anonymous
  • anonymous
\[ \cos(x) + \cos(y) - \cos (x+y) = \frac{3}{2}\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Then what?
anonymous
  • anonymous
solve for y, sorry
anonymous
  • anonymous
gonna read this later, bbl 2 hours *-*

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

lgbasallote
  • lgbasallote
<--will reply in 2 hours
RadEn
  • RadEn
i think not enough information for solve it
mathslover
  • mathslover
@nphuongsun93 What you tried? (just asking)
mathslover
  • mathslover
\(\cos x + \cos y - \cos x \cos y + \sin x \sin y=\frac{3}{2}\)
mathslover
  • mathslover
@asnaseer sir any ideas?
mathslover
  • mathslover
@hartnn
anonymous
  • anonymous
i thought of maybe using cos(a) = [e^(ia) + e^(-ia)] / 2 but i dont think it works either
asnaseer
  • asnaseer
Maybe try this:\[\cos(x)+\cos(y)=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})\]and:\[\cos(x+y)=2\cos^2(\frac{x+y}{2})-1\]That will lead to:\[\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})-\cos^2(\frac{x+y}{2})=\frac{1}{4}\]
asnaseer
  • asnaseer
which is a quadratic in \(\cos(\frac{x+y}{2})\) - although it also contains a \(\cos(\frac{x-y}{2})\)
asnaseer
  • asnaseer
Solving the quadratic leads to:\[\cos(\frac{x+y}{2})=\frac{\cos(\frac{x-y}{2})\pm\sqrt{\cos^2(\frac{x-y}{2})-1}}{2}\]
asnaseer
  • asnaseer
Which simplifies to:\[\cos(\frac{x+y}{2})=\frac{\cos(\frac{x-y}{2})\pm\sin(\frac{x-y}{2})}{2}\]
asnaseer
  • asnaseer
Don't know if this is helpful or not but I have to go now - good luck!
anonymous
  • anonymous
Thanks a lot @asnaseer i get it, that's just what i needed
mathslover
  • mathslover
@nphuongsun93 , what and how did you get the answer?
anonymous
  • anonymous
\[\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})-\cos^2(\frac{x+y}{2})-0.25 = 0\] then let \[a = \cos(\frac{x+y}{2}) \space and \space b = \cos ( \frac{x-y}{2})\] \[ab - a^2 - 0.25=0\] 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((x-y)/2) is 1 (angles), so the only real solution is b =1 or b^2 -1 = 0 \[b= \cos(\frac{x-y}{2})=1 \space or \space \frac{x-y}{2}=0 \space or \space x=y\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.