Can you explain whats going on here? is it some trig identity i dont know or something else?

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closed as off-topic by Shaun, Mohammad Riazi-Kermani, Xander Henderson, JMP, choco_addicted Mar 16 '18 at 6:48

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  • $\begingroup$ You're getting a lot of downvotes because you haven't supplied us with an attempt on your part. $\endgroup$ – Shaun Mar 16 '18 at 1:17
  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Mar 16 '18 at 1:17

Factor $R^2$ and note that

$$1- \sin ^2 (x) = \cos ^2 (x).$$

  • $\begingroup$ but if i put (5-5sin(90))^0.5 in a calculator is not the same as 5 (1-1sin(90))^0.5 $\endgroup$ – tgmjack Mar 16 '18 at 1:11
  • $\begingroup$ is the 5-5sin(90) not the two terms? $\endgroup$ – tgmjack Mar 16 '18 at 1:13
  • $\begingroup$ @tgmjack $\sqrt{5-5\sin^2(90^\circ)} = \sqrt{5(1-\sin^2(90^\circ))}$ $\endgroup$ – Andrew Li Mar 16 '18 at 1:15

Try this:

$ \sqrt{R^2-R^2(\sin\theta)^2} = \sqrt{R^2(1-\sin^2 \theta)} $

And because

$ 1-\sin^2 (\theta) = \cos^2 (\theta) $

You can get

$ \sqrt{R^2\cos^2 (\theta)} = R\cos(\theta) $

  • $\begingroup$ yeah but it equals 2R^2*cos^2(x) not 2√R^2*cos^2(x) where does the root go? if the power cancels the roots $\endgroup$ – tgmjack Mar 16 '18 at 1:31
  • $\begingroup$ the root cancels the powers $\endgroup$ – Dashi Mar 16 '18 at 3:14

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