1. If sinA + sinB = 2 what is the value of sin(A+B) ?
2. If cos (\beta - \gamma ) + cos (\gamma - \alpha )+cos(\alpha - \beta )= - 3 / 2 show that cos\alpha + cos\beta + cos\gamma = 0 and sin \alpha + sin \beta + sin \gamma = 0
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3 Answers
62the first one is as simple as saying that both sin A and sin B are 1...
now try?
1Thanks 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) + (sin2α + cos2α) + (sin2β + cos2β) + (sin2γ + cos2γ) = 0
or, (sin2α + sin2β + sin2γ + 2.sinα.sinβ + 2.sinβ.sinγ + 2.sinγ.sinα )
+(cos2α + cos2β + cos2γ + 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....
11. sin Amax + sin Bmax= 2
1+1=2
sin A=sin B=1
A=B=90°
so, sin(A+B)=sin 180°=0