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Solve the following.
If sin⁻¹ x + sin⁻¹ y + sin⁻¹ z = π, show that x² - y² - z² + 2yz√(1 - x²) = 0
If $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi$, show that $x^2 - y^2 - z^2 + 2yz\sqrt{1 - x^2} = 0$
Given: $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi$
To show: $x^2 - y^2 - z^2 + 2yz\sqrt{1 - x^2} = 0$
Answer
Answer
To show $x^2 - y^2 - z^2 + 2yz\sqrt{1-x^2} = 0$ when $\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \pi$.
- Let $A = \sin^{-1}x$, $B = \sin^{-1}y$, $C = \sin^{-1}z$, so $A + B + C = \pi$ and $B + C = \pi - A$.
- Take cosine of both sides: $\cos(B + C) = \cos(\pi - A) = -\cos A$.
- $\cos B\cos C - \sin B\sin C = -\cos A$, by the cosine addition formula.
- $\sin A = x,\ \sin B = y,\ \sin C = z$, so $\cos A = \sqrt{1-x^2},\ \cos B = \sqrt{1-y^2},\ \cos C = \sqrt{1-z^2}$.
- $\sqrt{1-y^2}\sqrt{1-z^2} - yz = -\sqrt{1-x^2}$, so $\sqrt{1-y^2}\sqrt{1-z^2} = yz - \sqrt{1-x^2}$.
- Square both sides: $(1-y^2)(1-z^2) = y^2z^2 - 2yz\sqrt{1-x^2} + 1 - x^2$.
- $1 - y^2 - z^2 + y^2z^2 = y^2z^2 - 2yz\sqrt{1-x^2} + 1 - x^2$.
- Cancel $1$ and $y^2z^2$: $-y^2 - z^2 = -2yz\sqrt{1-x^2} - x^2$.
- Hence $x^2 - y^2 - z^2 + 2yz\sqrt{1-x^2} = 0$.