Question

Let ABC be a triangle. If cos2A + cos2B + cos2C = -1 then which one of the following is correct?
1. sin A sin B sin C = 0
2. sin A sin B cos C = 0
3. cos A sin B sin C = 0
4. cos A cos B cos C = 0

Answer 1

Correct Answer - Option 4 : cos A cos B cos C = 0

Formula used:

cos x + cos y = \(\rm 2 cos \frac{x + y}{2} cos \frac{x - y}{2}\)

1 + cos 2x = 2 cos²x

ABC is a trianlge, ∠A + ∠B + ∠C = 180° 

cos(A + B) + cos(A - B) = 2 cos A. cos B

Calculation:

cos 2A + cos 2B + cos 2C = -1

⇒ cos 2A + cos 2B + (1 + cos 2C) = 0

⇒ 2 cos(A + B) cos(A - B) + 2 cos² C = 0     ----(i)

From formula used ∠A + ∠B = 180° - ∠C  

cos (A + B) = cos (180° - C) = - cos C     ----(ii)

From equation (i) and (ii), we get 

⇒ 2(-cos C) cos(A - B)  + 2 cos² C = 0

⇒ -2 cos C [cos(A - B) -  cos C] = 0     ----(iii)

From (ii) and (iii), we get

⇒ -2 cos C [cos(A - B) + cos(A + B)] = 0

⇒ -2 cos C [cos(A + B) + cos(A - B)] = 0

⇒ -2 cos C  (2 cos A. cos B) = 0

∴ cos A cos B cos C = 0.