9 cos(pi/12), 9 sin(pi/12) ) #

#pi/12 = 180^circ/12 = 15^circ #

We can get the trig functions of #15^circ# from the half angle formula or the difference angle formula.

#cos 15^circ = cos( 1/2 (30^circ)) = cos(45^circ - 30^circ)#

The difference angle form avoids nested square roots; let's use that.

#cos 15^circ = cos 45^circ cos 30^circ + sin 45^circ sin 30^circ #

#cos 15^circ = (sqrt{2}/2)(sqrt{3}/2) + (sqrt{2}/2)(1/2) #

#cos 15^circ = 1/4 (sqrt{6} + sqrt{2})#

Similarly,

#sin 15^circ = sin 45^circ cos 30^circ - cos 45^circ sin 30^circ #

#sin 15^circ = 1/4( \sqrt{6} - sqrt{2})#

Putting it together, #P(9,{1π}/12)# in rectangular form is:

#( 9 cos(pi/12), 9 sin(pi/12) ) = ( 9/4 (sqrt{6} + sqrt{2}), 9/4 (sqrt{6} - sqrt{2}) ) #