I have found formula's for the sine & cosine of an angle
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ...
and
cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! ...
Is there a similar formula for the tangent(x)?
OK. That is a lot more complicated than the formula I know for it.
cosine of angle C=c^2 - a^2 - b^2 + 2ab
Where a, b, and c are the lengths of the sides of the triangle.
----------------
Noah Moses
"And, when he shall die, take him and cut him out in little stars, and he will make the face of heaven so fine that all the world will be in love with night, and pay no worship to the garish sun."--William Shakespeare
For angles of a triangle:
Using the law of sines (a / sin(A) = b / sin(B) = c / sin(C)) you can derive something to the tune of: sin(A) = (a sin(B)) / b
Using the law of cosines (a^2 = b^2 + c^2 - 2bc cos(A)) you can derive: cos(A) = (-a^2 + b^2 + c^2) / (2bc)
And, finally, with a trig identity (tan(A) = sin(A) / cos(A)) you can substitute and get: tan(A) = [ (a sin(B)) / b ] / [ (-a^2 + b^2 + c^2) / (2bc) ]
Or you could simply use a calculator.
If you're just using the value of the angle, you can always fall back on the unit circle. Seems a much easier way than calculating a bunch of factorials.
Kal-El: May I ask where you found these formulas and if there's any proof supplied? They seem intuitively flawed, the cosine one pops out first. If you take x^(2/2!) (or x^[2/(2!)]) you end up with x. 1 - x...if x is greater than one, you get a negative number. Then, if it's meant to be x^[n/(n!)], you can never get past zero again, thus a negative cosine and you might not even be out of the first quadrant.
OK. That is a lot more complicated than the formula I know for it.
