1. ## PreCal questions..

can someone explain how to do this kind of problem using the unit circle and all in explicit detail please?

1: sin(arctan -1)
2: cos(arc csc 13/5)

just problems like that because i have no idea how to solve them.

Thanks.

2. Hi

You can start with $\displaystyle tan(Arctan \theta) = \theta$

$\displaystyle \frac{sin(Arctan \theta)}{cos(Arctan \theta)} = \theta$

$\displaystyle \frac{sin^2(Arctan \theta)}{cos^2(Arctan \theta)} = \theta^2$

$\displaystyle \frac{sin^2(Arctan \theta)}{1 - sin^2(Arctan \theta)} = \theta^2$

$\displaystyle \frac{1 - sin^2(Arctan \theta)}{sin^2(Arctan \theta)} = \frac{1}{\theta^2}$

$\displaystyle \frac{1}{sin^2(Arctan \theta)} -1 = \frac{1}{\theta^2}$

$\displaystyle \frac{1}{sin^2(Arctan \theta)} = 1 + \frac{1}{\theta^2} = \frac{1 + \theta^2}{\theta^2}$

$\displaystyle {sin^2(Arctan \theta)} = \frac{\theta^2}{1 + \theta^2}$

$\displaystyle {sin^2(Arctan (-1))} = \frac{1}{2}$

Arctan(-1) is an angle between $\displaystyle -\frac{\pi}{2}$ and 0
Therefore sin(Arctan(-1)) is negative

$\displaystyle {sin(Arctan (-1))} = -\frac{\sqrt{2}}{2}$