Question

Check my math.

Answer 1

1. \[\cos (\cos^{-1} \frac{ 1 }{ 2 })=\frac{ 1 }{ 2 }\]

Answer 2

cos and cos-1 are inverses of each other. inverses cancel out. so yes 1 is correct

Answer 3

2.\[\cos (\cos^{-1} 2)=2 \]

Answer 4

same reasoning as 1.

inverses cancel out...

Answer 5

But I recall that my teacher told me that this is not the case in certain circumstances.

Answer 6

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

Answer 7

I'll ask him tomorrow.

Answer 8

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.

Answer 9

the very defining feature of an inverse is (x,y) equals (y,x) in it's inverse.

therefore, they will always cancel.

Answer 10

cos^-1(2) is undefined, so cos(cos^-1(2)) is also undefined.

Answer 11

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.

Answer 12

\[
\cos^{-1}:[-1,1]\mapsto [0,\pi]\\
\sin^{-1}:[-1,1]\mapsto [-\pi/2,\pi/2]
\]

Answer 13

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.

Answer 14

you don't have to calculate arccos of 2. is my entire reasoning here.

Answer 15

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.

Answer 16

See what I mean: http://www.wolframalpha.com/input/?i=arccos+2+
I rest my case.

Answer 17

woops yeah i was definitely wrong about that part

i blame it on st paddy day stupor

Answer 18

WA works with complex numbers as a default. Many people don't.

Answer 19

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.