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 August 4th, 2019, 02:53 AM #1 Newbie   Joined: Aug 2019 From: Hong Kong Posts: 4 Thanks: 0 Domain Of Validity For Trigonometric Identities Does anyone know how to find the domain of validity for trigonometric identities? There doesnt seem to be much tutorials on this topic on youtube/online. Can anyone explain to me step by step since I have absolutely no clue how to find it?
 August 4th, 2019, 03:39 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Can you figure out where $\cos\theta$, $\sin\theta$ are equal to 0?
August 4th, 2019, 03:43 AM   #3
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 Originally Posted by greg1313 Can you figure out where $\cos\theta$, $\sin\theta$ are equal to 0?

cos(x) = 0 in pi/2, 3pi/2, 5pi/2, so the multiples of pi/2?

sin(x) = 0 in pi, 2pi/ 3pi, so the multiples of pi ?

is this right ?

so csc(x) = 1/sin(x), meaning the domain of validity is all real numbers except for the multiples of pi? The correct answer is all real numbers except for pi/2 though, why is that so?

Last edited by Tsepten; August 4th, 2019 at 03:47 AM.

August 4th, 2019, 06:05 AM   #4
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 Originally Posted by Tsepten cos(x) = 0 in pi/2, 3pi/2, 5pi/2, so the multiples of pi/2? sin(x) = 0 in pi, 2pi/ 3pi, so the multiples of pi ? is this right ? so csc(x) = 1/sin(x), meaning the domain of validity is all real numbers except for the multiples of pi? The correct answer is all real numbers except for pi/2 though, why is that so?
re-read choice D ... does it say what you stated above?

August 4th, 2019, 06:21 AM   #5
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 Originally Posted by skeeter re-read choice D ... does it say what you stated above?
Oh I meant, what I got was 'domain of validity is all real numbers except for multiples of pi' (which is B), but D is the correct answer.

If csc(x) = 1/sin(x), and sin(x) = 0 in multiples of pi, shouldn't the answer be: domain of validity is all real numbers except for multiples of pi?

Last edited by skipjack; August 4th, 2019 at 11:03 AM.

August 4th, 2019, 07:21 AM   #6
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 Originally Posted by Tsepten Oh I meant, what I got was 'domain of validity is all real numbers except for multiples of pi' (which is B), but D is the correct answer. If csc(x) = 1/sin(x), and sin(x) = 0 in multiples of pi, shouldn't the answer be: domain of validity is all real numbers except for multiples of pi?
$\cos{x} \ne 0 \implies x \text{ cannot be an odd multiple of }\frac{\pi}{2}$

$\sin{x} \ne 0 \implies x \text{ cannot be a whole multiple of }\pi = \text{ even multiples of }\frac{\pi}{2}$

therefore, $x \text{ cannot be any multiple of }\frac{\pi}{2}$, which matches up with choice D.

Last edited by skipjack; August 4th, 2019 at 11:04 AM.

August 4th, 2019, 04:34 PM   #7
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 Originally Posted by skeeter $\cos{x} \ne 0 \implies x \text{ cannot be an odd multiple of }\frac{\pi}{2}$ $\sin{x} \ne 0 \implies x \text{ cannot be a whole multiple of }\pi = \text{ even multiples of }\frac{\pi}{2}$ therefore, $x \text{ cannot be any multiple of }\frac{\pi}{2}$, which matches up with choice D.
But I don't know how that's related to the question, Do we look at the 1/cos and the cos/sine which is in the middle and avoid having the denominator = 0?

Last edited by skipjack; August 4th, 2019 at 07:20 PM.

August 4th, 2019, 06:01 PM   #8
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 Originally Posted by Tsepten But I don't know how that's related to the question, Do we look at the 1/cos and the cos/sine which is in the middle and avoid having the denominator = 0?
Yes.

Last edited by skipjack; August 4th, 2019 at 07:20 PM.

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