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anonymous

  • one year ago

I desperately need someone to help me with Trig!!! please!!! FAN AND MEDAL!!

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  1. triciaal
    • one year ago
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    review some basics |dw:1432906516081:dw|

  2. anonymous
    • one year ago
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    That's a good graph. I get that for the most part...but I have no idea how to verify identities

  3. anonymous
    • one year ago
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    for example csc(theta) - sin(theta) = cot(theta) cos(theta) I have to verify the identity and find the domain

  4. mathmate
    • one year ago
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    @B_O_R_E_D Are you familiar with the basic identities?

  5. anonymous
    • one year ago
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    sort of-ish...they are confusing

  6. mathmate
    • one year ago
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    Can you post what you know? We'll see if you have everything you need.

  7. anonymous
    • one year ago
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    hang on

  8. anonymous
    • one year ago
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  9. mathmate
    • one year ago
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    Sorry, I was expecting you to type what YOU know, not what's online.

  10. anonymous
    • one year ago
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    its not online its just a word doc...I can type if youd like

  11. mathmate
    • one year ago
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    Sorry, I was expecting you to type what YOU know, not what's online. Besides, I only open/display .jpg files.

  12. anonymous
    • one year ago
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    okay...hang on I'll type it

  13. mathmate
    • one year ago
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    Yes, it would be nice. Because typing them out also helps you getting familiar with them.

  14. triciaal
    • one year ago
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    |dw:1432907494538:dw|

  15. mathmate
    • one year ago
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    @triciaal what you posted is a very good review of the basics, before even the question was posted. Genial!

  16. anonymous
    • one year ago
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    csc(theta) = 1/sin(theta) sin(theta) = 1/csc(theta) sec(theta) = 1/cos(theta) cos(theta) = 1/sec(theta) tan(theta) = 1/cot(theta) cot(theta) = 1/tan(theta)

  17. anonymous
    • one year ago
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    reciprocal identities

  18. mathmate
    • one year ago
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    Did you miss out anything? There is at least one more that you need.

  19. anonymous
    • one year ago
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    tan(theta) = sin(theta)/cos(theta) (Tan identity) cot(theta) = cos(theta)/sin(theta) (Cot Identity)

  20. mathmate
    • one year ago
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    Good, we'll need these and... one more...

  21. anonymous
    • one year ago
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    that's all I've learned

  22. mathmate
    • one year ago
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    |dw:1432904305976:dw| This is extracted from @triciaal 's post. Can you use Pythagoras Theorem to write down the missing identity?

  23. anonymous
    • one year ago
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    ummm sin(theta)/tan(theta)?

  24. mathmate
    • one year ago
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    Think Pythagoras's theorem!

  25. anonymous
    • one year ago
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    SOHCAHTOA

  26. anonymous
    • one year ago
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    ???

  27. mathmate
    • one year ago
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    |dw:1432904481376:dw| Perhaps I missed out the 90 deg. angle didn't help. Pythagoras theorem says that the sum of the squares of the two legs of a right triangle equals the square of the hypotenuse. Can you figure out something?

  28. mathmate
    • one year ago
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    Hint: use the diagram I extracted from @Triciaal 's post.

  29. anonymous
    • one year ago
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    hang on

  30. anonymous
    • one year ago
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    I really have no idea... ummm did I miss cos?

  31. mathmate
    • one year ago
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    more hint: |dw:1432904684712:dw| Pythagoras theorem says \(a^2+b^2=c^2\)

  32. anonymous
    • one year ago
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    yes I know that

  33. mathmate
    • one year ago
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    ok, use the diagram below and use Pythagoras theorem: |dw:1432904785881:dw| It would be \(sin^2(x)+cos^2(x) = 1\) This is the missing identity. Have you seen that before?

  34. anonymous
    • one year ago
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    ohhhhh no I haven't....They have just introduced trig to me in this last unit for the year...everything I've done this year has been alg2

  35. mathmate
    • one year ago
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    But now you understand why this is true?

