# Thread: [SOLVED] Rearrange I = f(A) to A = f(I)

1. ## [SOLVED] Rearrange I = f(A) to A = f(I)

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

2. Hello, James!

I have a way . . . but it's really ugly . . .

The formula is: . $\tan I \:= \:\frac{\sin A}{M + \cos A}$

Can anyone tell me how to get $A$ as a function of $I$?

We have: . $\tan I(M + \cos A) \:=\:\sin A$

.From $\sin^2\!\!A + \cos^2\!\!A \:=\:1$, we have: . $\sin A \:=\:\pm\sqrt{1 - \cos^2\!\!A}$

Substitute: . $\tan I(M + \cos A)\:=\:\pm\sqrt{1 - \cos^2\!\!A}$

Square both sides: . $\tan^2\!I\left(M^2 + 2M\cos A + \cos^2\!\!A\right) \:=\: 1 - \cos^2\!\!A$

This simplifies to the quadratic: . $(\sec^2\!I)\cos^2\!\!A + (2M\tan^2\!I)\cos A + (M^2\tan^2\!I - 1)\:=\:0$

Quadratic Formula: . $\cos A \;=\;\frac{-2M\tan^2\!I \pm \sqrt{(2M\tan^2\!I)^2 - 4(\sec^2\!I)(M^2\tan^2\!I - 1)}}{2(\sec^2\!I)}$

. . . Good luck!

3. 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)

4. Originally Posted by AlvinCY
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)
This solution implies that there exists an I such that $cos(I) = M + cos(A)$, which may not be true. (If |M| > 2 there isn't one.) It is valid as a test, but may not produce all (or even any) solutions.

-Dan