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Babynini

  • one year ago

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

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  1. Babynini
    • one year ago
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    em I got 2cos(theta)sin(theta)-2cos(theta)-sin(theta)+1=0 yeah?

  2. freckles
    • one year ago
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    If a*b=0, then either a=0 or b=0 or both=0. This means you have: \[2\cos(\theta)-1=0 \text{ or } \sin(\theta)-1=0\]

  3. freckles
    • one year ago
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    isolate the trig function

  4. freckles
    • one year ago
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    \[\cos(\theta)=\frac{1}{2} \text{ or } \sin(\theta)=1 \]

  5. freckles
    • one year ago
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    try to solve both of those equations

  6. Babynini
    • one year ago
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    cos(theta) = 1.047 sin(theta) = 1.57

  7. freckles
    • one year ago
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    oh you mean: \[\theta \approx 1.05 \text{ you should also be able to obtain } \theta \approx 5.24 \text{ from } \cos(\theta)=\frac{1}{2}\] and \[\theta \approx 1.57 \text{ from the other equation } \sin(\theta)=1 \]

  8. Babynini
    • one year ago
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    Yes, that is what I meant :) how did you get 5.24?

  9. freckles
    • one year ago
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    \[\cos(\theta)=\cos(-\theta)=\cos(-\theta+2 \pi ) \\ \cos(1.05) \approx \cos(-1.05+2\pi) \approx \cos(-1.05+2(3.14)) \\ \cos(1.05) \approx \cos(-1.05+6.28)=\cos(5.23) \\ \text{ but honestly I used the unit circle }\]

  10. freckles
    • one year ago
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    You see the points on the circle. The first number of every pair is the output for the cos(the angle there). The second number of every pair is the output for the sin(the angle there). I was looking for when the x-coordinate (the first number aka the cos number) was 1/2. I see this was happening at the pi/3 mark and the 5pi/3 mark.

  11. freckles
    • one year ago
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    And your question wanted these values rounded to the nearest hundredth.

  12. freckles
    • one year ago
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    \[\frac{\pi}{3} \approx 1.05 \\ \frac{5 \pi}{3} \approx 5.24\]

  13. Babynini
    • one year ago
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    ooo I see! that makes sense! haha why do we not want to do the othe pi/3 s?

  14. freckles
    • one year ago
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    at 2pi/3 and 4pi/3 you should see the cos number (the x-coordinate) is negative

  15. freckles
    • one year ago
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    1/2 is not a negative number

  16. Babynini
    • one year ago
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    at pi/3 it is not negative either. That was another question I have why do we make it -1.05 instead of adding 2pi to 1.05

  17. freckles
    • one year ago
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    cos is an even function \[\cos(\theta)=\cos(-\theta)\]

  18. freckles
    • one year ago
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    sin is an odd function \[\sin(\theta)=-\sin(-\theta)\]

  19. Babynini
    • one year ago
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    ooh yes.

  20. freckles
    • one year ago
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    and then inside you can add as many 2pi as you like because we just wind up at the same place

  21. freckles
    • one year ago
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    but I add one 2pi since that would give me another solution in [0,2pi)

  22. freckles
    • one year ago
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    Pretend we want to solve this: \[\cos(\theta)=.23 , \theta \in [0,2\pi)\] This .23 is not on the unit circle. But using the unit circle and other identities helps me to solve this equation. First I can do arccos( ) on both sides to find one solution. \[\theta=\arccos(.23) \] another solution would be given by: \[-\theta+2\pi=\arccos(.23) \\ \theta=-\arccos(.23)+2\pi\] and if we wanted to approximate these solutions by rounding to the nearest hundredth we would have: \[\theta \approx 4.94 , 1.34 \]

  23. freckles
    • one year ago
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    The second solution I used the fact that cos was even and had a period of 2pi.

  24. Babynini
    • one year ago
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    Hmm ok.

  25. Babynini
    • one year ago
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    http://openstudy.com/users/babynini#/updates/55823438e4b07028ea611c18 so for this one, how do I do that?

  26. Babynini
    • one year ago
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    Thank you for the help on this one, btw. :)

  27. freckles
    • one year ago
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    np

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