anonymous
  • anonymous
Cos^2(x)+cos^2(y)+cos^2(z)=1 prove the triangle is right
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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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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