  36. anonymous
    • one year ago
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    yeah more or less

  37. anonymous
    • one year ago
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    im leaning toward less

  38. anonymous
    • one year ago
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    okay...so I know we have to use a reciprocal identity

  39. mathmate
    • one year ago
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    The identity comes from the definition of sin and cos. With a hypotenuse of 1, the identity is just an application of Pythagoras. In any case, you only have to know it, most of the time you're not expected to prove it.

  40. anonymous
    • one year ago
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    okay....

  41. mathmate
    • one year ago
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    Now back to the question. csc(theta) - sin(theta) = cot(theta) cos(theta) In verifying identities, we usually start from one side. In this case, we can start from the left.

  42. mathmate
    • one year ago
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    If we can show that the left side is identical to the right, we're done.

  43. anonymous
    • one year ago
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    okay.. so I think I need to get like similar terms or something so would I replace sin(theta) for 1/csc(theta) ???

  44. mathmate
    • one year ago
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    Exactly, that's an excellent start. A lot of the time, this strategy works!!!

  45. anonymous
    • one year ago
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    yay!! okay so csc(theta) - 1/csc(theta) = cot(theta) cos(theta)

  46. mathmate
    • one year ago
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    I don't see your reasoning, can you explain?

  47. anonymous
    • one year ago
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    I replaced sin...so I was just filling that in the equation

  48. mathmate
    • one year ago
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    If we start with the left, we don't usually write the right hand side, because equality is not yet proven. We write (I'll use x instead of \(\theta\), saves me time.) as your first step: csc(x)-sin(x) = 1/sin(x) - sin(x)

  49. mathmate
    • one year ago
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    Are you at ease with adding fractions, such as 1/3 + 3?

  50. anonymous
    • one year ago
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    oh! I put it on the wrong side yeah those are fine

  51. mathmate
    • one year ago
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    1/3-3 = 1/3 - 3^2/3 = (1-3^2)/3 Does this make sense to you?

  52. anonymous
    • one year ago
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    okay.... oh yes! I see what you did okay

  53. mathmate
    • one year ago
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    Now you need to do the same with 1/sin(x) and continue the following \( csc(x)-sin(x) = \frac{1}{sin(x)} - sin(x)\)

  54. anonymous
    • one year ago
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    umm okay...

  55. mathmate
    • one year ago
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    R U stuck or R U working on it?

  56. anonymous
    • one year ago
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    trying to work on it...but im still stuck

  57. mathmate
    • one year ago
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    Well... it's similar to the 1/3 example: \(\large csc(x)-sin(x) = \frac{1}{sin(x)} - sin(x)=\frac{1-sin^2(x)}{sin(x)}\)

  58. anonymous
    • one year ago
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    okay....so that's how you verify it?

  59. mathmate
    • one year ago
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    or \(\large csc(x)-sin(x) = \frac{1}{sin(x)} - sin(x)=\frac{1}{sin(x)}-\frac{sin(x)}{sin(x)}sin(x)=\frac{1-sin^2(x)}{sin(x)}\)

  60. mathmate
    • one year ago
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    With what you know in terms of trigonometric identities, you should be able to finish the job, can you not?

  61. anonymous
    • one year ago
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    csc(x) - sin(x) = 1?

  62. mathmate
    • one year ago
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    Anyway, from the identity \(sin^2(x)+cos^2(x)=1\) we conclude: \(1-sin^2(x)=cos^2(x)\)

  63. mathmate
    • one year ago
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    Does that help?

  64. anonymous
    • one year ago
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    ohhhh...oops sorry... OHHH!!! I get it! yes that helps a lot!

  65. anonymous
    • one year ago
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    Thank you so much for your help/time!!

  66. mathmate
    • one year ago
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    Actually, looking back, you have been asked to "verify" the identity. What we did was to prove it. Verification means putting numbers on each side and show that the results are the same. You need to put different numbers, but make sure theta does not equal zero (because the expression becomes infinite, that's where the domain comes in). Sorry for having gone through long steps, but this will also prepare for your future topic on identities.

  67. anonymous
    • one year ago
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    okay thank you. I think I can do that... I have to go, but thank you again!

  68. mathmate
    • one year ago
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    You're welcome! :)

  69. triciaal
    • one year ago
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    sorry system was down @mathmate thank you

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