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Grazes
 2 years ago
Check my math.
Grazes
 2 years ago
Check my math.

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Grazes
 2 years ago
Best ResponseYou've already chosen the best response.01. \[\cos (\cos^{1} \frac{ 1 }{ 2 })=\frac{ 1 }{ 2 }\]

funinabox
 2 years ago
Best ResponseYou've already chosen the best response.0cos and cos1 are inverses of each other. inverses cancel out. so yes 1 is correct

Grazes
 2 years ago
Best ResponseYou've already chosen the best response.02.\[\cos (\cos^{1} 2)=2 \]

funinabox
 2 years ago
Best ResponseYou've already chosen the best response.0same reasoning as 1. inverses cancel out...

Grazes
 2 years ago
Best ResponseYou've already chosen the best response.0But I recall that my teacher told me that this is not the case in certain circumstances.

funinabox
 2 years ago
Best ResponseYou've already chosen the best response.0that doesn't really make sense. in all circumstances, an equation and it's inverse will cancel. it is a property of inverse equations.

funinabox
 2 years ago
Best ResponseYou've already chosen the best response.0unless 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.

funinabox
 2 years ago
Best ResponseYou've already chosen the best response.0the very defining feature of an inverse is (x,y) equals (y,x) in it's inverse. therefore, they will always cancel.

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0cos^1(2) is undefined, so cos(cos^1(2)) is also undefined.

funinabox
 2 years ago
Best ResponseYou've already chosen the best response.0http://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.

wio
 2 years ago
Best ResponseYou've already chosen the best response.0\[ \cos^{1}:[1,1]\mapsto [0,\pi]\\ \sin^{1}:[1,1]\mapsto [\pi/2,\pi/2] \]

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0No 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.

funinabox
 2 years ago
Best ResponseYou've already chosen the best response.0you don't have to calculate arccos of 2. is my entire reasoning here.

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0So 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
 2 years ago
Best ResponseYou've already chosen the best response.0See what I mean: http://www.wolframalpha.com/input/?i=arccos+2+ I rest my case.

funinabox
 2 years ago
Best ResponseYou've already chosen the best response.0woops yeah i was definitely wrong about that part i blame it on st paddy day stupor

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0WA works with complex numbers as a default. Many people don't.

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0How 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 \(e^{iz}\): \((e^{iz})^2+1=4e^{iz} \Leftrightarrow (e^{iz})^24e^{iz}+1=0\) Solve it with the Quadratic Formula: \(e^{iz}=\dfrac{4 \pm \sqrt{164}}{2}=2\pm\sqrt{3}\). \(iz=\ln(2\pm\sqrt{3})\), 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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