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PoveHi how can prove cos-1(-x)=-cos-1x Sent from my iPhone using Tapatalk |

If you cannot use Latex, at least use "^" to indicate the inverse function. You want to prove that cos^(-1)(x)= -cos^(-1)(-x) ($\displaystyle cos^{-1}(x)= -cos^{-1}(-x)$. Your basic problem is that you can't prove that, it isn't true!For example, if $\displaystyle x= \frac{\sqrt{2}}{2}$ then $\displaystyle cos^{-1}(x)= \frac{\pi}{4}$ radians (45 degrees). But the definition of the function $\displaystyle cos^{-1}(x)$ restricts the range to between 0 and $\displaystyle \pi$ (180 degrees). $\displaystyle cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)= \frac{3\pi}{4}$ (135 degrees), NOT $\displaystyle -\frac{\pi}{4}$ radians or -45 degrees.If you were to change "$\displaystyle cos^{-1}$" to "$\displaystyle sin^{-1}$" then it would be true. |

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$\cos^{-1}(-x) = \pi - \cos^{-1}(x)$ |

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Ok how can l prove it Sent from my iPhone using Tapatalk |

$\cos^{-1}(-x) = -\cos^{-1}(x)$ $-x = cos(-\cos^{-1}(x))$ the cosine is an even function so $\cos(-x)=\cos(x)$ thus $-x = \cos(\cos^{-1}(x)) = x$ and thus as Country Boy keenly noted your statement isn't true. |

evaluate the cosine of each side ... $\cos[\cos^{-1}(-x)] = -x$ $\cos[\pi - \cos^{-1}(x)] = \cos(\pi)\cos[\cos^{-1}(x)] + \sin(\pi)\sin[\cos^{-1}(x)] = (-1)(x) + 0 = -x$ |

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