Proving that $a \cos A+b\cos B+c \cos C$ is the semi-perimeter of $\triangle ABC$

Question

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.

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I started by applying the rule of sines:

1. $a=k\sin A$
2. $b=k\sin B$
3. $c=k \sin C$
   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?

Answer 1

What is true is the relation $a\cos A+b \cos B+c\cos C=\dfrac{2rs}{R}$ where $r$ is the inradius, $R$ is the circumradius and $s$ is the semi perimeter of the triangle.

By the sine rule, we can write the length of the sides as $a=2R\sin A,b=2R\sin B,c=2R\sin C$ therefore we have $$a\cos A+b \cos B+c\cos C=R(\sin 2A+\sin 2B+\sin 2C)$$

It is well known that in a triangle, we have $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B \sin C$

So we have $a\cos A+b \cos B+c\cos C=4R\sin A\sin B \sin C$

Again, by the sine rule, re writing $\sin A=\dfrac{a}{2R}$ and similarly for other sides, we obtain $a\cos A+b \cos B+c\cos C=\dfrac{abc}{2R^2}$

It is also well known that $abc$ and $\Delta$ are related by the relation $\Delta=\dfrac{abc}{4R}$, so substituting the value into the equation we obtain the value as $\dfrac{2\Delta}{R}$, but since $\Delta=rs$, we have $$a\cos A+b \cos B+c\cos C=\dfrac{2rs}{R}$$

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Helpful: Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle

Answer 2

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 $s$