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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- January 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 - January 30th 2007, 05:14 AMSoroban
Hello, James!

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

Quote:

The formula is: .

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

We have: .

.From , we have: .

Substitute: .

Square both sides: .

This simplifies to the quadratic: .

Quadratic Formula: .

. . . Good luck!

- February 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) - February 1st 2007, 08:48 AMtopsquark