How do you find θ, if #0 < θ < 360# and #cot theta =-1 # and theta in in QIV?

1 Answer
Apr 24, 2018

Answer:

#theta = 315^circ#

Explanation:

I've been doing trig homework all week and almost all the problems use multiples of #30^circ# or #45^circ.# This one is the latter.

Cotangent is reciprocal slope. It's pretty much the unit the Egyptians used to build the pyramids. They liked the seked , which was the number of palms horizontal per cubit vertical, a cubit being seven palms. Thus a seked is seven times the cotangent.

The line through the origin with reciprocal slope #-1# is of course the line through the origin with the slope #-1#, namely #y=-x#. That's two rays, two angles, #theta=-45^circ# and #-45+180=135^circ#.

The former, #theta=-45^circ# is the one in the fourth quadrant.


Let's say we didn't know that and just needed to solve

#cot theta = -1 #

The trick to all of these is to work them into the form #cos x= cos a# which has solutions #x = pm a + 360^circ k,# integer #k#. Linear combinations of cosines and sines are scaled and phase shifted cosines.

# cos theta / sin theta = -1 #

#cos theta = - sin theta #

# cos theta + sin theta = 0 #

Since #cos 45^circ = sin 45^circ#,

# cos 45^circ cos theta + sin 45^circ sin theta = 0 #

# cos(theta - 45^circ) = cos 90^circ #

# theta - 45^circ = \pm 90^circ + 360^circ k quad# for integer # k#

#theta = 45 \pm 90^circ + 360 ^ circ k #

#theta = -45^circ + 180^circ k#

That's the same answer we got by knowing cotangent meant reciprocal slope, and we choose the one in the fourth quadrant as requested.

OK, we're asked for a positive value in the fourth quadrant which would be #theta = -45^circ + 360^circ =315^circ#