A formula involving the sum of cosines of the angles of a triangle

[brotherbobby]

Homework Statement

For a triangle ##\text{ABC}##, prove that $$\boxed{\cos A+\cos B+\cos C-1=\dfrac{a^2b+b^2c+c^2a+b^2a+c^2b+a^2c-a^3-b^3-c^3-2abc}{2abc}}$$

Relevant Equations

1. ##\cos\frac{A}{2}= \sqrt{\dfrac{s(s-a)}{bc}}## and its cyclic counterparts.
2. ##\sin\frac{A}{2}= \sqrt{\dfrac{(s-b)(s-c)}{bc}}## and its cyclic counterparts.
3. ##s=\dfrac{a+b+c}{2}##, the semi perimeter of a triangle.

Problem Statement : The statement appeared on a website where a different problem was being solved. I got stuck at the (first) statement in the solution that I posted above . Here I copy and paste that statement from the website, which I cannot show :

Attempt : To save time typing, I write out and paste the solution using Xournal++, hoping am not violating anything.

Issue : As evidence by the coefficients that I have marked in red, my answer is not the same as that shown for the problem.

A hint as to where I went wrong would be welcome.

 

[fresh_42]

Homework Statement:: For a triangle ##\text{ABC}##, prove that $$\boxed{\cos A+\cos B+\cos C-1=\dfrac{a^2b+b^2c+c^2a+b^2a+c^2b+a^2c-a^3-b^3-c^3-2abc}{2abc}}$$
Relevant Equations:: 1. ##\cos\frac{A}{2}= \sqrt{\dfrac{s(s-a)}{bc}}## and its cyclic counterparts.
2. ##\sin\frac{A}{2}= \sqrt{\dfrac{(s-b)(s-c)}{bc}}## and its cyclic counterparts.
3. ##s=\dfrac{a+b+c}{2}##, the semi perimeter of a triangle.

Problem Statement : The statement appeared on a website where a different problem was being solved. I got stuck at the (first) statement in the solution that I posted above . Here I copy and paste that statement from the website, which I cannot show : View attachment 323036

Attempt : To save time typing, I write out and paste the solution using Xournal++, hoping am not violating anything.

View attachment 323039Issue : As evidence by the coefficients that I have marked in red, my answer is not the same as that shown for the problem.

A hint as to where I went wrong would be welcome.

I cannot see what you have done in every step except the first one, so let's see.

And btw. here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

\begin{align*}
\cos(A)&= \dfrac{s(s-a)}{bc}-\dfrac{(s-b)(s-c)}{bc}=\dfrac{s^2-as-s^2+bs+cs-bc}{bc}\\&=\dfrac{s(-a+b+c)}{bc}-1=\dfrac{s}{abc}(-a^2+ab+ac)-1
\end{align*}
hence
\begin{align*}
\cos(A)+&\cos(B)+\cos(C)=\dfrac{s}{abc}(-a^2+ab+ac-b^2+bc+ba-c^2+ca+cb)-3\\
&=-\dfrac{s}{abc}(a^2+b^2+c^2-2ab-2ac-2bc)-\dfrac{6abc}{2abc}\\
&=-\dfrac{1}{2abc}\cdot ((a+b+c)(a^2+b^2+c^2-2ab-2ac-2bc)+6abc)\\
&=-\dfrac{1}{2abc}(a^3+ab^2+ac^2+a^2b+b^3+ac^2+a^2c+b^2c+c^3)+\ldots\\
&+\dfrac{1}{2abc}(2a^2b+2a^2c+2abc+2ab^2+2abc+2b^2c+2abc+2ac^2+2bc^2-6abc)\\
&=\dfrac{1}{2abc}(-a^3-b^3-c^3+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2)
\end{align*}
and thus
\begin{align*}
\cos(A)+&\cos(B)+\cos(C)-1=\dfrac{1}{2abc}(-a^3-b^3-c^3+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2-2abc)
\end{align*}
which is the correct answer.

I assume you made a sign error somewhere but I do not see exactly where since I cannot trace your calculation.

