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hypotenuse=0?


ydoaPs

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i was trying to make a trig chart for the complex plane and accindently found 0>1. at pi/4, the hypotenuse is 0, but the real leg is 1. how is that? imaginary trig must be fun.

 

edit: at [math]\theta=\frac{\pi}{4}[/math], i got:

 

sin=undefined

cos=undefined

tan=i

csc=0

sec=0

cot=-i

 

it is in the first quadrant and cot is negative...something is wonky

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When you start dealing with imaginary/complex numbers, it's no longer true that sin=opp/hyp, cos=adj/hyp... etc...

 

This is a complex sin...

For any real x and y, sin(x+iy)=sin(x)cosh(y)+i.cos(x)sinh(y).

 

Sorce (aside from my father)

 

Also, the "imaginary" triangle doesn't have a true right angle in it so sin=/=opp/hyp.

 

Additionally, when sin=undef... if you're getting the hypotinuse from sin=opp/hyp, where sin=1/0. To get the denomitator by itself, you multiply both sides by it, in this case 0.

 

This is the same type of thing that I described in a thread a while ago about singularities, where D=m/v. where v=0. You can't find the mass because to get the mass, you have to multiply both sides by 0, which gives you the equation 0=0. Which is true, but irrelevent. The same concept is true here...

 

where sin(x)=o/h, h=o/sin(x) WHERE sin(x)=/=0, and h=/=0.

 

I don't know if any of that made any sence to you, but I'm sure a math expert will help out any further if needed.

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i drew a triangle on the complex plane to find the tring functions for an angle. i started with 45 degrees. the hypotenuse was zero.
I can't make any sense of this. Can you attach a picture ?
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  • 1 month later...

In circuits we worked with cos(x)+isin(x). The link for that sin(x+iy) is reputable I;m not sure why the expanded function the showed woul dbe called the complex sin function unless there is acoplete suite of six complex trig functions taken over the real valued trig and hyperbolic trig functions.

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