# formula for getting the normlized X and Y values of a given degrees from a linear function

I have a number which we'll call α in degrees that represent the angle of a linear function with the X axis. for example when α is 0 the linear function is on the X axis , when α is 360 the linear function is on the X axis. when α is 90 the linear function is on the Y axis. and so on. I want to get from the linear function

y = αx


the X of y = 1, and the Y of x = 1. and I am not sure how can I do that. can anyone post a quick forumla for calc such a thing? does it have a name ? I know trigo may be needed to transfer degrees for a slope that represent the value of y for X = 1.

• When $\alpha =90°$, it doesn't necessarily mean that the linear function is on the y-Axis. It rather shows that it is parallel to the y-axis – Dr. Mathva Dec 9 '18 at 20:41
• Is $\alpha$ given, or do you get the function and have to calculate it? – Dr. Mathva Dec 9 '18 at 20:42
• alpha is given. – tomer zeitune Dec 9 '18 at 20:55
• What's the problem then? You would have: $$\text{X of y=1}=\frac{1}{\alpha}$$ and $$\text{Y of x=1}=\alpha$$ Or am I missing something? – Dr. Mathva Dec 9 '18 at 20:57
• Alpha is in degrees how do I transfer that to a slope (Y value for X = 1) ? – tomer zeitune Dec 9 '18 at 21:49

If $$\alpha$$ is the inclination angle of a linear function, then the slope of the function is $$\arctan \alpha$$
Let $$\Delta ABC$$ be the slope triangle of the function, with $$[AB]$$ lying on the x-axis. Denote by $$m$$ the slope of the function and by $$\alpha$$ the angle of the linear function with the x-axis. Since $$\angle CBA=90°$$, we can tell $$\tan(\angle BAC)=\tan(\alpha)=\frac{BC}{AB}=m$$
Let then simply $$X=\frac{1}{\tan(\alpha)}$$ and $$Y=\tan(\alpha)$$