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Babynini

  • one year ago

Solve each equation in the interval [0,2pi) rounded to two decimal points a) 3sin(theta)-1=0

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  1. Babynini
    • one year ago
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    I got sin(theta)=1/3 theta = 19.47 but this is not correct, am I meant to be even solving for theta?

  2. anonymous
    • one year ago
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    maybe you are supposed to be answering in radians, not degrees who knows?

  3. Babynini
    • one year ago
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    hrrm yeah doing it in rad gives me the correct answer =.= but they never specified that. sigh lol

  4. anonymous
    • one year ago
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    http://www.wolframalpha.com/input/?i=arcsin%281%2F3%29

  5. Babynini
    • one year ago
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    yeah the answers they give are 0.34 and 2.80 where is the 2.80 from?

  6. freckles
    • one year ago
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    for a second pretend we want to solve this one on [0,2pi) \[\sin(\theta)=\frac{1}{2} \] The unit circle gives the solutions as: \[\theta=\frac{\pi}{6}, \frac{5\pi}{6}\] The pi/6 can be obtain by taking arcsin( ) of both sides \[\theta=\arcsin(\frac{1}{2}) =\frac{\pi}{6} \approx 0.52 \\ \text{ now how we can get the other solution without looking at the unit circle } \\ \\ \text{ well remember } \sin \text{ is odd } \sin(\theta)=-\sin(-\theta) \\ \sin(\frac{\pi}{6})=-\sin(-\frac{\pi}{6}) \\ \text{ also recall \sum identity for } \sin \\ \sin(\pi-\frac{\pi}{6}) =\sin(\pi)\cos(\frac{\pi}{6})-\sin(\frac{\pi}{6})\cos(\pi) \\ =\sin(\pi-\frac{\pi}{6})=0 \cdot \cos(\frac{\pi}{6})-(-1)\sin(\frac{\pi}{6}) \\ =\sin(\pi-\frac{\pi}{6})=-(-1)\sin(\frac{\pi}{6}) \\ \text{ now recall } -\sin(\frac{\pi}{6})=\sin(-\frac{\pi}{6}) \\ \text{ so we have } \sin(\frac{\pi}{6})=\sin(\pi-\frac{\pi}{6}) =\sin(\frac{6\pi}{6}-\frac{\pi}{6})=\sin(\frac{5\pi}{6})\] so the other solution we can said is given by the equation: \[-\theta+\pi= \arcsin(\frac{1}{2}) \\ \theta-\pi=-\arcsin(\frac{1}{2}) \\ \theta=\pi-\arcsin(\frac{1}{2})=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\] so what I am saying for you: you have the following two solutions: \[\theta=\arcsin(\frac{1}{3}) \text{ or } -\theta+\pi=\arcsin(\frac{1}{3}) \\ \theta=\arcsin(\frac{1}{3}) \text{ or } \theta=\pi-\arcsin(\frac{1}{3}) \\ \theta \ \]

  7. freckles
    • one year ago
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    To help guide me I kinda look at the unit circle sometimes to remember me of the equations I need to solve things like this

  8. Babynini
    • one year ago
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    oh gosh this is going to take a while to get through to me haha but yeah it makes sense :)

  9. Babynini
    • one year ago
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    adding two pi to theta = ___ -arcsin(1/3) would be out of restrictions, correct? so that's why we only add pi?

  10. freckles
    • one year ago
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    looking back at theta=___ -arcsin(1/2) we know arcsin(1/2) is pi/6 so we are looking at theta=___-pi/6 now putting 2pi in that blank would actually take us back to -pi/6 instead of pi/6

  11. freckles
    • one year ago
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    look at the unit circle for a second let's look at the pi/6 and -pi/6 sin(pi/6)=1/2 where as sin(-pi/6)=-1/2 but this isn't totally bad because if you stay on that line at -pi/6 ....and move your finger on that line where -pi/6 is ...you will see the desire solution of sin being 1/2 again which would be at 5pi/6 and guess what the angle created by a line is 180 deg or you would say pi and pi-pi/6 would be another name for that 5pi/6

  12. freckles
    • one year ago
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    so to solve: \[\sin(\theta)=a \text{ for } \theta \text{ pretending } a \text{ is within range of } \sin( ) \\ \theta=\arcsin(a) \text{ will give one solution in } [0,2\pi] \\ \theta=\pi-\arcsin(a) \text{ will give another solution in } [0,2\pi] \\ \text{ this will always be two solutions if } \arcsin(a) \text{ isn't} \frac{\pi}{2} \text{ or } \frac{-\pi}{2} \\ \text{ because in those cases there will only be one solution in } [0,2\pi]\]

  13. Babynini
    • one year ago
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    hmm hrm. Ok. I'll have to do a few more and play around with it, but i'm kindd of getting the hang of it o.o

  14. freckles
    • one year ago
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    so you understand that sin(u)=2 does not satisfy what I said above right?

  15. freckles
    • one year ago
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    do you know why?

  16. Babynini
    • one year ago
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    It's out of domain? o_o

  17. freckles
    • one year ago
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    hells yes sin ranges from -1 to 1 this would I mean pretending a is within in the range of sin( ) very good

  18. freckles
    • one year ago
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    \[\cos(\theta)=a \text{ pretending } a \text{ is within range of } \cos( ) \\ \theta=\arccos(a) \text{ will give one solution in } [0,2\pi] \\ \theta=-\arccos(a)+2\pi \text{ will give another solution in } [0,2\pi ] \\ \text{ this will always be two solutions if } \arccos(a) \text{ isn't } \pi \\ \text{ because this will give only one solution}\]

  19. freckles
    • one year ago
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    All of the generalizations I made are only for [0,2pi]

  20. Babynini
    • one year ago
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    ....haha ai .

  21. Babynini
    • one year ago
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    oh. right right.

  22. Babynini
    • one year ago
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    Phew haha dude, you're amazing. You've stuck with me and my ignorance for so long xD

  23. freckles
    • one year ago
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    it is kinda of fun :) we could easily extend what I have said so you all the real solutions in the whole world

  24. freckles
    • one year ago
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    \[\cos(\theta)=a \text{ where } -1 \le a \le 1 \text{ has solutions } \\ \theta=\arccos(a)+2 \pi n \\ \theta=-\arccos(a)+2 \pi n \\ \text{ where } n \text{ is integer } \\ \sin(\theta)=a \text{ where} -1 \le a \le 1 \text{ has solutions } \\ \theta=\arcsin(a)+2 \pi n \\ \theta=\pi-\arcsin(a)+2 \pi n =-\arcsin(a)+\pi(1+2n) \\ \text{ where} n \text{ is integer }\]

  25. freckles
    • one year ago
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    you just add 2 pi n to the solutions since both sin and cos have period 2pi :)

  26. Babynini
    • one year ago
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    it would go on forever, infinite answers haha

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