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    Re: Using logs.
    From: Frank Reed
    Date: 2008 Apr 23, 21:20 -0400

    coralline algae writes:
    
    "at page 38 is the start of log tables for sin, cos and tan
    and what was done was to add the number 10 to the
    log10 result so no logarithms are negative.  I havent
    actually tried to use these tables as yet but I am
    guessing that this is another workaround as discussed
    in this thread?"
    
    That solves a different problem, also important. Let's consider a simple
    trig calculation. We want to work out the product x = tan(a)*sin(b) where
    a=-60 degrees and b=+15 degrees. Work it out with modern tools, a calculator
    or in software, and you will find that x is
     -0.448288. But let's do it with logarithms...
    
    The problem x=tan(a)*sin(b) is equivalent to x=invlog[logtan(a)+logsin(b)]
    where the function "invlog(z)" means 'find the number whose logarithm is
    equal to z' and is equivalent to 10^z. First, we look up the logtangent of
    60 degrees (ignoring the sign of the argument). That's 0.2385607. Next we
    look up the logsine of 15 degrees. That should be -0.5870038. Now here's the
    trick. Instead of dealing with subtraction, which is generally a little more
    error-prone than addition, we just add ten to this negative logarithm making
    it 9.4129962. That is, the folks who constructed the tables add this 10 so
    that we never see the negative value of the logarithm at all. If you think
    about it, this is equivalent to multiplying the original problem by ten
    billion (10 to the tenth power). We can get away with this trick as long a
    number this large would never appear in any of the calculations we might do.
    Then, we add the logarithms. The result is 9.6515569. If the result had been
    greater than ten, we would "drop the tens" (a common instruction in old
    navigation manuals) which we would today refer to as taking the result "mod
    10". We would then "seek" this logarithm in a table of standard logarithms
    and read out the result: 0.448288. And finally, we have to remember that
    negative sign that we ignored at the very beginning. So far we've done the
    calculation as if the arguments were both positive. But since the argument
    of the tangent was actually -60 degrees, the factor itself would be negative
    and therefore, so is the final answer: -0.448288. You can see that fussing
    over signs could get to be fairly complicated. The instructions for solving
    this problem in a traditional source might read something like this: "if the
    argument of the tangent is negative or the argument of the sine is negative
    (but not both negative), then the final result should be marked negative,
    otherwise positive". These conditional statements could get to be difficult
    to follow so practical mathematicians always looked for solutions to
    problems with a minimal number of possible signs.
    
     -FER
    
    
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