- #1

mathmari

Gold Member

MHB

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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)