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Re:Why is e^(pi i) = -1? by sophacles on Friday October 30, 2009 @10:20AM (#912031)
So to broadcast my math ignorance to all:

Why does cos(pi) + i sin(pi) == -1 ??

Why does i disappear in this?

Edit: Thanks to the replies below.

Re:Why is e^(pi i) = -1? by rottencupcakes on Friday October 30, 2009 @10:24AM (#912042)
We verify in a quick Google search: http://www.google.com/search?hl=en&safe=off&q=sin(pi...

that sin(pi) = 0.

Re:Why is e^(pi i) = -1? by travisjeffery on Friday October 30, 2009 @10:27AM (#912048)
The formula is:

e^(ix) = cos(x) - i sin(x)

Since sin(π) == 0, i * 0 == 0 so you have cos(π) left which is -1.

So in sum:

e^(i π) = cos(π) - i sin(π) == -1 - 0 == -1

Re:Why is e^(pi i) = -1? by catzaa on Friday October 30, 2009 @10:42AM (#912076)
If you draw a Sine graph, you will see that sin(x) is zero at 0, pi, 2*pi, ...

If you draw a cosine graph you will see that it is -1 at pi. Therefore in the above sum the sine part just falls away.

It is easy to see that the above is true when you recall that cosine(theta) = adjacent side/(hypotenuse). Since pi=180 degrees, adjacent will be -1 hypotenuse 1 and opposite side 0.

Re:Why is e^(pi i) = -1? by amalcon on Friday October 30, 2009 @10:24AM (#912045)
sin(pi)=0

I wouldn't feel ignorant or anything. The article left out that information seemingly on purpose, presumably so that people would try to figure out where the i went.