finding the slope of a line connecting two given Terminal points Say that Point X is the terminal point of Angle $\alpha$  and Point Y is the terminal point of $\beta$ on the unit circle. 
"Terminal point", in this case, is the point that is obtained by rotating counterclockwise from point (1,0) on the unit circle.
Point X is in the first quadrant. Point Y is in the second quadrant.  
$\tan \alpha = 1$ and $\tan \beta = -7$. Find the slope of $\overline{XY}$.
What I have so far: 
$\alpha$ is $\frac{\pi}{4}$. since cos x = the x coordinate of the terminal point, would I have to find $cos \frac{\pi}{4}$ to get $\frac{\sqrt{2}}{2}$? 
I am not sure what $\beta$ is in reduced terms because doing $arc tan(-7)$ results in -1.4288999... but finding the cos of that is $\frac{1}{5\sqrt{2}}$ (or is it $-\frac{1}{5\sqrt{2}}$?). 
I am assuming that they both will have the same y coordinate, so would the slope be $\frac{\sqrt{2}}{2} + \frac{1}{5\sqrt{2}} = \frac{3\sqrt{2}}{5}$?
 A: Since $X$ and $Y$ are on the unit circle, we have
$$
X=(\cos(\alpha),\sin(\alpha))
$$
and
$$
Y=(\cos(\beta),\sin(\beta))
$$
The slope of $\overline{XY}$ would be
$$
\frac{\sin(\beta)-\sin(\alpha)}{\cos(\beta)-\cos(\alpha)}
$$
Since $\tan^2(x)+1=\sec^2(x)=\frac1{\cos^2(x)}$, we have
$$
\cos(x)=\frac{\pm1}{\sqrt{1+\tan^2(x)}}
$$
Also $\sin(x)=\tan(x)\cos(x)$. Therefore,
$$
\sin(x)=\frac{\pm\tan(x)}{\sqrt{1+\tan^2(x)}}
$$
The choice of $\pm$ depends on the quadrant. Use $+$ in quadrants I and IV and $-$ in quadrants II and III.
A: The secant ( not tangent) joining X and Y intersects x-axis at an angle $ > \pi/2 $ but $ < \pi $ in second quadrant in an anti-clockwise direction.  Hence its tangent should be negative. The angular bisector of these directions is perpendicular to this secant line.The bisector makes $ (\alpha + \beta )/2 $ to x-axis.
$$ slope = \tan ( \pi/2 +(\alpha + \beta )/2) = -\cot ((\alpha + \beta )/2) $$
To further simplify use tan of half angle with Trig formula of double angle $(\alpha + \beta ) $ on right hand side:
$$ tan\, x = \dfrac {(1 - cos\, 2 x)}{sin\, 2 x} $$
to take it further from there.
