Thread: Inverse Tangent
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Unread 25-01-2003, 00:05
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ArcTan

I did some searching and found this http://www.mcs.surrey.ac.uk/Personal...pi.html#arctan

arctan( t ) = t – (t^3)/3 + (t^5)/5 - (t^7)/7 + (t^9)/9 - (t^11)/11 .........

-1< t < 1

I tried it out its pretty close to what the real value is

x = tan (.5)
arctan(x) should equal .5

x – (x^3)/3 + (x^5)/5 - (x^7)/7 + (x^9)/9 - (x^11)/11 = 0.4999763883

But Surprisingle enough

x – (x^3)/3 + (x^5)/5 - (x^7)/7 + (x^9)/9 = 5.000939788

The More terms you do the more accurate your answer is

!The only problem you might run into in pbasic is memory and speed!

Cipher X
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