anonymous
  • anonymous
Cos^2(x)+cos^2(y)+cos^2(z)=1 prove the triangle is right
Mathematics
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anonymous
  • anonymous
Cos^2(x)+cos^2(y)+cos^2(z)=1 prove the triangle is right
Mathematics
schrodinger
  • schrodinger
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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.

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Loser66
  • Loser66
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anonymous
  • anonymous
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 \(\frac{\pi}{2}\)
anonymous
  • anonymous
you were not here, and I gave the answer directly, sry

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anonymous
  • anonymous
@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
anonymous
  • anonymous
Here
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anonymous
  • anonymous
Sorry i cut a part cause it didnt fit
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anonymous
  • anonymous
Start with the 2nd
anonymous
  • anonymous
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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