solving trigonometrical equations
posted by Jen .
arctan(tan(2pi/3)
thanks.
arctan(tan(2pi/3) = pi/3
since arctan and tan are inverse operations, the solution would be 2pi/3
the number of solutions to arctan(x) is infinite, look at its graph.
generally, unless a general solution is asked for, or a domain is specified,
arc(trigratio) answers are given in the domain 0 <> 2pi
Using the CAST rule, two solutions are then possible, 2pi/3 and 5pi/3
when punched in a calculator, your machine will blindly give you pi/3, because it cannot do the analysis, and is programmed to give the closest answer to zero.
I have to disagree with Reiny here. The arctan function is conventionally defined (and you have to choose some definition) such that arctan(tan(x) = x when x is between pi/2 to pi/2.
Given this definition of arctan, the answer is pi/3. Note that you are applying some given function to the tan and that function does not know what went in the tan.
A similar case is that of the squareroot function. One conventionally defines the squareroot to be the positve root, not the negative root. So, it's wrong to say that
sqrt[(1)^2] = 1
The inverse trignometric functions, square roots etc. can all be expressed as logarithms. Once a definition of the logarithm is chosen, e.g. by putting the branch cut on the negative real axis, all the other functions are defined.
It is very easy to make mistakes by using inconsistent definitions of the inverse trignometric function. There is a well known example in theoretical physics were some anomalous behavior was predicted for some capture cross section. Years later it transpired that it was all due to a simple artifact of wrongly defined inverse functions.
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