Kal-El Posted November 24, 2002 Report Share Posted November 24, 2002 I have found formula's for the sine & cosine of an anglesin(x) = x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ...andcos(x) = 1 - x^2/2! + x^4/4! - x^6/6! ... Is there a similar formula for the tangent(x)? Link to comment Share on other sites More sharing options...

Wannabe Genius Posted December 7, 2002 Report Share Posted December 7, 2002 I dont know who posted that formula, but they must be insane, break it down: sin(x) = x - x^3/3 (1) + x^ 5/5 (1) - x^7/7 (1) + x^9/9 (1)......... = x -1 + 1 - 1 + 1 etc etc... All its saying is that your angle (x) = x! its not giving you an answer, or even attempting to. The cosine is the same........... I'll see if I still have my notes on sine, cosine and tangents handy a bit later and post the proper versions. Martin Link to comment Share on other sites More sharing options...

Noah Posted December 7, 2002 Report Share Posted December 7, 2002 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. Link to comment Share on other sites More sharing options...

paperclip Posted December 7, 2002 Report Share Posted December 7, 2002 For angles of a triangle:Using the law of sines (a / sin(A) = b / sin(B) = c / sin©) 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. -paperclip Link to comment Share on other sites More sharing options...

Wannabe Genius Posted December 7, 2002 Report Share Posted December 7, 2002 This saves me digging through my notes...........I have found this site no end of help even with some pretty advanced math stuff http://mathworld.wolfram.com/Trigonometry.html Have fun Martin Link to comment Share on other sites More sharing options...

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