math question: arccos4
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math question: arccos4
can someone explain to me how to evaluate expressions like arccos4 and arcsin2?
I don't mean on my calculator, i know how to do that, i want to know how to
answer an examination question that expects me to show my working.
The questions are like:
Find the following in the form a+ib (where a and b are real):
1) arccos 4
2) arcsin 2
3) arcsin i
The answers given in my text-book (with no explanations) are:
1) 2m*pi +- i arcosh 4
2) (4m + 1) * pi/2 +- i arcosh 2
3) n*pi + i arsinh[(-1)^n]
I've tried doing the sums as follows (for number 1):
let arccos 4 = z
therefore, cos z = 4
or cosh iz = 4
or iz = arcosh 4
or z = -i*arcosh 4
>From here, i can put the solution into the formula for the general solution of
cos A = C, and that seem to give me the books answer.
But if i do this for sum no. 2, i get
z = -i * arsinh(-2i), which when put into the general solution of sin A = C,
gives me
z = n*pi +[(-1)^n]*(-i * arsinh(-2i)), which is not the books answer.
What should i do? the exam is in 14 days...
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