Question

Given that: $(1+\cos \alpha )(1+\cos \beta )(1+\cos \gamma )=(1-\cos \alpha )(1-\cos \beta )(1-\cos \gamma )$. Show that one of the values of each member of his equality is $\sin \alpha \sin \beta \sin \gamma .$

Answer 1

We have: $(1+\cos \alpha )(1+\cos \beta )(1+\cos \gamma )$

$(1-\cos \alpha )(1-\cos \beta )(1-\cos \gamma )$

Multiplying both sides by

$(1+\cos \alpha )(1+\cos \beta )(1+\cos \gamma )$, we get

$(1+\cos \alpha {)}^2(1+\cos \beta {)}^2(1+\cos \gamma {)}^2$

$(1-\cos \alpha )(1-\cos \beta )(1-\cos \gamma )(1+\cos \alpha )(1+\cos \beta )(1+\cos \gamma )$

$\Rightarrow (1+\cos \alpha {)}^2(1+\cos \beta {)}^2(1+\cos \gamma {)}^2$

$=(1-{\cos }^2\alpha )(1-{\cos }^2\beta )(1-{\cos }^2\gamma )$

$\Rightarrow (1+\cos \alpha {)}^2(1+\cos \beta {)}^2(1+\cos \gamma {)}^2={\sin }^2\alpha {\sin }^2\beta {\sin }^2\gamma$

$\Rightarrow (1+\cos \alpha )(1+\cos \beta )(1+\cos \gamma )=\pm \sin \alpha \sin \beta \sin \gamma$

Hence, one of the values of $(1+\cos \alpha )(1+\cos \beta )(1+\cos \gamma )$ is $\sin \alpha \sin \beta \sin \gamma$