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anonymous

  • one year ago

please help... FAN AND MEDAL

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @freckles

  3. freckles
    • one year ago
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    which one do you looking at?

  4. freckles
    • one year ago
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    all of them?

  5. anonymous
    • one year ago
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    yes all

  6. freckles
    • one year ago
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    ok well lets look at: \[\cos(2x)=\frac{1}{\sqrt{2}} \\ \text{ or also known as } \\ \cos(2x)=\frac{\sqrt{2}}{2}\] Before we solve this one, do you know how to solve: \[\cos(\theta)=\frac{\sqrt{2}}{2}\] also are we solving in a specific interval ?

  7. anonymous
    • one year ago
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    the interval is between 0 and 360

  8. anonymous
    • one year ago
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    45 and 315

  9. freckles
    • one year ago
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    ok if we have \[0<x<360 \\ \text{ and we \let } theta=2x \\ \text{ then } x=\frac{\theta}{2} \\ \text{ and so we have } \\ 0<\frac{\theta}{2}<360 \\ \text{ multiplying both sides by 2 we have } \\ 0< \theta<720 \\ \text{ so you said } \cos(\theta)=\frac{\sqrt{2}}{2} \text{ has solutions } \\ \theta=45^o,315^o, 45^o+360^o,315+360^o \\ \text{ after simplifying } \\ \theta=45^o,315^o,405^o,675\]

  10. freckles
    • one year ago
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    now recall \[\theta=2x \\ \text{ so we have } \\ 2x=45^o ,315^o,405^o,675^o\]

  11. freckles
    • one year ago
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    just divide both sides by 2

  12. anonymous
    • one year ago
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    so 22.2?

  13. anonymous
    • one year ago
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    157.5?

  14. freckles
    • one year ago
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    ok and two more solutions

  15. freckles
    • one year ago
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    \[x=\frac{45^o}{2},\frac{315}{2}^o,\frac{405^o}{2},\frac{675^o}{2}\]

  16. freckles
    • one year ago
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    we had to solve cos(theta)=sqrt(2)/2 in (0,720) since x was between 0 and 360

  17. freckles
    • one year ago
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    theta was 2 times more than x

  18. anonymous
    • one year ago
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    ahh gotcha.. and thats the answer?

  19. freckles
    • one year ago
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    yes those 4 solutions right there at mentioned are the answers

  20. anonymous
    • one year ago
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    is that considered double identity angles?

  21. freckles
    • one year ago
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    we didn't use the double angle identity but we could have... \[\cos(2x)=\cos^2(x)-(1-\cos^2(x)) \\ \cos(2x)=2\cos^2(x)-1 \\ \text{ so we have } 2\cos^2(x)-1=\frac{\sqrt{2}}{2} \\ 2 \cos^2(x)=\frac{\sqrt{2}}{2}+1 \\ 2\cos^2(x)=\frac{2+\sqrt{2}}{2} \\ \cos^2(x)=\frac{2+\sqrt{2}}{4} \] but yucky this looks a whole bunch uglier

  22. anonymous
    • one year ago
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    ooo yeah definitely better the way you showed me.. thank you.. next one?

  23. anonymous
    • one year ago
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    btw i like how you actually explain whats going on its greatly appreciated

  24. freckles
    • one year ago
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    take the square root of both sides you will end solving two equations

  25. freckles
    • one year ago
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    \[\sin^2(x)=\frac{1}{2} \\ \text{ implies you have } \\ \sin(x)=\frac{1}{\sqrt{2}} \text{ or } \sin(x)=\frac{-1}{\sqrt{2}}\] find the solutions to both equations and the solution will be the union of the sets of solutions you found

  26. anonymous
    • one year ago
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    45?

  27. anonymous
    • one year ago
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    and -45?

  28. freckles
    • one year ago
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    earlier you said you wanted the solutions between 0 and 360 \[\sin(x)=\frac{\sqrt{2}}{2} \text{ has solutions } x=45^o \text{ or } x=? \\ \sin(x)=-\frac{\sqrt{2}}{2} \text{ has solutions } x=-45^o+360^o \text{ or } x=?\] you still have 2 more solutions (1 for each equation)

  29. anonymous
    • one year ago
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    315

  30. freckles
    • one year ago
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    well you already have the solution

  31. freckles
    • one year ago
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    that was the -45+360 one

  32. anonymous
    • one year ago
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    i found 45, 315, 135 225

  33. freckles
    • one year ago
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    awesome sauce

  34. freckles
    • one year ago
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    number 59 is very similar to 57

  35. freckles
    • one year ago
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    number 60 is a quadratic try solve 2u^2-1-u=0 ( notice I just replaced cos(X) with u)

  36. freckles
    • one year ago
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    solve for u you can use the quadratic formula if you want

  37. freckles
    • one year ago
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    then replace u with cos(x) and then solve for x

  38. freckles
    • one year ago
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    i will be back in like 30 minutes (sorry)

  39. anonymous
    • one year ago
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    OMGGGG I GET IT! THANK YOU SO MUCH!!!!!!!! (hope the pie tastes bomb!)

