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Figure $1$

Given $\triangle ABC$, we have$$c=a\cos B+b\cos A\tag{1}$$

Where $a=BC$, $b=AC$ and $c=AB$.

Question: How do you prove the relation?

I started with$$\begin{align*} & CD=a\cos B\\ & CD=b\cos A\end{align*}$$ So adding them up gives$$2CD=a\cos B+b\cos A$$However, what's confusing is how $2CD=c$ for the equation $(1)$ to hold.

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    looks like you have got your cos & sin muddled ? – 2017-02-18
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    Maybe $CD = a\sin B$? – 2017-02-18
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    The correct relation is $CD=a\sin B=b\sin A$, which gives the law of sines: $\frac{a}{\sin A}=\frac{b}{\sin B}$. – 2017-02-18
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    This triangle relationship looks complicated... – 2017-02-18

3 Answers 3

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You are mistaken. We have $$\begin{align*} & CD=a\sin B\\ & CD=b\sin A\end{align*}$$

Please review the definition of the trigonometric functions, as provided in this link, to see why this is so.

This is why you're not getting an answer. To prove the property, we use $$\begin{align*} & BD=a\cos B\\ & AD=b\cos A\end{align*}$$ In the diagram provided. Adding the two, we have $$c=a\cos B+b\cos A$$

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    Wow... How did I make that mistake?!? – 2017-02-18
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    @Frank To be frank with you, I have made those kinds of mistakes before too. It's fine. Just be sure to never make it again. – 2017-02-18
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    **NOTE** If you're wondering why I did not divide the cases of when the angle is obtuse and acute seperately, I was assuming the diagram was given. In the triangle, – 2017-02-18
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    @S.C.B. isn't it a bit harsh to say "Just be sure to never make it again"? :P – 2017-02-18
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    @SimplyBeautifulArt Is that harsh? I have no idea. – 2017-02-18
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    Well, it has a final tone, like a "do or die and be disgraced forever" type of tone. – 2017-02-18
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    @S.C.B lol I get your joke!! :P – 2017-06-22
2

Let $B>90^{\circ}$.

Hence, $$c=b\cos A-a\cos(180^{\circ}-B)=b\cos A+a\cos B$$ The case $B\leq90^{\circ}$ is obvious.

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Using the Law of cosines : $c=a\cos B+b\cos A \Leftrightarrow c=a \frac {a^2 + c^2 - b^2}{2 ac}+b\frac {b^2 + c^2 -a^2}{2bc} \Leftrightarrow 2c^2=2c^2$