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# Tangent Formula?

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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)?

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• 2 weeks later...

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

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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.

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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

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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