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anonymous

  • one year ago

Emergency help on an Algebra II exam! Prove: sin θ - sin θ•cos2 θ = sin3 θ. You must show all work.

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  1. nincompoop
    • one year ago
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    it looks like this is an exam

  2. anonymous
    • one year ago
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    It's a practice exam, but it's worth a good bit of points and while I can understand the proofs written out in my lessons, this particular problem is stumping me. I've been working on these tests for hours :(

  3. anonymous
    • one year ago
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    So I know that identity of sin^2theta + cos^2theta = 1, or...something pretty close to that...

  4. nincompoop
    • one year ago
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    show me the attempts you have made.

  5. anonymous
    • one year ago
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    Okay. So the sin theta should cancel, right? cos^2 theta = sin^3 theta ****(that 3 is a cube, not sin(3))**** I don't know what to do with that!

  6. nincompoop
    • one year ago
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    how does it cancel?

  7. anonymous
    • one year ago
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    Bc it's just subtracted. Like 1 - 1 = 0.

  8. anonymous
    • one year ago
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    sin theta - sin theta = 0.

  9. nincompoop
    • one year ago
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    no

  10. nincompoop
    • one year ago
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    sin is multiplied to cos sin θ - (sin θ•cos2 θ) = sin3 θ ORDERS of operation

  11. anonymous
    • one year ago
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    Parentheses first, then

  12. nincompoop
    • one year ago
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    btw, is that \(cos^2 \theta \) or \(cos (2 \theta) \)

  13. anonymous
    • one year ago
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    The first, it's squared.

  14. anonymous
    • one year ago
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    Okay, so...then do we distribute the negative into the parentheses??

  15. anonymous
    • one year ago
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    I'm thinking maybe we subtract sin theta from both sides to give us (sin theta * cos^2 theta) = sin^2 theta

  16. nincompoop
    • one year ago
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    is that \(sin^3 \theta \) or \(sin (3 \theta) \)

  17. anonymous
    • one year ago
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    Cubed, the first one.

  18. nincompoop
    • one year ago
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    you need to start typing these things appropriately because they mean different things

  19. anonymous
    • one year ago
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    I'm sorry, I didn't realize it before I posted bc I just copy and pasted it to preserve the theta symbols :(

  20. nincompoop
    • one year ago
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    like this? PROVE \(sin ~\theta - (sin~ \theta ~cos^2 \theta) = sin^3 \theta \)

  21. anonymous
    • one year ago
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    There weren't any parentheses, but otherwise, yes.

  22. nincompoop
    • one year ago
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    so tell me what should be the best identity we can use?

  23. anonymous
    • one year ago
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    The only one I know of is the one I typed above, the sin^2 theta + cos^2 theta = 1.

  24. anonymous
    • one year ago
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    We didn't focus a lot on identities, so I suppose I may be forgetting some, but they aren't in my notes =/

  25. nincompoop
    • one year ago
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    what did you focus on?

  26. nincompoop
    • one year ago
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    let us use trigonometric expansion and make use of some algebra techniques like factoring, yes?

  27. nincompoop
    • one year ago
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    do you think you can factor the left-hand-side (LHS) of the equation for me?

  28. anonymous
    • one year ago
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    Sorry! Lost connection!

  29. anonymous
    • one year ago
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    I can tryyy, hold on

  30. nincompoop
    • one year ago
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    |dw:1440734256689:dw|

  31. anonymous
    • one year ago
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    sin theta(1 - cos^2 theta) = sin^3 theta

  32. nincompoop
    • one year ago
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    we can skip writing the theta for now and just add it in the end so it is easier, agreed?

  33. anonymous
    • one year ago
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    Okay! sin(1 - cos^2) = sin^3

  34. nincompoop
    • one year ago
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    do you have a note for your pythagorean identities? if not it is very simple \(\large sin^2 + cos^2 = 1 \) keep in mind for future reference that \(sin = y\) and \(cos = x \)

  35. nincompoop
    • one year ago
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    |dw:1440734748047:dw|

  36. anonymous
    • one year ago
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    Yep, I remember that part from the unit circle section we just did.

