I like doing stuff wrong. Wasting everyone's time is also a hobby of mine. A:=arctan(i*3/5) If you do the expansion with 'e' is fun. You'll learn about hyperbolic trig functions. So, there is a dual function going on here. It's kind of top-secret cool stuff. In a situation where you have a right-angle triangle, you use the hypotenuse in the denominator instead of the other leg. sin^2(A/2)=((1-(cos(A))^2)/2)=-1/8 sin(A/2)=isqrt(2)/4 cos^2(A/2)=(1+(cos(A))^2)/2=9/8 cos(A/2)=3sqrt(2)/4 I suppose I'm not being careful. I'm going to ignore the whole -1 thing on the x-axis and check the distance. And I'll remove the 'i'. If I don't, the distance is '1' and it's boring. sqrt((sqrt(2)/4)^2+(3sqrt(2)/4)^2) = sqrt(5)/2 So, initially I did arctan wrong. Instead of two legs of a rational 90-degree triangle, I used a leg over the hypotenuse. So, it was a 3 4 5 triangle in the unit circle. I think this only works with rational triangles. Something to do with the dual functionality of the rational trig functions and that complex number. cosh(arctanh(3/5)/2) = 3sqrt(2)/4 sinh(arctanh(3/5)/2) = sqrt(2)/4 ** ** Ignore the non-existent imaginary value at your own peril. You can test it. tanh(arctanh(3/5)/2)=i/3 (i/3+i/3)/(1-(i/3)^2)= i3/5 i3/5 is what we started with. So, we can make a 90-degree triangle with these three values (sqrt(2)/4. 3sqrt(2)/4, sqrt(5)/2). The altitudes are 3sqrt(2)/4, sqrt(2)/4, and 3sqrt(5)/20 The area is 3/16. Now for cosine law. cos(C)=(a^2+b^2-c^2)/(2*a*b), yeah I also did that wrong. ((sqrt(2)i/4)^2+(3sqrt(2)/4)^2-(sqrt(5)/2)^2)/(2(sqrt(2)i/4)(3sqrt(2)/4)) = i/3 I haven't bother to check if the tangent half angle is always that yet. If you were to do relativistic velocity addition and you wanted to add c/3 to c/3 you would go at c3/5 m/s.
Thansk alot for this explanation!
this figure alone is very powerful to derive many trig fromula, i loved this
Exactly what I needed. Thanks!
Great to hear!
thank you so much !
thanks👍
I like doing stuff wrong. Wasting everyone's time is also a hobby of mine.
A:=arctan(i*3/5)
If you do the expansion with 'e' is fun. You'll learn about hyperbolic trig functions.
So, there is a dual function going on here. It's kind of top-secret cool stuff. In a situation where you have a right-angle triangle, you use the hypotenuse in the denominator instead of the other leg.
sin^2(A/2)=((1-(cos(A))^2)/2)=-1/8
sin(A/2)=isqrt(2)/4
cos^2(A/2)=(1+(cos(A))^2)/2=9/8
cos(A/2)=3sqrt(2)/4
I suppose I'm not being careful. I'm going to ignore the whole -1 thing on the x-axis and check the distance. And I'll remove the 'i'. If I don't, the distance is '1' and it's boring.
sqrt((sqrt(2)/4)^2+(3sqrt(2)/4)^2) = sqrt(5)/2
So, initially I did arctan wrong. Instead of two legs of a rational 90-degree triangle, I used a leg over the hypotenuse. So, it was a 3 4 5 triangle in the unit circle. I think this only works with rational triangles. Something to do with the dual functionality of the rational trig functions and that complex number.
cosh(arctanh(3/5)/2) = 3sqrt(2)/4
sinh(arctanh(3/5)/2) = sqrt(2)/4 **
** Ignore the non-existent imaginary value at your own peril.
You can test it.
tanh(arctanh(3/5)/2)=i/3
(i/3+i/3)/(1-(i/3)^2)= i3/5
i3/5 is what we started with.
So, we can make a 90-degree triangle with these three values (sqrt(2)/4. 3sqrt(2)/4, sqrt(5)/2).
The altitudes are 3sqrt(2)/4, sqrt(2)/4, and 3sqrt(5)/20
The area is 3/16.
Now for cosine law. cos(C)=(a^2+b^2-c^2)/(2*a*b), yeah I also did that wrong.
((sqrt(2)i/4)^2+(3sqrt(2)/4)^2-(sqrt(5)/2)^2)/(2(sqrt(2)i/4)(3sqrt(2)/4)) = i/3
I haven't bother to check if the tangent half angle is always that yet.
If you were to do relativistic velocity addition and you wanted to add c/3 to c/3 you would go at c3/5 m/s.
Nice nice !!
Hi, I have a question please. Can you please explain how you arrived at 1 + cos(x)=AH and how 2+2cos(x)= (AC)^2. Thanks!
AO=1 and OH=1• cos(x) so AH=1+cos(x)
The other is from a property of right triangle AC^2 = AH • AB = 2+2 cos(x)
triangle ACB is not a right triangle. ACH is a right triangle.
Any triangle formed between a point on the semicircumference and the diameter will be right. Thale’s theorem