Hi
I need to rearrange the following formula from I = f(A) to A = f(I)
The formula is :
tan(I) = sin(A)/(M+cos(A))
Can anyone tell me how I rearrange this to get A as a function of I.
Thanks
James
Hello, James!
I have a way . . . but it's really ugly . . .
The formula is: .$\displaystyle \tan I \:= \:\frac{\sin A}{M + \cos A}$
Can anyone tell me how to get $\displaystyle A$ as a function of $\displaystyle I$?
We have: .$\displaystyle \tan I(M + \cos A) \:=\:\sin A$
.From $\displaystyle \sin^2\!\!A + \cos^2\!\!A \:=\:1$, we have: .$\displaystyle \sin A \:=\:\pm\sqrt{1 - \cos^2\!\!A}$
Substitute: .$\displaystyle \tan I(M + \cos A)\:=\:\pm\sqrt{1 - \cos^2\!\!A}$
Square both sides: .$\displaystyle \tan^2\!I\left(M^2 + 2M\cos A + \cos^2\!\!A\right) \:=\: 1 - \cos^2\!\!A$
This simplifies to the quadratic: .$\displaystyle (\sec^2\!I)\cos^2\!\!A + (2M\tan^2\!I)\cos A + (M^2\tan^2\!I - 1)\:=\:0$
Quadratic Formula: .$\displaystyle \cos A \;=\;\frac{-2M\tan^2\!I \pm \sqrt{(2M\tan^2\!I)^2 - 4(\sec^2\!I)(M^2\tan^2\!I - 1)}}{2(\sec^2\!I)}$
. . . Good luck!
tan(I) = sin(A)/(M+cos(A))
we know for sin(X)/cos(X)=sin(Y)/cos(Y), then Y = X is one solution.
tan(I) = sin(A)/(M+cos(A)) is the same as
sin(I)/cos(I) = sin(A)/(M+cos(A))
so one solution by my guess would be.
I = inv.cos (M + cos A)
cos A + M = cos I
A = inv. cos (cos I - M)