I saw on a website, law of sines are transposed. Example:
c^2=a^2+b^2-2abcos(C)
Transposed: cos(C)= a^2+b^2-c^2/2ab
I need a proof how it became this way.

You are dealing with transposition and cross-multiplication.
$c^2 = a^2 + b^2 - 2ab \, \cos C$

$2ab \cos C = a^2 + b^2 - c^2$

$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$

Long answer:

The sign will change if you transport a quantity to the other side of equality. Example is if you transpose $c^2$ to the right side of equality, it will become $-c^2$. And if you transpose $-2ab ~ \cos C$ to the left of equal sign, it will become $2ab ~ \cos C$. Hence, $2ab ~ \cos C = a^2 + b^2 - c^2$

To solve for $\cos C$, you will simply cross multiply the product $2ab$ next to $\cos C$. The process of cross multiplication is straight forward. Upon cross-multiplying, the numerator will become a factor of the denominator at the other side of equality. Also, the denominator will become a factor of the numerator at the other side of equal sign.
$\dfrac{2ab \, \cos C}{1} = \dfrac{a^2 + b^2 - c^2}{1}$

You are dealing with transposition and cross-multiplication.

$c^2 = a^2 + b^2 - 2ab \, \cos C$

$2ab \cos C = a^2 + b^2 - c^2$

$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$

Long answer:To solve for $\cos C$, you will simply cross multiply the product $2ab$ next to $\cos C$. The process of cross multiplication is straight forward. Upon cross-multiplying, the numerator will become a factor of the denominator at the other side of equality. Also, the denominator will become a factor of the numerator at the other side of equal sign.

$\dfrac{2ab \, \cos C}{1} = \dfrac{a^2 + b^2 - c^2}{1}$

$\dfrac{\cos C}{1} = \dfrac{a^2 + b^2 - c^2}{2ab}$

Hence,

$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$

I am not sure if this long answer suits you but that's it.

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