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anonymous

  • one year ago

Cos^2(x)+cos^2(y)+cos^2(z)=1 prove the triangle is right

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  1. Loser66
    • one year ago
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    *

  2. anonymous
    • one year ago
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    For a triangle \(x+y+z=\pi\) and we get\[\cos^2 x+\cos^2 y+\cos^2 z=1 \\ \cos^2 x+\cos^2 y+\cos^2 (\pi-x-y)=1 \\ \cos^2 x+\cos^2 y+\cos^2 (x+y)=1 \\ \cos^2 x+\cos^2 y=1-\cos^2 (x+y) \\ \cos^2 x+\cos^2 y=\sin^2 (x+y)=(\sin x \cos y+\cos x \sin y)^2 \\ \cos^2 x+\cos^2 y=\sin^2 x \cos^2 y+\cos^2 x \sin^2 y + 2 \sin x \sin y \cos x \cos y \\ \cos^2 x (1-\sin^2 y)+\cos^2 y(1-\sin^2 x)=2 \sin x \sin y \cos x \cos y \\ \cos^2 x \cos^2 y= \sin x \sin y \cos x \cos y \\ \cos x \cos y(\cos x \cos y-\sin x \sin y)=0 \\ \cos x \cos y \cos (x+y)=0 \\ \cos x \cos y \cos (\pi-x-y)=0 \\ \cos x \cos y \cos z=0\]so one of the angles must be equal to \(\frac{\pi}{2}\)

  3. anonymous
    • one year ago
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    you were not here, and I gave the answer directly, sry

  4. anonymous
    • one year ago
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    @mukushla no problem thx for the answer but man yesterday after posting it i continued thinking of it and after itching my mind alott yessss i knew it I was really happy 😂😂 look thats my prove

  5. anonymous
    • one year ago
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    Here

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  6. anonymous
    • one year ago
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    Sorry i cut a part cause it didnt fit

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  7. anonymous
    • one year ago
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    Start with the 2nd

  8. anonymous
    • one year ago
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    Quite right! Note that in the last part you cannot divide by \(2 \cos x \cos y\), see last part of my solution.

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