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If cosx+cosy-cos(x+y)=3/2 then find the value of SinxSiny where x>0, y

If cosx+cosy-cos(x+y)=3/2 then find the value of SinxSiny where x>0, y
 

Grade:11

2 Answers

noogler
489 Points
5 years ago
cosx+cosy-cos(x+y)=3/2 is true if x=y=60o
cos60o+cos60o-cos120o=1/2+1/2-(-1/2)=3/2
SinxSiny=sin60o sin60o=3/4
sinx siny=3/4
 
approve if it is r8 and useful to u..........
Lab Bhattacharjee
121 Points
5 years ago
 
⟹22cos(x+y)/2cos(x-y)/2-[2cos^2(x+y)/2-1]=3/2
 
implies 4cos^2(x+y)/2-4cos(x+y)/2cos(x-y)/2+1=0
 
As cos(x+y)/2 is real, the discriminant  (4cos(x-y)/2)^2-4^2=-16sin^2(x-y)/2 must be >=0
 
So,  sin^2(x-y)/2=0 implies sin(x-y)/2=0 implies(x-y)/2=n pi, x-y=2n pi where n is an integer
 
So, (2cos(x+y)/2-1)^2=0, cos(x+y)/2=1/2=cos pi/3 implies  (x+y)/2=2m pi +/- pi/3 implies x+y=4n pi +/- 2pi/3
 
Solve for  x,y

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