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Heron formula for area of a triangle

Let ABC be a triangle, and a=length(BC), b=length(CA) and c=length(AB).Let $ p=\frac{a+b+c}{2} $
Then area S of the triangle is

$$S=\sqrt{p(p-a)(p-b)(p-c)}$$

We know that

$$\cos A=\frac{b^2+c^2-a^2}{2bc}$$

so

$$2\sin^2\frac{A}{2}=1-\cos A=1-\frac{b^2+c^2-a^2}{2bc}=\frac{a^2-(b-c)^2}{2bc}=\frac{(a-b+c)(a+b-c)}{2bc}=2\frac{(p-b)(p-c)}{bc}$$

Similar

$$2\cos^2\frac{A}{2}=1+\cos A=1+\frac{b^2+c^2-a^2}{2bc}=\frac{(b+c)^2-a^2}{2bc}=\frac{(a+b+c)(-a+b+c)}{2bc}=2\frac{p(p-a)}{bc}$$

so we have

$$\sin\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{bc}}$$

and

$$\cos\frac{A}{2}=\sqrt{\frac{p(p-a)}{bc}}$$

Inverse of Fourier transform

Let $ \^{f}(\xi)=\int_{R^n}f(x)e^{-i<x,\xi>}dx $ be the Fourier transform of a function f in $ \mathcal{S}(R^n) $. We shall prove that

$$f(x)=(2\pi)^{-n}\int_{R^n}\^{f}(\xi)e^{i<x,\xi>}d\xi$$

Let g be another function in $ \mathcal{S}(R^n) $. We have

$$\int_{R^n}\^{f}(\xi)g(\xi)e^{i<x,\xi>}d\xi=\int_{R^n}\int_{R^n}f(y)e^{-i<y,\xi>}dy\;g(\xi)e^{i<x,\xi>}d\xi=$$
$$\int_{R^n}f(y)\int_{R^n}g(\xi)e^{-i<y-x,\xi>}d\xi\;dy=<br />
\int_{R^n}f(y)\^{g}(y-x)dy$$

From this relation for $ g_{\epsilon}(\xi)=g(\epsilon\xi) $

Second order partial differential of implicite function

Let $ G(x, y, z)=F(\frac{x}{z},\frac{y}{z})=0 $ an equation defined by a function $ F\in \mathcal{C}^2 $
Suppose we can apply the implicit function theorem in a neighborhood of a given point (a,b,c).
So we suppose

$$G(a,b,c)=F(\frac{a}{c},\frac{b}{c})=0$$

and

$$\frac{\partial G}{\partial z}(a,b,c)=\frac{\partial F}{\partial u}(\frac{a}{c},\frac{b}{c})(-\frac{a}{c^2})+\frac{\partial F}{\partial v}(\frac{a}{c},\frac{b}{c})(-\frac{b}{c^2})\ne 0$$

Then there is a function z=f(x,y) with f(a,b)=c; G(x,y,f(x,y))=0 or

$$ F(\frac{x}{f(x,y)},\frac{y}{f(x,y)})=0$$

We have

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