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MelindaR Group Title

Can somebody help me solve this one? sinx*cos³x + sin³x*cosx = 1/2 I've got: cos²x(sinx*cosx) + sin²x(sinx*cosx) = 1/2 (sin²x + cos²x)(sinx*cosx) = 1/2 Since sin²x + cos²x = 1 sinx*cosx = 1/2 How should I proceed now?

  • one year ago
  • one year ago

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  1. MelindaR Group Title
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    P.S. \[0 \le x < 2\]

    • one year ago
  2. tcarroll010 Group Title
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    Use identity of 2sinxcosx = sin2x

    • one year ago
  3. MelindaR Group Title
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    I tried, but then I got sin 2x = 1/2, so what?

    • one year ago
  4. MelindaR Group Title
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    I mean sin2x = 1

    • one year ago
  5. tcarroll010 Group Title
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    First, you would get sin2x =1, not 1/2. Agreed?

    • one year ago
  6. MelindaR Group Title
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    Yes

    • one year ago
  7. tcarroll010 Group Title
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    Ok, so you know where sin m = 1, and that's at pi/2 + 2kpi. Good so far?

    • one year ago
  8. MelindaR Group Title
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    Okay but keep in mind that \[0 \le x < 2\pi\]

    • one year ago
  9. tcarroll010 Group Title
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    np. Your FIRST value is pi/4 because sin m = sin 2x where m = pi/2 so x = pi/4 You SECOND and only other value is at x = (5/4)pi because when that is doubled, you are right back at pi/2

    • one year ago
  10. tcarroll010 Group Title
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    So, those are the only 2 values for x.

    • one year ago
  11. MelindaR Group Title
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    Hmmm, let me see if I get this

    • one year ago
  12. tcarroll010 Group Title
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    Go ahead, ask questions. I'll try to clarify, but it looks like you already understand most of this.

    • one year ago
  13. MelindaR Group Title
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    I don't understand why x = (5/4)pi is also a solution

    • one year ago
  14. MelindaR Group Title
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    3/2*pi should also be a value for m, right?

    • one year ago
  15. MelindaR Group Title
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    Hmmm nevermind my last question

    • one year ago
  16. tcarroll010 Group Title
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    np. After I show you, I think you'll realize that you already knew the answer, which is good, because you'll get the answers on your own then. If you take (5/4)pi as "x" and multiply by 2, you will get (5/2)pi, which is 2pi + pi/2, so you "cycled" back to pi/2 by going around the circle once and then went pi/2. (3/2)pi will not be an answer because doubled it's 3pi which is 180 degrees and you have sin 180 = 0.

    • one year ago
  17. MelindaR Group Title
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    And how did you find out 5/4pi is an answer? You just guessed?

    • one year ago
  18. tcarroll010 Group Title
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    The trick is why did I work "backwards" and select (5/4)pi and then double it to "magically" arrive at pi/2, going once around the circle. If that's how I worked it, THAT would be confusing, so I'll explain how I did it. Sometimes you can work "backwards" to get an intuitive idea of where to go, but that doesn't give you the "how to get the answer". You can be intuitive and work backward to "see what you want" and what is really there at the same time. You then have to work forward, the logical way, so that you can "prove" what needs to be proved. So here goes...

    • one year ago
  19. tcarroll010 Group Title
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    We know that we have the constraint of 0 <= x <= 2pi. Good. However, our final equation is sin2x = 1. The left side 2x tells us that as x goes from 0 to 2pi, then 2x goes from 0 to 4 pi. Lightbulb time! We go around the circle TWICE! That's the key! So, before resloving x and 2x, we know we have sin (something) = 1 at pi/2 and 5pi/2 because we supposed to twice around. Again, that's the key. Twice around. So, look for halves of those values because x is doubled. So you start intuitively, but back it all up logically. Ask more questions.

