If $(1+ \cos A) (1+ \cos B) (1+\cos C)= y = (1- \cos A) (1-\cos B) (1-\cos C)$ then prove that $y = \pm \sin A \sin B \sin C$

Question

If $$(1+ \cos A) (1+ \cos B) (1+\cos C)= y = (1- \cos A) (1-\cos B) (1-\cos C)$$ then prove that $$y = \pm \sin A \sin B \sin C$$

My Work: If $y=\prod(1+\cos A)=\prod(1-\cos A)$

$y^2=y\cdot y=\prod(1+\cos A)\cdot\prod(1-\cos A)$

How, should I move further? I am not getting any idea.

Answer 1

You've essentially solved it. Note $\prod(1+\cos A)\cdot\prod(1-\cos A) = \prod((1+\cos A) (1 - \cos A)) = \prod (1 - \cos^2 A) = \prod \sin^2 A$.

Therefore, you can finish by showing $y^2 = \prod \sin^2 A$, and the claim follows.

Hope this helps!

Answer 2

Hint: $$\sin \theta = \pm \sqrt{1- \cos^2 \theta}$$