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

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- Jan 30th 2007, 04:29 AMjamesparker_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 - Jan 30th 2007, 05:14 AMSoroban
Hello, James!

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

Quote:

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!

- Feb 1st 2007, 06:12 AMAlvinCY
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) - Feb 1st 2007, 08:48 AMtopsquark