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anonymous
 4 years ago
Fool's problem of the day,
If \( \cos x +\cos y +\cos z =0 \) and \(\sin x +\sin y +\sin z =0 \), then find the value of \( \cos(xy) + \cos (yz) + \cos (zx) \)
PS: This is high school level trigonometry. Enjoy!
anonymous
 4 years ago
Fool's problem of the day, If \( \cos x +\cos y +\cos z =0 \) and \(\sin x +\sin y +\sin z =0 \), then find the value of \( \cos(xy) + \cos (yz) + \cos (zx) \) PS: This is high school level trigonometry. Enjoy!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\cos{x}+\cos{y}+\cos{z}=0\quad,\quad\sin{x}+\sin{y}+\sin{z}=0\]\[\cos{x}+\cos{y}=\cos{z}\quad,\quad\sin{x}+\sin{y}=\sin{z}\]\[(\cos{x}+\cos{y})^2=\cos^2{z}\quad,\quad(\sin{x}+\sin{y})^2=\sin^2{z}\]\[(\cos{x}+\cos{y})^2+(\sin{x}+\sin{y})^2=\cos^2{z}+\sin^2{z}=1\]\[2+2\cos{(xy)}=1\]\[\Rightarrow\quad\cos{(xy)}=\cos{(yz)}=\cos{(zx)}=1/2\]\[\Rightarrow\quad\cos{(xy)}+\cos{(yz)}+\cos{(zx)}=3/2\]
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