sine and inverse sine
Posted: Sun Nov 16, 2025 5:02 am
I am just looking over my notes on trig, when I found that I did solve the problem of the inverse sign.
Here is what I found.
If one works the triangle from the circumference of a circle, the sine produced by the algebra, is always correct. However, the inverse sine is not always correct. In this configuration, in order for the inverse sine to always be correct would require logical operators, so that is a non-starter.
However, if you work every angle from the center of the circle, instead of the circumference, the sine always works algebraically, and so does the inverse sine. So, we had two choices of how to do the trig algebraically, and the figure itself told us which configuration was to be used.
What is the difference? Working from the center of the circle you are always working with equilateral triangles. You can now always get both sine and inverse sine simply by working the angle in the equilateral configuration.
For any sine any triangle, no right angle required. For any inverse sign, any equilateral triangle, again, no right angle required.
One can look at it like this, working from the circumference we can easily derive any sine of any angle, To get the inverse sine of any on those angles, from an equilateral construct.
I got work to do here.
Here is what I found.
If one works the triangle from the circumference of a circle, the sine produced by the algebra, is always correct. However, the inverse sine is not always correct. In this configuration, in order for the inverse sine to always be correct would require logical operators, so that is a non-starter.
However, if you work every angle from the center of the circle, instead of the circumference, the sine always works algebraically, and so does the inverse sine. So, we had two choices of how to do the trig algebraically, and the figure itself told us which configuration was to be used.
What is the difference? Working from the center of the circle you are always working with equilateral triangles. You can now always get both sine and inverse sine simply by working the angle in the equilateral configuration.
For any sine any triangle, no right angle required. For any inverse sign, any equilateral triangle, again, no right angle required.
One can look at it like this, working from the circumference we can easily derive any sine of any angle, To get the inverse sine of any on those angles, from an equilateral construct.
I got work to do here.