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GrazesBest ResponseYou've already chosen the best response.0
1. \[\cos (\cos^{1} \frac{ 1 }{ 2 })=\frac{ 1 }{ 2 }\]
 one year ago

funinaboxBest ResponseYou've already chosen the best response.0
cos and cos1 are inverses of each other. inverses cancel out. so yes 1 is correct
 one year ago

GrazesBest ResponseYou've already chosen the best response.0
2.\[\cos (\cos^{1} 2)=2 \]
 one year ago

funinaboxBest ResponseYou've already chosen the best response.0
same reasoning as 1. inverses cancel out...
 one year ago

GrazesBest ResponseYou've already chosen the best response.0
But I recall that my teacher told me that this is not the case in certain circumstances.
 one year ago

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

funinaboxBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

ZeHanzBest ResponseYou've already chosen the best response.0
cos^1(2) is undefined, so cos(cos^1(2)) is also undefined.
 one year ago

funinaboxBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

ZeHanzBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

ZeHanzBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

funinaboxBest ResponseYou've already chosen the best response.0
woops yeah i was definitely wrong about that part i blame it on st paddy day stupor
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.0
WA works with complex numbers as a default. Many people don't.
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.0
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 \(e^{iz}\): \((e^{iz})^2+1=4e^{iz} \Leftrightarrow (e^{iz})^24e^{iz}+1=0\) 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.
 one year ago
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