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period of sin x

Hi guys,
I'm trying to find the values of t in the equation sin(t)=0 -2pi<t<2pi (the sign has equal on it couldn't type it here), the first one i've found using calculator was 0 by taking sin^-1 of both sides and then for the others solutions i decided to add 2pi and -2pi to zero which was obviously the answers but the text me book also has -pi and pi as answer as well which confuses me a bit as i thought the period of sin(x) is 2pi so you either add 2pi or subtrac from your first answer but here i can't find out how the text book got pi and -pi as well. any help please?
(edited 8 years ago)

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for sin, you get your first two solutions by doing sin^-1(...) and pi-sin^-1(...). Then add and subtract 2pi from those. Thus you get 0 and pi for your first two solutions, then 2pi, -pi, and -2pi (the other three solutions) by adding and subtracting.
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Zacken
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Original post by Alen.m
Hi guys,
I'm trying to find the values of t in the equation sin(t)=0 -2pi<t<2pi (the sign has equal on it couldn't type it here), the first one i've found using calculator was 0 by taking sin^-1 of both sides and then for the others solutions i decided to add 2pi and -2pi to zero which was obviously the answers but the text me book also has -pi and pi as answer as well which confuses me a bit as i thought the period of sin(x) is 2pi so you either add 2pi or subtrac from your first answer but here i can't find out how the text book got pi and -pi as well. any help please?


Have you heard of the cast method for finding solutions? Look it up.

If sin⁡x=a\sin x = a for 0<a≤10 < a \leq 1 then x=arcsin⁡a+2nπx = \arcsin a + 2n\pi and x=π−a+2nπx = \pi - a + 2n\pi

If sin⁡x=b\sin x = b for −1≤b<0-1 \leq b < 0 then x=2π−arcsin⁡∣b∣+2nπx = 2\pi - \arcsin |b| +2n\pi or x=π+arcsin⁡∣b∣+2nπx = \pi + \arcsin |b| + 2n\pi

If sin⁡x=0\sin x = 0 then it's both the above cases.
(edited 8 years ago)
ohhh, and here I was expecting something completely different
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Alen.m
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Original post by constellarknight
for sin, you get your first two solutions by doing sin^-1(...) and pi-sin^-1(...). Then add and subtract 2pi from those. Thus you get 0 and pi for your first two solutions, then 2pi, -pi, and -2pi (the other three solutions) by adding and subtracting.


that's the part that i have problem i thought i need to subtract the first solution from 2pi instead of pi because the period of sin x is 2pi
yes, but sin x also has symmetry about x=pi/2, 3pi/2, etc, hence why you also need to subtract form pi.
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Alen.m
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Original post by constellarknight
yes, but sin x also has symmetry about x=pi/2, 3pi/2, etc, hence why you also need to subtract form pi.


so basically for sin questions always subtract the first solution from pi and keep adding 2pi from those two till you find answers in the given interval but for cos you need to take away the first answer from 2pi and then keep adding or subtracting 2pi to the answer is that right?
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Zacken
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Have you heard of the cast method for finding solutions? Look it up.

If sin⁡x=a\sin x = a for 0<a≤10 < a \leq 1 then x=arcsin⁡a+2nπx = \arcsin a + 2n\pi and x=π−a+2nπx = \pi - a + 2n\pi

If sin⁡x=b\sin x = b for −1≤b<0-1 \leq b < 0 then x=2π−arcsin⁡∣b∣+2nπx = 2\pi - \arcsin |b| +2n\pi or x=π+arcsin⁡∣b∣+2nπx = \pi + \arcsin |b| + 2n\pi

If sin⁡x=0\sin x = 0 then it's both the above cases.

...

Original post by Alen.m
so basically for sin questions always subtract the first solution from pi and keep adding 2pi from those two till you find answers in the given interval but for cos you need to take away the first answer from 2pi and then keep adding or subtracting 2pi to the answer is that right?
sin and cos have the same period because they live together.
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Alen.m
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Original post by Zacken
...


that's bit confusing can i just say that the first answers need to be taken away from symmetry points of sin(if it's a sin question) or symmetry points of cos(if it's a cos question) then keep adding and subtracting 2pi from those?and for tan keep adding or subtracting pi from the answers which was taken away from symmetry points of tan ?
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Original post by EricPiphany
sin and cos have the same period because they live together.

do you think this is right
the first answers need to be taken away from symmetry points of sin(if it's a sin question) or symmetry points of cos(if it's a cos question) then keep adding and subtracting 2pi from those?and for tan keep adding or subtracting pi from the answers which was taken away from symmetry points of tan ?
Original post by Alen.m
do you think this is right
the first answers need to be taken away from symmetry points of sin(if it's a sin question) or symmetry points of cos(if it's a cos question) then keep adding and subtracting 2pi from those?and for tan keep adding or subtracting pi from the answers which was taken away from symmetry points of tan ?


for cos take the negative. For sin take pi - solution. Then add 2pi to taste.
for tan you're right. Actually for tan just add pi to the original number to taste.
(edited 8 years ago)
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Zacken
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Original post by Alen.m
that's bit confusing can i just say that the first answers need to be taken away from symmetry points of sin(if it's a sin question) or symmetry points of cos(if it's a cos question) then keep adding and subtracting 2pi from those?and for tan keep adding or subtracting pi from the answers which was taken away from symmetry points of tan ?


Watch this: https://www.youtube.com/watch?v=tFBQ2YMdfhU
Original post by sleepysnooze
ohhh, and here I was expecting something completely different



Me too. :tongue:

Was preparing for some serious debauchery. :ahee:
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Alen.m
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Original post by EricPiphany
for cos take the negative. For sin take pi - solution. Then add 2pi to taste.
for tan you're right. Actually for tan just add pi to the original number to taste.

What you mean take the negarive for cos?
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Alen.m
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Original post by Zacken


Thanks but that explained cast diagram which im familiar with it but i was trying to find a way to find solutions without cast diagrams so it'd be quicker
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Zacken
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Original post by Alen.m
Thanks but that explained cast diagram which im familiar with it but i was trying to find a way to find solutions without cast diagrams so it'd be quicker


The CAST diagram is exceedingly quick. How much quicker do you want?
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Alen.m
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[QUOTE="Zacken;63007453"]The CAST diagram is exceedingly quick. How much quicker do you want?[/QUOTE
I struggle with it when it comes to find radian angles for example sin(t)=o would go to what part of cast diagram? The part that sin is positive?
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Zacken
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Original post by Alen.m
The CAST diagram is exceedingly quick. How much quicker do you want?[/QUOTE
I struggle with it when it comes to find radian angles for example sin(t)=o would go to what part of cast diagram? The part that sin is positive?


Like I said, when it's 0 - then all four quadrants apply.
Reply 19
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Alen.m
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Original post by Zacken
Like I said, when it's 0 - then all four quadrants apply.


well in that case pi/2-o will be the solution as well isn't or 3pi/2-o

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