I need the inverse of $$ ±tan(x) = -\frac{(x+8010)(x^2+32400)}{((x-90)(x+90)(x^2-180x+40500))}. $$ This being the arctangent using Bhaskara I's sine approximation formula.
Need the inverse of the approximation of $\tan$
1
$\begingroup$
geometry
trigonometry
-
0Just guessed, but this formula seems not very invertible in an elementary way. Or are you talking about an aproximations of $\tan^{-1}$? – 2017-02-23
-
0Yes in short but I need the equation to be no more complicated than square roots if possible. – 2017-02-23
-
0Your formula is in degrees, all right ? First thing: why do you write for example 8010 and not $90 (= 8010-22*360)$ degrees ? Second thing: you should turn to radians (the natural unit, as agreed now by all mathematicians), at least for the time you (try to) inverse your expression, and hopefuly switch to degrees at the end, if it makes sense ? – 2017-02-23
-
0I am sorry but you have to check your formula : under its present form it is **completely erroneous**, giving $\tan(0^{\circ})=0.791...$ instead of $0$ and $\tan(45^{\circ})=1.326$ instead of $1$ ... – 2017-02-24
-
0Yes it's very possible I made an error. I used the guys sin approx to calculate cos(x)=sin(90-x) and did tan(x)=sin(x)/cos(x) and simplified it and that's what I have above. – 2017-02-28