if cos-1x + cos-1y + cos-1z = pi prove that x2 + y2 + z2 +2xyz = 1 - Maths - Inverse Trigonometric Functions

Question

if cos-1x + cos-1y + cos-1z =
pi
prove that x2 + y2 + z2 +2xyz =
1

Ajanta Trivedi · Accepted Answer

the given equation is :

$\cos^{- 1} x + \cos^{- 1} y + \cos^{- 1} z = π$ ...........(1) $- 1 \leq x, y, z \leq 1$

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

since $\cos^{- 1} A + \cos^{- 1} B = \cos^{- 1} [A B - \sqrt{1 - A^{2}} \cdot \sqrt{1 - B^{2}}]$ where $- 1 \leq A, B \leq 1$

therefore

$\cos^{- 1} [x y - \sqrt{1 - x^{2}} \cdot \sqrt{1 - y^{2}}] = π - \cos^{- 1} z \cos {\cos^{- 1} [x y - \sqrt{1 - x^{2}} \cdot \sqrt{1 - y^{2}}]} = \cos {π - \cos^{- 1} z} x y - \sqrt{1 - x^{2}} \cdot \sqrt{1 - y^{2}} = - \cos {\cos^{- 1} z} [∵ \cos {π - θ} = - \cos θ] x y - \sqrt{1 - x^{2}} \cdot \sqrt{1 - y^{2}} = - z$

i.e $x y + z = \sqrt{1 - x^{2}} \cdot \sqrt{1 - y^{2}}$

squaring on both sides:

$(x y + z)^{2} = (1 - x^{2}) \cdot (1 - y^{2}) x^{2} y^{2} + z^{2} + 2 x y z = 1 - x^{2} - y^{2} + x^{2} y^{2} x^{2} + y^{2} + z^{2} + 2 x y z = 1$

hope this helps you.

cheers!!