MHB Show that the determinant is equal to 0

[mathmari]

Hey!

Let $\alpha, \beta, \gamma$ be internal angles of an arbitrary triangle.

I want to show that $$\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=0$$

We have the following:
\begin{align*}&\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=\cos \beta \cdot \det \begin{pmatrix} -1 & \cos\alpha \\ \cos\gamma & \cos\beta\end{pmatrix}-\cos \alpha \cdot \det \begin{pmatrix} \cos\gamma & \cos\alpha \\ -1 & \cos\beta\end{pmatrix}+(-1)\cdot \det \begin{pmatrix} \cos\gamma & -1 \\ -1 & \cos\gamma \end{pmatrix} \\ & = \cos \beta \cdot \left (-\cos \beta-\cos\gamma\cdot \cos\alpha\right )-\cos \alpha \cdot \left (\cos\gamma\cdot \cos\beta-(-1)\cdot \cos\alpha\right )-\left (\cos^2 \gamma -(-1)^2\right ) \\ & =-\cos^2 \beta-\cos\gamma\cdot \cos\alpha\cdot \cos \beta-\cos\alpha\cdot \cos\gamma\cdot \cos\beta- \cos^2\alpha-\cos^2 \gamma +1 \\ & =1-2\cdot \cos\alpha\cdot \cos\beta\cdot \cos\gamma - \cos^2\alpha-\cos^2 \beta-\cos^2 \gamma \end{align*}

Do we use here the Law of cosines?
\begin{align*}&c^2=a^2+b^2-2ab\cdot \cos \gamma \Rightarrow \cos \gamma=\frac{a^2+b^2-c^2}{2ab} \\ &b^2=a^2+c^2-2ac\cdot \cos \beta \Rightarrow \cos \beta=\frac{a^2+c^2-b^2}{2ac} \\ &a^2=b^2+c^2-2bc\cdot \cos \alpha \Rightarrow \cos\alpha=\frac{b^2+c^2-a^2}{2bc}\end{align*}

Or how could we continue? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

That should work I think although it looks like the expression will get pretty complicated first.
Alternatively we might try to use $α+β+γ=\pi$.
That is, substitute $γ=\pi-α-β$. (Thinking)

 

[mathmari]

That should work I think although it looks like the expression will get pretty complicated first.
Alternatively we might try to use $α+β+γ=\pi$.
That is, substitute $γ=\pi-α-β$. (Thinking)

Do you mean $\cos \gamma=\cos (\pi -\alpha-\beta)=-\cos (\alpha+\beta )$ ? (Wondering)

 

[I like Serena]

Do you mean $\cos \gamma=\cos (\pi -\alpha-\beta)=-\cos (\alpha+\beta )$ ?

That is what I was thinking yes.
I didn't try to calculate it manually though. It's just an alternative approach.
However, I did find that it is sufficient for Wolfram to conclude that it's zero, so it seems likely that filling it in by hand is feasible. (Thinking)

Another alternative approach is to try to find and find a geometric reason why the vectors are linearly dependent.
I didn't find one yet though. (Thinking)

 

[Greg]

$$1-2\cdot \cos\alpha\cdot \cos\beta\cdot \cos\gamma - \cos^2\alpha-\cos^2 \beta-\cos^2 \gamma$$

Doesn't this expression have cyclic symmetry? Implying there is an extremum at $\cos\alpha=\cos\beta=\cos\gamma$? Then, applying La Grange multipliers to the general case $\alpha+\beta+\gamma=\pi$, we arrive at the desired result.