whenever you have problems like this, draw a right triangle. In this case, you want a triangle with one angle x. If the sides are a,b,c with b opposite angle x, and c the hypotenuse, then
sin(x) = b/c
Now, the angle whose cosine is b/c is the other acute angle of the triangle, namely (π/2 - x).
y=arccos(sin(x)), find dy/dx and sketch it's graph(I guess I can do this on wolframalpha after I'm done solving the question). And by arcsin I mean inverse of the expression(written like cos^-1(sin(x)), but is not 1/cos(sinx) but the inverse of cos).
I am not getting this inverse stuff very well, please give a step by step solution. Thanks in advance!!
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Okay, what you say makes sense, but what do I do with the (Pi/2)?
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