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Check my math.

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1. \[\cos (\cos^{-1} \frac{ 1 }{ 2 })=\frac{ 1 }{ 2 }\]
cos and cos-1 are inverses of each other. inverses cancel out. so yes 1 is correct
2.\[\cos (\cos^{-1} 2)=2 \]

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Other answers:

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