Re-arranging an complicated equation

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The discussion revolves around the difficulty of rearranging a complicated equation to make BETA the subject. The user has shared an image of the equation but noted the poor quality of the scan. After some back-and-forth, it is clarified that the two answers presented are actually the same, resolving the confusion. The user ultimately realizes the solution was apparent all along. The conversation highlights the common challenges faced in algebraic manipulation and the importance of careful examination of details.
stevemilw
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Hello

Im having some trouble making BETA the subject (thats the B)

I have scanned in my problem and posted it below.
Last time i did something similar to this, it was obviously and right infront of me, so let's hope its as easy this time.

Thanks


http://postimage.org/image/2qanh4plw/

http://postimage.org/image/2qanh4plw/

sorry for the bad quality
 
Last edited by a moderator:
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stevemilw said:
Hello

Im having some trouble making BETA the subject (thats the B)

I have scanned in my problem and posted it below.
Last time i did something similar to this, it was obviously and right infront of me, so let's hope its as easy this time.

Thanks


http://postimage.org/image/2qanh4plw/


sorry for the bad quality
The two answers are not "wildly different." They are the same.
 
Last edited by a moderator:
please explain ?
 
oh i see it now... knew it would be right infront of me!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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