Question

What other special right triangles are there?

Besides the regular 30 60 90, 45 45 90, 90 90 0, (180 0 0)?

So using the double angles theorem should be able to give us even more:

15 75 90 should give out a fraction with a root on top most likely. There's also 22.5 67.5 90, and by using 15/2 15/4... 15/2^x.

Do these even count as "special" anymore? They have full calculable root values, so why not? Maybe just because we really only need these two to create any other type and that let's them be special. Maybe there should be a little more to this by using the theorem behind `cos(A-B)` and it's variants, like it should allow 15 (45-30), 75 (45-(-30)).

Also, since calculators are able to find sin(random number) so fast, maybe they are just using a guess and check and plug in method, or a huge table. Or maybe they have a set path to calculating weird things like sin(21) is some really huge fraction with a lot of roots. Maybe this all goes back to where we got the original special triangles?

Answer 1

A few thoughts...

Explanation:

If you are talking about triangles with a whole number of degrees, then I would suggest that any multiples of `3^@` make special right triangles, since any multiple of `3^@` has trigonometric values that can be expressed in terms of rational numbers and square roots.

Any other whole number of degrees does not have such an expression.

Starting from `45^@` and `30^@` you can find tidy expressions for trigonometric functions `15^@` using the difference formulas.

You can also find decent expressions for trigonometric values of `18^@` by considering the roots of `x^5-1=0`.

Hence you can use the difference formulas again to find a slightly more messy expression for `3^@ = 18^@-15^@`.

From the values for `3^@` you can get to trigonometric values of any multiple of `3^@`.

Answer 2

y-intercept is `sqrt`prime.

Explanation:

Let the odd number p be a prime and `I_(p)` = integer in p/2= [ p/2 ].

Then p = 2`I_(p) + 1`.

From A( `I_(p)`, 0 ), use compass to mark B on the y-axis, at the distance `I_(p)` + 1.then,

OB = `sqrt p`

Graph for y-,intercept `sqrt (19)`, with `I_(19)` = 9

graph{((x-9)^2+ y^2-100)((x-9)^2+ y^2-0.01)(x^2+(y-sqrt 19)^2-0.01)(x/9+y/sqrt(19)-1)=0[-1 11 -1 5]}.

This is an excerpt, from one of my documents, prepared jointly,

with two others .