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sin(sin-1π) = ? i do not remember our lesson on this from friday.
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first off, you don't need to bother with the one in front of pi, and \[\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\] keep that one handy. so use this expansion
sorry that isn't a one, thats inverse sin
oh well when you have \[sinx = a \] how do you get the value of x? you take the inverse, so that means \[\arcsin(sinx) =sina\] \[acrsin(sinx)=x = sina\]
so it = π?
it is not in -π/2 to π/2 though
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yes it is :) from -pi/2 to pi/2 is a full rotation, which is going to include pi
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