Question

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

Answer 1

`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.

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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`