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Grazes

  • 3 years ago

Check my math.

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  1. Grazes
    • 3 years ago
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    1. \[\cos (\cos^{-1} \frac{ 1 }{ 2 })=\frac{ 1 }{ 2 }\]

  2. funinabox
    • 3 years ago
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    cos and cos-1 are inverses of each other. inverses cancel out. so yes 1 is correct

  3. Grazes
    • 3 years ago
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    2.\[\cos (\cos^{-1} 2)=2 \]

  4. funinabox
    • 3 years ago
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    same reasoning as 1. inverses cancel out...

  5. Grazes
    • 3 years ago
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    But I recall that my teacher told me that this is not the case in certain circumstances.

  6. funinabox
    • 3 years ago
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    that doesn't really make sense. in all circumstances, an equation and it's inverse will cancel. it is a property of inverse equations.

  7. Grazes
    • 3 years ago
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    I'll ask him tomorrow.

  8. funinabox
    • 3 years ago
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    unless he was referring to restricting domain and range. in order to create an inverse, it must pass the horizontal line test. that means for every x, there is a unique f(x). in this sense, equations and there inverses are different.

  9. funinabox
    • 3 years ago
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    the very defining feature of an inverse is (x,y) equals (y,x) in it's inverse. therefore, they will always cancel.

  10. ZeHanz
    • 3 years ago
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    cos^-1(2) is undefined, so cos(cos^-1(2)) is also undefined.

  11. funinabox
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=cos%28arccos%282%29 doesn't matter @ZeHanz the cos and it's inverse cancel out before you even do anything. imagine they aren't even there, basically.

  12. wio
    • 3 years ago
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    \[ \cos^{-1}:[-1,1]\mapsto [0,\pi]\\ \sin^{-1}:[-1,1]\mapsto [-\pi/2,\pi/2] \]

  13. ZeHanz
    • 3 years ago
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    No that's not true. See image attached. You cannot calculate the inverse cosine of 2, which is what you would have to do here. WolframAlpha can do it, because they use complex numbers, not real numbers.

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  14. funinabox
    • 3 years ago
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    you don't have to calculate arccos of 2. is my entire reasoning here.

  15. ZeHanz
    • 3 years ago
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    So the reason they would cancel is not that they do beforehand, but it is because you CAN calculate the inverse cosine of 2, when complex numbers (imaginary or real) are allowed. If that is the case, there is nothing wrong.

  16. ZeHanz
    • 3 years ago
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    See what I mean: http://www.wolframalpha.com/input/?i=arccos+2+ I rest my case.

  17. funinabox
    • 3 years ago
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    woops yeah i was definitely wrong about that part i blame it on st paddy day stupor

  18. ZeHanz
    • 3 years ago
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    WA works with complex numbers as a default. Many people don't.

  19. ZeHanz
    • 3 years ago
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    How to calculate \(\cos^{-1}2?\) Solve \(\cos z=2\). Now \(\cos z=\dfrac{e^{iz}+e^{-iz}}{2}=2\), so \(e^{iz}+e^{-iz}=4\). Multiply with \(e^{iz}\): \((e^{iz})^2+1=4e^{iz} \Leftrightarrow (e^{iz})^2-4e^{iz}+1=0\) Solve it with the Quadratic Formula: \(e^{iz}=\dfrac{4 \pm \sqrt{16-4}}{2}=2\pm\sqrt{3}\). \(iz=\ln(2\pm\sqrt{3})\), so \(z=\frac{1}{i}\ln(2\pm\sqrt{3})=-i\ln(2\pm\sqrt{3})\). One of these simplifies to 1.316957897i, just as the solution of WA.

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