 

[brotherbobby]

And btw. here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

I was trying to save work, typing equations out. However, I will now type out the answer and remedy my mistake above.

To prove : For a triangle ABC, show that ##\boxed{\cos A+\cos B+\cos C-1=\dfrac{a^2b+b^2c+c^2a+b^2a+c^2b+a^2c-a^3-b^3-c^3-2abc}{2abc}}##.

Solution : Starting from the left hand side,
$$\begin{equation*}
\begin{aligned}
\cos A+\cos B+\cos C-1 & =\cos^2A/2-\sin^2 A/2+\cos^2B/2-\sin^2 B/2+\cos^2C/2-\sin^2 C/2-1\\
&=\frac{s(s-a)}{bc}-\frac{(s-b)(s-c)}{bc}+\frac{s(s-b)}{ca}-\frac{(s-c)(s-a)}{ca}+\frac{s(s-c)}{ab}-\frac{(s-a)(s-b)}{ab}-1\\
&=\frac{as(b+c-a)-abc+bs(c+a-b)-abc+cs(a+b-c)-abc-abc}{abc}\\
&=\frac{(a+b+c)(ab+ac-a^2)+(a+b+c)(bc+ab-b^2)+(a+b+c)(ac+ab-c^2)-8abc}{2abc}\\
\end{aligned}
\end{equation*}$$

The numerator simplifies to the required ##\mathcal{N} = -a^3-b^3-c^3+a^2b+ab^2+b^2c+bc^2+c^2a+ca^2-2abc##.

Thank you for your help.

 

[Charles Link]

I think you could have begun with the law of cosines: e.g. ## c^2=a^2+b^2-2ab \cos{C} ##, rather than a formula that most of us probably wouldn't ever memorize, or maybe even recognize. With a little algebra with the 3 cosines, the result follows very routinely.

 

[neilparker62]

I think you could have begun with the law of cosines: e.g. ## c^2=a^2+b^2-2ab \cos{C} ##, rather than a formula that most of us probably wouldn't ever memorize, or maybe even recognize. With a little algebra with the 3 cosines, the result follows very routinely.

This was certainly my first thought! I don't even know why they included the " - 1 " at the end.

 

[brotherbobby]

I think you could have begun with the law of cosines: e.g. ## c^2=a^2+b^2-2ab \cos{C} ##, rather than a formula that most of us probably wouldn't ever memorize, or maybe even recognize. With a little algebra with the 3 cosines, the result follows very routinely.

Yes, thank you. I totally overlooked the law of cosines.

Attempt :
\begin{equation*}

\begin{aligned}

\cos A+\cos B+\cos C-1 & =\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}-1\\

&=\frac{a(b^2+c^2-a^2)+b(c^2+a^2-b^2)+c(a^2+b^2-c^2)-2abc}{2abc}\\

&=\boxed{\frac{a^2b+ab^2+b^2c+bc^2+c^2a+ca^2-a^3-b^3-c^3-2abc}{2abc}}\\

\end{aligned}

\end{equation*}

Thank you.

 

[brotherbobby]

This was certainly my first thought! I don't even know why they included the " - 1 " at the end.

Yes, it looks unusual. But that is because it is a portion of a different problem.

I posted requiring to show that $$\cos A+\cos B+\cos C-1=\boxed{\dfrac{a^2b+b^2c+c^2a+b^2a+c^2b+a^2c-a^3-b^3-c^3-2abc}{2abc}}$$

The numerator of the expression in the box factorises to $$\boxed{a^2b+b^2c+c^2a+b^2a+c^2b+a^2c-a^3-b^3-c^3-2abc} = \boldsymbol{(a+b-c)(b+c-a)(c+a-b)}$$
The factorisation is involved.

You can see that, for the triangle, the factorised expressed in bold simplifies readily to $$8(s-a)(s-b)(s-c),$$ furthering the solution to original problem I didn't show. $$\text{Show that, for a triangle}\,\cos A+\cos B+\cos C = 1+\frac{r}{R}$$

I have, with blood, sweat and tears, done the problem.