I was in class today and we were proving identities, my teacher told me I was correct, but I should be careful as sin x doesn't always equal 1/cosec x
Why is this
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$\begingroup$ Perhaps when cosec x = 0 $\endgroup$– arberavdullahuOct 19, 2016 at 17:49
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$\begingroup$ @arberavdullahu $\csc(x) = 0$ doesn't have any solutions. $\endgroup$– AdnanOct 19, 2016 at 17:50
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2 Answers
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$\sin(x) = \frac{1}{\csc(x)}$ in any case.
It's can be a logical statement if you're worried about the $\frac10$ case, but it is false.
Perhaps your teacher was worried about the $\csc(x) = 0$ case, but there is no such $x$ for which this applies.
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$\begingroup$ does this mean it's a valid statement for an x $\endgroup$ Oct 19, 2016 at 18:00
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$\begingroup$ @user376253 Yes, since $\sin a = \cos (\frac\pi2-a) = \frac1{\csc a}$ using the identities for goniometric functions. $\endgroup$– AdnanOct 19, 2016 at 18:04
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$\begingroup$ Hi, I emailed my teacher and he said that I'm mistaken. He claims he said cot x doesn't always equal 1/tan x $\endgroup$ Oct 19, 2016 at 18:37
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$\begingroup$ @user376253 Ah, in that case you're teacher is right. That's because $\tan(x) = 0$ is possible. Namely for $x = \pi$, you get a division by zero if you do that. Both $\frac{1}{\tan(\pi)}$ and $\cot(\pi)$ are undefined. $\endgroup$– AdnanOct 19, 2016 at 18:43
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The purely symbolic relationship $sin(x) = 1 / csc(x)$ always holds. As I learned it, this is because $csc(x)$ is defined as $1 / sin(x)$. However, as functions they are not equal for every value of x: wherever $sin(x)$ is zero, $csc(x)$ is undefined (due to divide-by-zero). That is what the teacher is referring to.