    • one year ago
  20. MelindaR Group Title
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    Hmmm but actually the constraint is 0<= x < 2pi

    • one year ago
  21. tcarroll010 Group Title
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    That's fine. You'll notice that x = pi/4 and x = 5pi/4 and both of those ARE less than 2pi. It doesn't matter that 2x = 5pi/2 > 2pi. The restriction is on x which we are working with. The restriction is not on 2x.

    • one year ago
  22. tcarroll010 Group Title
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    The "restriction" for 2x is that 2x has to be less than 4pi, not 2pi.

    • one year ago
  23. tcarroll010 Group Title
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    Believe me, I can understand how this could be a little disturbing at first. Keep going over it, and you will see it. You are not expected to see it immediately, but I'm positive about this. If it helps, think of 2x as another variable of think of 2x as a f(x). Or think "twice around" for 2x is "once around" for x.

    • one year ago
  24. MelindaR Group Title
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    Hmmm I think I get it. But how would I go about writing that? I left at sin 2x = 1. So let 2x = m and we have m = pi/2. Therefore, 2x = pi/2 and x = pi/4. What about explaining how I got 5/4pi and 5/2pi? I'd need to prove that in an exam, right?

    • one year ago
  25. tcarroll010 Group Title
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    The way I'd think about it if I put myself in the position of seeing it for the first time would be to say 1) where is sin m = 1? It's at pi/2 and all 2kpi additions to that. 2) I'd know to concentrate on pi/2 right off, but what about those other potential values? Are they "within range". 3) You'd have to relate 0 <= x <= 2pi and 0 <= 2x <= 4pi because that is like a function with y = 2x. You get to use all legitimate values of "Y". And you HAVE to or else you'll get only half credit for your answer. Think of how the sine function is periodic and repeats. That's your conceptual ace in the hole.

    • one year ago
  26. tcarroll010 Group Title
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    Another help: your first value was sin pi/2 = 1, but that's sin 2(pi/4) = 1 for your "x" value because you are working with 2x, so that is your intuitive inclination to consider other values. You had to double something a lot smaller.

    • one year ago
  27. tcarroll010 Group Title
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    You have to mull things like this over. But don't worry, they expand your mind, they really do. Just be comfortable with it and then it flows.

    • one year ago
  28. MelindaR Group Title
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    Wow, you're a great teacher. I get it now, thanks! I always have a little trouble considering values that aren't so apparent at first.

    • one year ago
  29. tcarroll010 Group Title
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    This problem is like my avatar. See how it is cyclic and goes offshing point? Very Zen!

    • one year ago
  30. tcarroll010 Group Title
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    that is off to a vanishing point.

    • one year ago
  31. tcarroll010 Group Title
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    You're a very good student, I can tell!

    • one year ago
  32. MelindaR Group Title
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    Thanks! The function part is very interesting, I didn't know it was okay to think of the restrictions like that!

    • one year ago
  33. tcarroll010 Group Title
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    Math is way cool. Intuitive and analytical at the same time!

    • one year ago
  34. MelindaR Group Title
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    That's true. I'm always thinking of how math is all about creativity.

    • one year ago
  35. tcarroll010 Group Title
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    That's it! There is an absolutely great biography about John Nash and his math exploits. It was made into a good movie called "A Beautiful Mind". The written account is way better though. All about the creative thinking process.

    • one year ago
  36. MelindaR Group Title
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    I'll definitely look it up! Thanks!

    • one year ago
  37. tcarroll010 Group Title
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    You'll love it I guarantee it!

    • one year ago
  38. tcarroll010 Group Title
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    Watch the movie first if you don't have a lot of time, though. Actually a great movie.

    • one year ago
  39. tcarroll010 Group Title
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    So gotta go, but wonderful working with you and thx for the recognition medal!

    • one year ago
  40. MelindaR Group Title
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    Goodbye! Thanks and keep in touch!

    • one year ago
  41. tcarroll010 Group Title
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    Will do!

    • one year ago
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