How do sine, cosine and tangent extend beyond right triangle ratios?

[Mark44]

I think the unit circle is taken not because of physical reasons but mathematical ones.
The main reason I think is because of the
$e^{i\theta}=\cos{\theta}+i\sin{\theta}$ (how do i add latex in my reply...I am new).
Suppose it had radius 2 then we would have
$2=e^{i2\pi}=e^{i\pi}e^{i\pi}=-2*-2$
which does not seem right

Correct - it's not right. $e^{i 2\pi} = cos(2\pi) + i sin(2\pi) = 1 + 0i = 1$
You can't just come along and set this expression to 2.

 

[BiGyElLoWhAt]

I think the unit circle is taken not because of physical reasons but mathematical ones.
The main reason I think is because of the
$e^{i\theta}=\cos{\theta}+i\sin{\theta}$ (how do i add latex in my reply...I am new).
Suppose it had radius 2 then we would have
$2=e^{i2\pi}=e^{i\pi}e^{i\pi}=-2*-2$
which does not seem right

You're not using euler formula correctly.
$e^{i\theta}=\cos{\theta}+i\sin{\theta}$
This is correct.

but this:
$2=e^{i2\pi}=e^{i\pi}e^{i\pi}=-2*-2$
is not.

$2 = e^{i*2\pi} + e^{-i*2\pi} = 2e^{i*2\pi} = cos(2\pi) + i*sin(2\pi) + cos(2\pi) - i*sin(2\pi)$
OR $=2[cos(2\pi) + i*sin(2\pi)$
the first expansion is a representation of the first eulers, and the second expansion is a representation of the second eulers.

You have $e^{i\theta} = \cos{\theta}+i\sin{\theta}$
and $e^{i*k\theta}= \cos{k\theta}+i\sin{k\theta}$
Where k can be any function.

 

[Fredrik]

You're not using euler formula correctly.
$e^{i\theta}=\cos{\theta}+i\sin{\theta}$
This is correct.

Actually, using the alternative definitions of sin and cos that piethein21 was considering there, it's not. His sin and cos are exactly 2 times the normal sin and cos. So the left-hand side and the right-hand side don't have the same absolute value, unless we also redefine the exponential function as 2 times the normal exponential function. Of course, if we do that, then $e^{x+y}=e^xe^y$ doesn't hold, so his calculation is still wrong.

but this:
$2=e^{i2\pi}=e^{i\pi}e^{i\pi}=-2*-2$
is not.

It's probably time to close this thread, since it has drifted off topic. The OP just wanted to know how to define sin and cos with a domain larger than $[0,2\pi]$. That was answered early in the thread. Most of the posts after that are based on misunderstandings and have nothing to do with the original topic.

 

[piethein21]

the time it was not properly used was to show it did not hold when rules where altered (so I am very aware that it was not correct). I agree with fredrik let's close the thread ... it is drifting...