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sqrt(cos a+1)/(1-cos a)

I did (not sure how to format the final answer but the numerator should be sqrt'd. $$\sqrt{\frac{{\cos a +1}}{1-\cos a}}=\sqrt{\frac{{\cos a +1}}{1-\cos a}\frac{{\cos a +1}}{1+\cos a}}=\left|\frac{1+\cos a}{\sin a}\right|$$ Thanks to gimusi for formatting.

Is that correct?

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  • $\begingroup$ Do you mean $\sqrt{\frac{{\cos a +1}}{1-\cos a}}$? $\endgroup$
    – user
    Feb 21, 2018 at 22:14
  • $\begingroup$ Yes I meant that. Thanks $\endgroup$
    – Bill
    Feb 21, 2018 at 22:14
  • $\begingroup$ @AndrewLi It doesn't make the denominator more rational, but when I rationalize the denominator it was equal to 1-cos^2a which is equal to sina. I may be wrong which is why I am asking for help :) $\endgroup$
    – Bill
    Feb 21, 2018 at 22:15

1 Answer 1

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It is a correct way to simplify, indeed $$\sqrt{\frac{{\cos a +1}}{1-\cos a}}=\sqrt{\frac{{\cos a +1}}{1-\cos a}\frac{{\cos a +1}}{1+\cos a}}=\left|\frac{1+\cos a}{\sin a}\right|$$

Note also that since for the existence of square root we need

$$\cos a\neq 1 \quad\land\quad \frac{{\cos a +1}}{1-\cos a}\ge0\implies -1\le\cos a<1$$

thus we can write

$$\sqrt{\frac{{\cos a +1}}{1-\cos a}}=\frac{1+\cos a}{|\sin a|}$$

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  • $\begingroup$ This is exactly what I did, except my answer the numerator is sqrtd $$\sqrt{\frac{{\cos a +1}}{1-\cos a}}=\sqrt{\frac{{\cos a +1}}{1-\cos a}\frac{{\cos a +1}}{1+\cos a}}=\left|\frac{1+\cos a}{\sin a}\right|$$ $\endgroup$
    – Bill
    Feb 21, 2018 at 22:18
  • $\begingroup$ @Bill indeed I noticed that it was a bit different as result but I wasn't sure about what you were looking for, try to write in Tex to help the work! $\endgroup$
    – user
    Feb 21, 2018 at 22:20
  • $\begingroup$ @RobertFrost Thanks for your ping but really I’ve decided to ignore this kind of bad behavior and enjoy the site following my way to enjoy with it. Thanks again, Bye $\endgroup$
    – user
    Jul 10, 2018 at 16:11
  • $\begingroup$ Good on you... Your answers are very helpful. $\endgroup$ Jul 10, 2018 at 16:14
  • $\begingroup$ @RobertFrost Thanks a lot for your kind appreciation, I’ll try to do my best! $\endgroup$
    – user
    Jul 10, 2018 at 16:17

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