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nphuongsun93
Group Title
\[ \cos(x) + \cos(y)  \cos (x+y) = \frac{3}{2}\]
 one year ago
 one year ago
nphuongsun93 Group Title
\[ \cos(x) + \cos(y)  \cos (x+y) = \frac{3}{2}\]
 one year ago
 one year ago

This Question is Closed

Zekarias Group TitleBest ResponseYou've already chosen the best response.0
Then what?
 one year ago

nphuongsun93 Group TitleBest ResponseYou've already chosen the best response.0
solve for y, sorry
 one year ago

nphuongsun93 Group TitleBest ResponseYou've already chosen the best response.0
gonna read this later, bbl 2 hours **
 one year ago

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

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

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

mathslover Group TitleBest 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

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

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@hartnn
 one year ago

Coolsector Group TitleBest 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

asnaseer Group TitleBest 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

asnaseer Group TitleBest 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

asnaseer Group TitleBest 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

asnaseer Group TitleBest 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

asnaseer Group TitleBest 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

nphuongsun93 Group TitleBest ResponseYou've already chosen the best response.0
Thanks a lot @asnaseer i get it, that's just what i needed
 one year ago

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

nphuongsun93 Group TitleBest 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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