I came across this question that asked to prove that in every triangle, this statement is true:
$$2(1+\cos(α))(1+\cos(β))(1+\cos(γ)) = (\sin(α)+\sin(β)+\sin(γ))^2$$
I tried rewriting gamma using alpha and beta and used trig identities, but the algebra got too complicated very quickly. I would like an elementary solution, preferably high school level, but of course all solutions are accepted. Visual or algebraic too, doesn't matter.
The left-hand side of your equation expands to:
$$L := 2(1 + \cos\alpha)(1 + \cos\beta)(1 + \cos\gamma)$$ $$= 2(1 + \cos\alpha + \cos\beta + \cos\alpha\cos\beta)(1 + \cos\gamma)$$ $$= 2(1 + \cos\gamma + \cos\alpha + \cos\alpha\cos\gamma + \cos\beta + \cos\beta\cos\gamma + \cos\alpha\cos\beta + \cos\alpha\cos\beta\cos\gamma)$$ $$=2 + 2\cos\alpha + 2\cos\beta + 2\cos\gamma + 2\cos\alpha\cos\beta + 2\cos\alpha\cos\gamma + 2\cos\beta\cos\gamma + 2\cos\alpha\cos\beta\cos\gamma$$
The right-hand side expands to:
$$R := (\sin\alpha + \sin\beta + \sin\gamma)^2$$ $$=\sin^2\alpha + \sin\alpha\sin\beta + \sin\alpha\sin\gamma + \sin\beta\sin\alpha + \sin^2\beta + \sin\beta\sin\gamma + \sin\gamma\sin\alpha + \sin\gamma\sin\beta + \sin^2\gamma$$ $$=\sin^2\alpha + \sin^2\beta + \sin^2\gamma + 2\sin\alpha\sin\beta + 2\sin\alpha\sin\gamma + 2\sin\beta\sin\gamma$$
The difference between the two sides is:
$$L - R$$ $$=2 + 2\cos\alpha + 2\cos\beta + 2\cos\gamma + 2\cos\alpha\cos\beta + 2\cos\alpha\cos\gamma + 2\cos\beta\cos\gamma + 2\cos\alpha\cos\beta\cos\gamma - \sin^2\alpha - \sin^2\beta - \sin^2\gamma - 2\sin\alpha\sin\beta - 2\sin\alpha\sin\gamma - 2\sin\beta\sin\gamma$$ $$=2 + 2\cos\alpha + 2\cos\beta + 2\cos\gamma + 2(\cos\alpha\cos\beta - \sin\alpha\sin\beta ) + 2(\cos\alpha\cos\gamma - \sin\alpha\sin\gamma) + 2(\cos\beta\cos\gamma - \sin\beta\sin\gamma) + 2\cos\alpha\cos\beta\cos\gamma - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$ $$=2 + 2\cos\alpha + 2\cos\beta + 2\cos\gamma + 2\cos(\alpha + \beta) + 2\cos(\alpha + \gamma) + 2\cos(\beta + \gamma) + 2\cos\alpha\cos\beta\cos\gamma - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$
But since all triangles have $\alpha + \beta + \gamma = \pi$, then $\cos(\alpha + \beta) = \cos(\pi - \gamma) = -\cos(\gamma)$, and similarly for the other angle sums.
$$L - R = 2 + 2\cos\alpha + 2\cos\beta + 2\cos\gamma - 2\cos\gamma - 2\cos\beta - 2\cos\alpha + 2\cos\alpha\cos\beta\cos\gamma - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$ $$=2 + 2\cos\alpha\cos\beta\cos\gamma - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$ $$=2 + (\cos(\alpha + \beta) + \cos(\alpha - \beta))\cos\gamma - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$ $$=2 + \cos(\alpha + \beta)\cos\gamma + \cos(\alpha - \beta)\cos\gamma - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$ $$=2 + \frac{1}{2}\cos(\alpha + \beta + \gamma) + \frac{1}{2}\cos(\alpha + \beta - \gamma) + \frac{1}{2}\cos(\alpha - \beta + \gamma) + \frac{1}{2}\cos(\alpha - \beta - \gamma) - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$ $$=2 + \frac{1}{2}\cos(\pi) + \frac{1}{2}\cos(\pi - 2\gamma) + \frac{1}{2}\cos(\pi - 2\beta) + \frac{1}{2}\cos(2\alpha - \pi) - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$ $$=2 - \frac{1}{2} - \frac{1}{2}\cos(2\gamma) - \frac{1}{2}\cos(2\beta) - \frac{1}{2}\cos(2\alpha) - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$ $$=\frac{3}{2} - \frac{1}{2}(1 - 2\sin^2\gamma) - \frac{1}{2}(1 - 2\sin^2\beta) - \frac{1}{2}(1 - 2\sin^2\alpha) - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$ $$=\frac{3}{2} - \frac{1}{2} + \sin^2\gamma - \frac{1}{2} + \sin^2\beta - \frac{1}{2} + \sin^2\alpha - \sin^2\alpha - \sin^2\beta - \sin^2\gamma$$ $$= 0$$
Since $L - R = 0$, $L = R$, Q.E.D.
