sin(x) = e^(ix)-e^(-ix)/(2i)

cos(x)= e^(ix)+e^(-ix)/2

tan(x) = (i(1-e^(2ix)))/(1+e^(2ix))

 

sec(x) = (2i)/e^(ix)-e^(-ix)

csc(x) = 2/e^(ix)+e^(-ix)

cot(x) = (1+e^(2ix))/(i(1-e^(2ix))

 

sinh(x) = e^(x)-e^(-x)/2

cosh(x)= e^(x)+e^(-x)/2

tanh(x) = (e^x-e^-x)/(e^-x+e^x)

 

 

Have fun thinking about why all this is true.

 

How about dave prove them for us?

Proving converging infinite series involving imaginary numbers is not a task for the faint of heart. I worked out a proof for one of these, it took 4 hours. But what else could have I been doing in calc1?

it's not nice.

 

but they are pretty, you can work out all kinds of stuff with them (like arcsin(2) or whatever).

Originally posted by fafalone

sinh(x) = e^(x)-e^(-x)/2

cosh(x)= e^(x)+e^(-x)/2

tanh(x) = (e^x-e^-x)/(e^-x+e^x)

 

 

Have fun thinking about why all this is true.

 

I think those are true because that's how hyperbolic functions are defined.

That's easy...

 

e^u = 1 + u + u^2/2! + u^3/3! + ... via taylor series

 

Then, exponentiation of a number to a complex number is defined through this so that:

 

e^(ix) = 1 + ix + (ix)^2/2! + (ix)^3/3! + ...

 

= 1 + ix - x^2/2! - ix^3/3! + ...

 

= (1 - x^2/2! + ... ) + i(x - x^3/3! + ... )

 

= cos(x) + i*sin(x)

 

Therefore, *by definition* e^(ix) = cos(x) + i*sin(x).

 

For example, if we want cos(x) in terms of some expression with e^(ix) and/or a variant of e^(-ix) we can do the following:

 

 

e^(ix) = cos(x) + i*sin(x)

e^(-ix) = cos(x) - i*sin(x) //since cos(-x) = cos(x), sin(-x) = -sin(x)

 

Then by adding the equations, [e^(ix) + e^(-ix)] = 2cos(x)

 

cos(x) = (1/2) * (e^(ix) + e^(-ix))

 

Hyperbolic functions are defined in their own right, the names given to them are similar to the trig functions due to the properties they share with each other

 

THESE FORUMS ROCK!!!! GO FAF!!!!

Thanks

 

Of course it's easy if you know about Taylor series... but if you don't, it's really interesting.

Well that's the thing: They are all the result of the _definition_:

 

e^(ix) = cos(x) + i*sin(x)

 

..Which came from using the taylor expansion of e^u while assuming it holds for complex u (e.g. u = ix)

 

So there isn't anything mysterious behind these alternate expressions for the trig functions.

 

How else would you have arrived at them?

Trigonometric function definitions that will blow, this thread proclaims to be on the front page.

Fixed. Increased max thread title shown to 60...

eps said in post #8 :

Well that's the thing: They are all the result of the _definition_:

 

e^(ix) = cos(x) + i*sin(x)

 

..Which came from using the taylor expansion of e^u while assuming it holds for complex u (e.g. u = ix)

 

Definitions don't "come from" other things.

 

The Taylor expansion you speak of is a theorem, and the complex exponential relation you cited is a special case of that. It is not a definition.

The function e^x is known to be analytic for real x, but the concept of finding a "meaning" for a real number raised to a complex exponential didn't exist until we decided to take the taylor series for e^u and go ahead and *define* it for non-real u, e.g. u = ix.

 

So in turn, e^(ix) = cos(x) + i*sin(x) due to the fact that we defined e^u to be analytic for all complex u