Problem on Compound Angles

Thanks Nishant bhaiyya for the hint...

Here's the solution of Q.2:

cos(β-γ) + cos(γ-α) + cos(α-β) = -32

or, 2cos(β-γ) + 2cos(γ-α) + 2cos(α-β) +3 = 0

or, (2cosβcosγ + 2sinβsinγ) + (2cosγcosα + 2sinγsinα) + (2cosαcosβ + 2sinαsinβ) + (1+1+1) = 0

or, (2cosβcosγ + 2sinβsinγ) + (2cosγcosα + 2sinγsinα) + (2cosαcosβ + 2sinαsinβ) + (sin²Î± + cos²Î±) + (sin²Î² + cos²Î²) + (sin²Î³ + cos²Î³) = 0

or, (sin²Î± + sin²Î² + sin²Î³ + 2.sinα.sinβ + 2.sinβ.sinγ + 2.sinγ.sinα )
+(cos²Î± + cos²Î² + cos²Î³ + 2.cosα.cosβ + 2.cosβ.cosγ + 2.cosγ.cosα) = 0

or, (sinα + sinβ +sinγ)² + (cosα + cosβ + cosγ)² = 0

Since a square quantity is always +ve or zero,
therefore,
(sinα + sinβ +sinγ)² = 0
AND
(cosα + cosβ + cosγ)² = 0

Kick the squares from the two equations and you have the required proof....

1. sin A_max + sin B_max= 2
1+1=2
sin A=sin B=1
A=B=90°
so, sin(A+B)=sin 180°=0