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