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

Apr 24, 2018

$\theta = {315}^{\circ}$

#### 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.

$\cos \frac{\theta}{\sin} \theta = - 1$

$\cos \theta = - \sin \theta$

$\cos \theta + \sin \theta = 0$

Since $\cos {45}^{\circ} = \sin {45}^{\circ}$,

$\cos {45}^{\circ} \cos \theta + \sin {45}^{\circ} \sin \theta = 0$

$\cos \left(\theta - {45}^{\circ}\right) = \cos {90}^{\circ}$

$\theta - {45}^{\circ} = \setminus \pm {90}^{\circ} + {360}^{\circ} k \quad$ for integer $k$

$\theta = 45 \setminus \pm {90}^{\circ} + {360}^{\circ} k$

$\theta = - {45}^{\circ} + {180}^{\circ} k$

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 $\theta = - {45}^{\circ} + {360}^{\circ} = {315}^{\circ}$