<s>[INST] Proving that $a \cos A+b\cos B+c \cos C$ is the semi-perimeter of $\triangle ABC$ 
With each symbol having its usual meaning, prove that in $\triangle ABC$,
$$a \cos A+b\cos B+c \cos C=s$$
where $s$ is the semi-perimeter.


I started by applying the rule of sines:

*

*$a=k\sin A$

*$b=k\sin B$

*$c=k \sin C$
where $k$ is a constant

so, we have:

$k(\sin A\cos A+\sin B\cos B+\sin C\cos C)$

manipulating this,  we get:

$\frac{k}{2}(\sin 2A+\sin 2B+\sin 2C)$


$\frac{k}{2}(2\sin(A+B)\cos(A-B) + \sin 2C)$


$\frac{k}{2}(2\sin C\cos(A-B)+\sin 2C)$


$\frac{k}{2}2\sin C(\cos(A-B)+\cos C)$

But this doesn't seem to be yielding desired results. Could someone help me out with this? Or maybe tell me a better approach to this problem?
 [/INST] The semi-perimeter formula you're trying to use here is not true
$$a\cos{A}+b\cos{B}+c\cos{C} \ne s $$
If each symbol has its usual meaning in $\Delta ABC$
$$\cos{A} = \frac{b^2+c^2-a^2}{2bc}$$
$$a \frac{b^2+c^2-a^2}{2bc} + b  \frac{a^2+c^2-b^2}{2ac} + c  \frac{a^2+b^2-c^2}{2ab} $$
$$\frac{ a^2(b^2+c^2-a^2)+b^2(a^2+c^2-b^2)+c^2(a^2+b^2-c^2) }{2abc}$$
$$\frac{ 2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}{2abc}$$
Simplifying this and checking it out doesn't yield the semi-perimeter $s$
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