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

In summary, the problem statement involves a different problem being solved on a website and the person got stuck at the first statement of the solution. They attempted to save time by using a program called Xournal++ but their answer did not match the one shown. They asked for help in identifying where they went wrong and another user provided a detailed explanation of the solution. The original poster then realized their mistake and provided the correct solution using the law of cosines. The conversation also includes a discussion about the unusual "-1" at the end of the original problem statement and how it is a portion of a larger problem. The factorization of the expression in the box is also mentioned.
  • #1
brotherbobby
588
150
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 :
1677664505352.png


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

triangle.png
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.
 
Physics news on Phys.org
  • #2
brotherbobby said:
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.
 
  • #3
fresh_42 said:
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.
 
  • #4
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.
 
Last edited:
  • Like
Likes Steve4Physics and neilparker62
  • #5
Charles Link said:
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.
 
  • Like
Likes mathwonk and Charles Link
  • #6
Charles Link said:
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.
 
  • Like
Likes SammyS and Charles Link
  • #7
neilparker62 said:
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.
 
  • Like
Likes neilparker62 and Charles Link

1. What is the formula for finding the sum of cosines in a triangle?

The formula for finding the sum of cosines in a triangle is cos(A) + cos(B) + cos(C) = 1, where A, B, and C are the angles of the triangle.

2. How is this formula derived?

This formula is derived from the Law of Cosines, which states that in a triangle with sides a, b, and c, the cosine of an angle is equal to the sum of the squares of the other two sides minus twice the product of those sides and the cosine of the included angle. By applying this to all three angles of the triangle and simplifying, we arrive at the formula cos(A) + cos(B) + cos(C) = 1.

3. What is the significance of this formula in geometry?

This formula is significant in geometry because it allows us to find the sum of the cosines of the angles in any triangle, regardless of the triangle's size or shape. It is also useful in solving problems involving triangles, such as finding missing angles or sides.

4. Can this formula be applied to other shapes besides triangles?

No, this formula is specific to triangles and cannot be applied to other shapes. The sum of cosines in other shapes may follow different formulas or rules.

5. How can this formula be used in real-world applications?

This formula can be used in various real-world applications, such as surveying, navigation, and engineering. It can also be applied in physics and astronomy to calculate the angles of celestial objects. Additionally, the sum of cosines can be used in computer graphics to create 3D models and animations.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
12
Views
679
  • Precalculus Mathematics Homework Help
Replies
7
Views
505
  • Precalculus Mathematics Homework Help
Replies
11
Views
1K
  • Precalculus Mathematics Homework Help
Replies
9
Views
529
  • Precalculus Mathematics Homework Help
Replies
12
Views
2K
  • Precalculus Mathematics Homework Help
Replies
1
Views
3K
  • Precalculus Mathematics Homework Help
Replies
19
Views
2K
  • Precalculus Mathematics Homework Help
Replies
14
Views
455
  • Precalculus Mathematics Homework Help
Replies
18
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
Back
Top