Question 17612: cosx+cosy-cos(x+y)=3/2
this is only possible if x=y=60
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! cosx+cosy-cos(x+y)=3/2..
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[actually from here you can straight away put the step
2cos((x+y)/2)cos((x-y)/2)-2[cos((x+y)/2))]^2-0.5=0...using cos 2a=2(cos a)^2-1,
and proceed further as shown below.but i am showing another elaboration ,in case ,this problem appears as one to prove in any triangle xyz,where sum of angles x+y+z=180...]
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cosx+cosy+cos(180-x-y)=3/2...
2cos((x+y)/2)cos((x-y)/2)+1-2[sin((180-x-y)/2)]^2-3/2=0
2cos((x+y)/2)cos((x-y)/2)-2[sin(90-((x+y)/2))]^2-0.5=0...
2cos((x+y)/2)cos((x-y)/2)-2[cos((x+y)/2))]^2-0.5=0...
for ease of simplification put cos((x+y)/2))=p and cos((x-y)/2))=b..we get
-2p^2+2pb-0.5=0..or dividing with -2 throughput p^2-pb+0.25=0
..taking this as a quadratic in p we find that the discriminant is b^2-1 which is to be positive to get real solution for p.but the highest value of........... b= cos((x-y)/2)) is 1 so the only possibility for getting a real solution is for b=1 or b^2-1=0..in such a case we get p=b/2.hence we have
b= cos((x-y)/2)=1 or (x-y)/2=0..or..x=y and
p=b/2=1/2=cos((x+y)/2)..putting x=y as obtained above,we get
coa((x+x)/2)=cos(x)=1/2...or x=60..so..x=y=60
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