Wasn't it James Taylor who wrote:
>"Christopher James Huff" <chr### [at] maccom> wrote in message
>> BTW, the usual solution using trig functions uses sin() and cos().
>
>if I remember right, tan()/sin() is cos()...

I always thought that sin()/tan() was cos().

  sin() = opposite/hypotenuse
  tan() = opposite/adjacent
  cos() = adjacent/hypotenuse

So sin()/tan(() = (o/h)/(o/a) = a/h = cos()

-- 
Mike Williams
Gentleman of Leisure

Post a reply to this message