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June 30th, 2019, 02:35 PM   #11
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What exactly does that prove?
They are identities. Try it yourself substitute x on the left hand side and on the right hand side and you will get the same angle.

The exercise is just to demonstrate this is true by following it through logically and step by step.

Last edited by skipjack; July 1st, 2019 at 04:44 AM.
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July 1st, 2019, 04:51 AM   #12
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I'm questioning that. Had you checked it out in detail (including any implicit assumptions), requiring precise definitions at all stages, you'd have found a flaw.
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July 1st, 2019, 09:32 AM   #13
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I'm questioning that. Had you checked it out in detail (including any implicit assumptions), requiring precise definitions at all stages, you'd have found a flaw.
I'm always happy to learn as it makes me stronger. Which flaw is there? The path I took is absolutely logically consistent with the fractions and algebra principles.

The path that greg1313 posted makes sense to me.

Please point out the flaw...

Last edited by skipjack; July 1st, 2019 at 12:20 PM.
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July 1st, 2019, 12:18 PM   #14
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Did greg1313 start with (1 + sin x)/cos x or did you simply assume that was the case?
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July 2nd, 2019, 09:53 AM   #15
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Did greg1313 start with (1 + sin x)/cos x or did you simply assume that was the case?
I think you can start from either the left or the right hand side as long as you can reconcile to the other side.
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July 2nd, 2019, 10:24 AM   #16
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(1+sin x)/cos x = cos x/(1-sin x)
$\dfrac{1+\sin{x}}{\cos{x}} \implies \cos{x} \ne 0 \implies x \ne (2k+1) \cdot \dfrac{\pi}{2} \, ; \, k \in \mathbb{Z}$


$\dfrac{\cos{x}}{1-\sin{x}} \implies \sin{x} \ne 1 \implies x \ne (4k+1) \cdot \dfrac{\pi}{2} \, ; \, k \in \mathbb{Z}$

... there are values of x where the right side equals 0, whereas the left side never equals 0
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