Question

Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)

Show that one of the values of each member of this equality is sin α sin β sin γ

Answer 1

Given (1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)

Let us assume that

(1 + cos α)(1 + cos β)(1 + cos γ) = (1 -cos α)(1 - cos β)(1 - cos γ) = L

Weknow that `sin^2 theta + cos^2 theta = 1`

Then, we have

L X L = (1 + cos α)(1 +_ cos β)(1 + cos γ) x (1 - cos α)(1 - cos β)(1 - cos γ)

=> :^2 = {(1 - cos α)(1 - cos α)}{(1 + cos β)(1 - cos β)}{(1 + cos γ)(1 - cos γ)}

`=> L^2 = (1 - cos^2 α )(1 - cos^2 β)(1 - cos^2 γ)`

`=> L^2 = sin^2 α sin^2 β sin^2 γ`

`=> L = +- sin α sin β sin γ`

Therefore, we have

`(1 + cos α)(1 + cos β)(1 + cos γ) = (1 - cos α)(1 - cos β)(1 - cos γ) = +- sin α sin β sin γ`

Taking the expression with the positive sign, we have

`(1  + cos α)(1 + cos β)(1 + cos γ) = (1 - cos α)(1 - cos β)(1 - cos γ) = sin α  sin β  sin γ`