It's difficult for me to be sure what function you are asking about, because of the limitations of answers.com.
I am considering y = -1 + csc (x)
csc (x) = 1/sin(x) and sin is a periodic function of x.
Ignoring what happens for negative values of sin(x), csc(x) is at local minima for maximal values of sin(x), which occur at x = (2k+1/2)pi for i any integer.
Putting the -1 into -1 + csc(x) simply 'lowers' the function without changing the positions of these minima.
PS: Incidentally, using GeoGebra can be a big help in solving problems like these. Free. Easy to use.
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sinx cscx = 1 is the same thing as sinx(1/sinx) = 1 which is the same as sinx/sinx = 1. This evaluates to 1=1, which is true.
Trig functions have their own special derivatives that you will have to memorize. For instance: the derivative of sinx is cosx. The derivative of cosx is -sinx The derivative of tanx is sec2x The derivative of cscx is -cscxcotx The derivative of secx is secxtanx The derivative of cotx is -csc2x
(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx
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According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.