  40. freckles
    • one year ago
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    l messed up on the crust but the filling should taste great.

  41. anonymous
    • one year ago
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    haha aww its okay.. tbh ive never had pecan pie.. but i have one last question

  42. freckles
    • one year ago
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    sure what is it

  43. anonymous
    • one year ago
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    just curious how old are you bc youre INCREDIBLE at math

  44. freckles
    • one year ago
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    There is some math superior to me (a lot of math actually) but I'm 30.

  45. anonymous
    • one year ago
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  46. anonymous
    • one year ago
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    ahh gotcha... that makes me feel ALOT better haha.. im only a senior in high school and if you were my age and that mathematically inclined id be awfully embarrasssed lol

  47. freckles
    • one year ago
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    I would write purely in terms of sin(x) which will result in a quadratic equation again: \[\cos(2x)=\cos^2(x)-\sin^2(x) \\ \cos(2x)=(1-\sin^2(x))-\sin^2(x) \\ \cos(2x)=1-2 \sin^2(x)\] you would be embarrassed?

  48. freckles
    • one year ago
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    oops I mean in terms of cos(x)

  49. freckles
    • one year ago
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    one sec

  50. anonymous
    • one year ago
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    yes i would bc compared to you im math illiterate...

  51. freckles
    • one year ago
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    \[\cos(2x)=\cos^2(x)-\sin^2(x) \\ \cos(2x)=\cos^2(x)-(1-\cos^2(x)) \\ \cos(2x)=2\cos^2(x)-1 \\ \text{ and } \sin^2(x)=1-\cos^2(x)\] I forgot about that one term being cos(x) :p

  52. freckles
    • one year ago
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    \[\sin^2(x)+\cos(2x)-\cos(x)=0 \\ \text{ is equivalent to } \\ 1-\cos^2(x)+2\cos^2(x)-1-\cos(x)=0 \\ \text{ I replaced } \sin^2(x) \text{ with } 1-\cos^2(x) \\ \text{ and I replaced } \cos(2x) \text{ with } 2\cos^2(x)-1\]

  53. freckles
    • one year ago
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    \[1-\cos^2(x)+2\cos^2(x)-1-\cos(x)=0 \\ \text{ combine like terms on the \left } \\ \cos^2(x)-\cos(x)+1-1=0 \\ \cos^2(x)-\cos(x)=0\] and actually this is tons easier to solve than using the quadratic formula this one is easily factorable on the the left hend side

  54. freckles
    • one year ago
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    \[a^2-a=a(a-1)\] \[\cos^2(x)-\cos(x)=\cos(x)(\cos(x)-1)\] you will have two equations to solve \[\cos(x)(\cos(x)-1)=0 \\ \text{ gives you } \cos(x)=0 \text{ or } \cos(x)-1=0\]

  55. anonymous
    • one year ago
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    soooo 0 90 180

  56. anonymous
    • one year ago
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    not 180 i meant 270

  57. freckles
    • one year ago
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    cos(x)=0 when x=90 or x=270 good job there the other equation can be written as cos(x)=1

  58. anonymous
    • one year ago
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    360

  59. anonymous
    • one year ago
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    for when cos(x)=1

  60. freckles
    • one year ago
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    one question is the interval in which we solve the equation (0,360) or [0,360) or (0,360] or [0,360]

  61. anonymous
    • one year ago
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    [0,360)

  62. anonymous
    • one year ago
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    so it would be 0 not 360

  63. freckles
    • one year ago
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    ok so [0,360) means we don't look at 360 just everything up to it and we do include 0 and things after you know until we get to 360 (which we do not include)

  64. freckles
    • one year ago
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    so instead of saying the solution to cos(x)=1 is 360 we will say 0

  65. freckles
    • one year ago
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    however if we were solving cos(x)=1 on [0,360] then we would say x=0 or x=360 since we can include both endpoints

  66. anonymous
    • one year ago
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    yesss gotcha! seriously thank you soooo much :)

  67. freckles
    • one year ago
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    np

  68. anonymous
    • one year ago
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    have a great night!

  69. freckles
    • one year ago
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    you too!

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