anonymous
  • anonymous
Check my math.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1. \[\cos (\cos^{-1} \frac{ 1 }{ 2 })=\frac{ 1 }{ 2 }\]
anonymous
  • anonymous
cos and cos-1 are inverses of each other. inverses cancel out. so yes 1 is correct
anonymous
  • anonymous
2.\[\cos (\cos^{-1} 2)=2 \]

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anonymous
  • anonymous
same reasoning as 1. inverses cancel out...
anonymous
  • anonymous
But I recall that my teacher told me that this is not the case in certain circumstances.
anonymous
  • anonymous
that doesn't really make sense. in all circumstances, an equation and it's inverse will cancel. it is a property of inverse equations.
anonymous
  • anonymous
I'll ask him tomorrow.
anonymous
  • anonymous
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.
anonymous
  • anonymous
the very defining feature of an inverse is (x,y) equals (y,x) in it's inverse. therefore, they will always cancel.
ZeHanz
  • ZeHanz
cos^-1(2) is undefined, so cos(cos^-1(2)) is also undefined.
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[ \cos^{-1}:[-1,1]\mapsto [0,\pi]\\ \sin^{-1}:[-1,1]\mapsto [-\pi/2,\pi/2] \]
ZeHanz
  • ZeHanz
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.
1 Attachment
anonymous
  • anonymous
you don't have to calculate arccos of 2. is my entire reasoning here.
ZeHanz
  • ZeHanz
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.
ZeHanz
  • ZeHanz
See what I mean: http://www.wolframalpha.com/input/?i=arccos+2+ I rest my case.
anonymous
  • anonymous
woops yeah i was definitely wrong about that part i blame it on st paddy day stupor
ZeHanz
  • ZeHanz
WA works with complex numbers as a default. Many people don't.
ZeHanz
  • ZeHanz
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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