Question 18605: I am terrible with algebraic manipulations. I need to solve the following equation for L. I do not want the answer to the question as I have the answer already from the back of the book - I simply need help in rearranging the formula to solve for L. I have the values for P2/P1, and X (I even have the value for L from the solution manual) I know I need to move cos to the other side and I should be solving for cos^-1; however, I am missing some "gene" that does not allow me to figure it out correctly. Any help you can give me would be greatly appreciated.
P2/P1 = cos^2 [(pi*L) / (2X) ]
Answer by mmm4444bot(95)   (Show Source):
You can put this solution on YOUR website! Hello There:
I'll take a stab at it.
P2/P1 = cos[(L*pi)/(2*X)]^2
Take the square root of both sides.
cos[(L*pi)/(2*X)] = +/- sqrt(P2/P1)
Use the inverse cosine function (arccos) to get L out of the input to cosine.
arccos[+/- sqrt(P2/P1)] = L*pi/(2*X)
Multiply both sides by (2*X)/pi.
L = [(2*X)/pi]*arccos[+/- sqrt(P2/P1)]
Hopefully, this matches the result in your book. You need to decide what to do about the +/- part. In other words, maybe only the positive one is applicable.
~ Mark
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