  37. nincompoop
    • one year ago
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    you probably already learned the Pythagorean Theorem which states that the length of the longest leg \(\large r \) is equal to the square root of the sum of the two shorter legs |dw:1440734959343:dw|

  38. nincompoop
    • one year ago
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    do you think we can manipulate our identity algebraically so it would look familiar as the problem you have?

  39. anonymous
    • one year ago
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    Let me see...

  40. anonymous
    • one year ago
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    y(1 - x^2) = y^3 when you say cos = x and sin = y...

  41. nincompoop
    • one year ago
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    |dw:1440735114209:dw|

  42. anonymous
    • one year ago
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    I'm sorry this is taking me so long to understand ._.

  43. nincompoop
    • one year ago
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    |dw:1440735291067:dw|

  44. nincompoop
    • one year ago
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    |dw:1440735446837:dw|

  45. nincompoop
    • one year ago
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    what part did I lose you?

  46. nincompoop
    • one year ago
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    |dw:1440735591245:dw|

  47. anonymous
    • one year ago
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    Where did the sin^3 go?

  48. nincompoop
    • one year ago
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    that's just the left-hand-side (LHS) since the board is too small for me to fit everything in

  49. anonymous
    • one year ago
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    So it's 1^2 + sin^3 on the right hand?

  50. nincompoop
    • one year ago
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    no, the right-hand-side stays the same since it is what we're trying to accomplish

  51. anonymous
    • one year ago
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    ...so then where is the sin^3 if you said it's not on the left hand? D:

  52. nincompoop
    • one year ago
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    what we are basically doing is trying to manipulate and re-arrange the left-hand-side of your problem so it would EQUAL to the right-hand side of the problem

  53. nincompoop
    • one year ago
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    |dw:1440735790511:dw|

  54. nincompoop
    • one year ago
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    |dw:1440735817436:dw|

  55. anonymous
    • one year ago
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    Okay, I'm mostly following

  56. nincompoop
    • one year ago
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  57. nincompoop
    • one year ago
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    from the pythagorean identity, we derived that \(1-cos^2 = sin^2 \) because from our original identity (derived from the unit circle) \(sin^2 + cos^2 = 1 \) we can re-arrange it to \(sin^2 + cos^2 \color{red}{-cos^2} =1 \color{red}{-cos^2}\) simplifying to: \(sin^2 = 1-cos^2 \)

  58. anonymous
    • one year ago
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    I don't think any of this is getting through anymore, it's really late here and I'm thinking my exam has probably timed out by now ._. I'm tracking but I don't see where this is actually leading, we keep ending up back at this identity and I'm not sure what to do with it

  59. nincompoop
    • one year ago
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    then it would suffice to say that \(sin - sin~cos^2 = sin^3 \) because \(sin (1-cos^2) = sin^3 \) because the left-hand side is equivalent to \(sin (sin^2) \) therefore \(sin^3 = sin^3 \)

  60. anonymous
    • one year ago
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    I'll definitely reference this when I retake. Thank you so so so much for all of your time and help <3

  61. anonymous
    • one year ago
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    You were very patient, I hope it comes back to you tenfold :x

  62. nincompoop
    • one year ago
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    you're not even telling me what part you are not following

  63. nincompoop
    • one year ago
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    up to what part you are able to grasp and what part seems a little fuzzy still

  64. anonymous
    • one year ago
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    I'm sorry, it's just because I don't really know where I'm lost, I'm just really tired >.<

  65. nincompoop
    • one year ago
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    come back when your brain is fully rested. there is no point in studying when your brain is not working to its potential.

  66. nincompoop
    • one year ago
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    go to sleep

  67. anonymous
    • one year ago
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    Exactly :( I hope you have a good night, thanks so much for the guidance, I'm sure it'l help a lot more in the morning

  68. anonymous
    • one year ago
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    You, too!

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