April 18th, 2018, 05:09 AM |
#1 |

Newbie Joined: Apr 2018 Posts: 2 Thanks: 0 | Pove
Hi how can prove cos-1(-x)=-cos-1x Sent from my iPhone using Tapatalk |

April 18th, 2018, 07:14 AM |
#2 |

Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 |
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. |

April 18th, 2018, 09:23 AM |
#3 |

Math Team Joined: Jul 2011 From: Texas Posts: 2,761 Thanks: 1416 | |

April 18th, 2018, 09:45 AM |
#4 |

Newbie Joined: Apr 2018 Posts: 2 Thanks: 0 | |

April 18th, 2018, 10:05 AM |
#5 |

Senior Member Joined: Sep 2015 From: USA Posts: 2,042 Thanks: 1065 |
$\cos^{-1}(-x) = -\cos^{-1}(x)$ $-x = cos(-\cos^{-1}(x))$ 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. |

April 18th, 2018, 10:10 AM |
#6 |

Math Team Joined: Jul 2011 From: Texas Posts: 2,761 Thanks: 1416 |
evaluate the cosine of each side ... $\cos[\cos^{-1}(-x)] = -x$ $\cos[\pi - \cos^{-1}(x)] = \cos(\pi)\cos[\cos^{-1}(x)] + \sin(\pi)\sin[\cos^{-1}(x)] = (-1)(x) + 0 = -x$ |