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Grade: 12th pass
        
Cos3A+Cos3B+Cos3C+Cos3π=0 then the least value of sum of two angles is
 
 
7 months ago

Answers : (1)

Arun
22761 Points
							

Dear student,

 

COS3B+COS3C+COS3A=1
COS3B+COS3C+COS3A-1=0
COS3(B+C)/2.COS3(B-C)/2-SIN^3A/2=0
SOLVING WE GET
COS3C/2.COS3B/2.COS3A/2=0
EITHER3C/2=pi
C= 2 pi/3     
 
Hence A+B = pi/3
 
Hope it helps
 
Regards
Arun (askIITians forum expert)
7 months ago
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