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Geometry Level 3

If cos 1 x + cos 1 y + cos 1 z = π \cos^{-1}x+\cos^{-1}y+\cos^{-1}z=\pi , then find the value of x 2 + y 2 + z 2 + 2 x y z x^{2}+y^{2}+z^{2}+2xyz .

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1 solution

Samrit Pramanik
Apr 25, 2015

cos 1 x + cos 1 y \cos^{-1}x\ + \cos^{-1}y = = π cos 1 z \pi\ - \cos^{-1}z

cos 1 ( x y 1 x 2 1 y 2 ) \cos^{-1}(xy - \sqrt{1 - x^2}\sqrt{1 - y^2}) = = π cos 1 z \pi\ - \cos^{-1}z

x y 1 x 2 1 y 2 xy - \sqrt{1 - x^2}\sqrt{1 - y^2} = = cos ( π cos 1 z ) \cos (\pi - \cos^{-1}z)

1 x 2 1 y 2 \sqrt{1 - x^2}\sqrt{1 - y^2} = = cos π \cos \pi cos ( cos 1 z ) \cos (\cos^{-1}z) - sin π \sin \pi sin ( cos 1 z ) \sin (\cos^{-1}z)

x y 1 x 2 1 y 2 = z xy - \sqrt{1 - x^2}\sqrt{1 - y^2}\ = - z

( x y + z ) 2 = ( 1 x 2 1 y 2 ) 2 (xy+z)^{2} = (\sqrt{1 - x^2}\sqrt{1 - y^2})^{2}

x 2 y 2 + 2 x y z + z 2 = 1 x 2 + x 2 y 2 y 2 x^2y^2 + 2xyz + z^2 = 1 - x^2 + x^2y^2 - y^2

x 2 + y 2 + z 2 + 2 x y z = 1 \boxed{x^2 + y^2 + z^2 + 2xyz = 1}

Your method is better than the one which i used :) nice!!

Prashant Kr - 6 years, 1 month ago

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Thank you !!!

Samrit Pramanik - 6 years, 1 month ago

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