Suppose $\alpha$, $\beta$, and $\gamma$ are the three angles of a triangle so that $\alpha + \beta + \gamma = \pi$. Show that: $$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2\cos \alpha \cos \beta \cos \gamma = 1$$

My thoughts

I've reduced the degree of the squared trig functions to one using the trig power reduction formula but that didn't work. Should I convert $1$ to $\sin^2x+\cos^2x$?


$$\cos^2α + \cos^2β + \cos^2γ=$$ $$=\frac{1+\cos 2 \alpha}2+\frac{1+\cos 2 \beta}2+\frac{1+\cos 2 \gamma}2=$$ $$=\frac32+\frac{\cos 2 \alpha+\cos 2 \beta}{2}+\frac{\cos(2\pi-(2 \alpha+2 \beta))}{2}=$$ $$=\frac32+\cos(\alpha+\beta)\cos(\alpha-\beta)+\frac{\cos2(\alpha+\beta)}{2}=$$ $$=\frac32+\cos(\alpha+\beta)\cos(\alpha-\beta)+\frac{2\cos^2(\alpha+\beta)-1}{2}=$$ $$=1+\cos(\alpha+\beta)\cos(\alpha-\beta)+\cos^2(\alpha+\beta)=$$ $$=1+\cos(\alpha+\beta)(\cos(\alpha-\beta)+\cos(\alpha-\beta))=$$ $$=1-2\cos\alpha\cos\beta\cos\gamma$$


Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,


Now $\cos(\alpha+\beta)=\cos(\pi-\gamma)=-\cos\gamma$

$$\implies\cos(\alpha+\beta)\cos(\alpha-\beta)+\cos^2\gamma$$ $$=-\cos\gamma\cos(\alpha-\beta)+\cos\gamma\cdot[-\cos(\alpha+\beta)]$$



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