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nphuongsun93Best ResponseYou've already chosen the best response.0
solve for y, sorry
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.0
gonna read this later, bbl 2 hours **
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
<will reply in 2 hours
 one year ago

RadEnBest ResponseYou've already chosen the best response.0
i think not enough information for solve it
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
@nphuongsun93 What you tried? (just asking)
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
\(\cos x + \cos y  \cos x \cos y + \sin x \sin y=\frac{3}{2}\)
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
@asnaseer sir any ideas?
 one year ago

CoolsectorBest ResponseYou've already chosen the best response.0
i thought of maybe using cos(a) = [e^(ia) + e^(ia)] / 2 but i dont think it works either
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
Maybe try this:\[\cos(x)+\cos(y)=2\cos(\frac{x+y}{2})\cos(\frac{xy}{2})\]and:\[\cos(x+y)=2\cos^2(\frac{x+y}{2})1\]That will lead to:\[\cos(\frac{x+y}{2})\cos(\frac{xy}{2})\cos^2(\frac{x+y}{2})=\frac{1}{4}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
which is a quadratic in \(\cos(\frac{x+y}{2})\)  although it also contains a \(\cos(\frac{xy}{2})\)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
Solving the quadratic leads to:\[\cos(\frac{x+y}{2})=\frac{\cos(\frac{xy}{2})\pm\sqrt{\cos^2(\frac{xy}{2})1}}{2}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
Which simplifies to:\[\cos(\frac{x+y}{2})=\frac{\cos(\frac{xy}{2})\pm\sin(\frac{xy}{2})}{2}\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.2
Don't know if this is helpful or not but I have to go now  good luck!
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.0
Thanks a lot @asnaseer i get it, that's just what i needed
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
@nphuongsun93 , what and how did you get the answer?
 one year ago

nphuongsun93Best ResponseYou've already chosen the best response.0
\[\cos(\frac{x+y}{2})\cos(\frac{xy}{2})\cos^2(\frac{x+y}{2})0.25 = 0\] then let \[a = \cos(\frac{x+y}{2}) \space and \space b = \cos ( \frac{xy}{2})\] \[ab  a^2  0.25=0\] taking this as a quadratic in p, b^2 1 has to be positive to get real solution for a, but the highest value of b = cos((xy)/2) is 1 (angles), so the only real solution is b =1 or b^2 1 = 0 \[b= \cos(\frac{xy}{2})=1 \space or \space \frac{xy}{2}=0 \space or \space x=y\]
 one year ago
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