Uncovering the Mystery of cos(-pi)=-1: Insights and Explanations

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Sorry I made a mistake in the title, I was thinking of something else I meant "Why is cos(-pi)=-1?"

cos(-pi)=-1, why is this so? I feel like maybe I'm missing the obvious, but I would think that it's equal to 1, considering cos(pi)=-1?
 
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The cosine function satisfies the identity

cos (-x) = cos (x).

Perhaps you are thinking of the sine function? It satisfies the identity

sin (-x) = -sin (x).
 
@Petek: Wow I feel slow, totally passed me that cosine is an even function lol. Thanks
 
Think about the unit circle !

Where is the angle π relative to the angle -π ?
 
As pointed out above, cosine is an even function.

Another way to think about it is that the period of cosine is 2π. So if cos(m)=A
Then cos(k2π+m) = A, where k is some integer. In your case, you say that cos(π) = -1, so it must be true that cos(π - 2π) = -1, ie: cos(-π) = -1
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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