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nphuongsun93

  • 2 years ago

\[ \cos(x) + \cos(y) - \cos (x+y) = \frac{3}{2}\]

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  1. Zekarias
    • 2 years ago
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    Then what?

  2. nphuongsun93
    • 2 years ago
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    solve for y, sorry

  3. nphuongsun93
    • 2 years ago
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    gonna read this later, bbl 2 hours *-*

  4. lgbasallote
    • 2 years ago
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    <--will reply in 2 hours

  5. RadEn
    • 2 years ago
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    i think not enough information for solve it

  6. mathslover
    • 2 years ago
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    @nphuongsun93 What you tried? (just asking)

  7. mathslover
    • 2 years ago
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    \(\cos x + \cos y - \cos x \cos y + \sin x \sin y=\frac{3}{2}\)

  8. mathslover
    • 2 years ago
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    @asnaseer sir any ideas?

  9. mathslover
    • 2 years ago
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    @hartnn

  10. Coolsector
    • 2 years ago
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    i thought of maybe using cos(a) = [e^(ia) + e^(-ia)] / 2 but i dont think it works either

  11. asnaseer
    • 2 years ago
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    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:\[\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})-\cos^2(\frac{x+y}{2})=\frac{1}{4}\]

  12. asnaseer
    • 2 years ago
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    which is a quadratic in \(\cos(\frac{x+y}{2})\) - although it also contains a \(\cos(\frac{x-y}{2})\)

  13. asnaseer
    • 2 years ago
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    Solving the quadratic leads to:\[\cos(\frac{x+y}{2})=\frac{\cos(\frac{x-y}{2})\pm\sqrt{\cos^2(\frac{x-y}{2})-1}}{2}\]

  14. asnaseer
    • 2 years ago
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    Which simplifies to:\[\cos(\frac{x+y}{2})=\frac{\cos(\frac{x-y}{2})\pm\sin(\frac{x-y}{2})}{2}\]

  15. asnaseer
    • 2 years ago
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    Don't know if this is helpful or not but I have to go now - good luck!

  16. nphuongsun93
    • 2 years ago
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    Thanks a lot @asnaseer i get it, that's just what i needed

  17. mathslover
    • 2 years ago
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    @nphuongsun93 , what and how did you get the answer?

  18. nphuongsun93
    • 2 years ago
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    \[\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})-\cos^2(\frac{x+y}{2})-0.25 = 0\] 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 \[b= \cos(\frac{x-y}{2})=1 \space or \space \frac{x-y}{2}=0 \space or \space x=y\]

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