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Atkinsoha

PreCalc & Trig Help! Simplify the expression using a sum or difference formula: sin(x+y)sec x sec y

  • one year ago
  • one year ago

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  1. Atkinsoha
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    ^opps, show all steps too please.

    • one year ago
  2. satellite73
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    \[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\] then multiply out

    • one year ago
  3. Atkinsoha
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    how do you get to that point? like, can you start from the beginning because I need to show all steps for full credit.

    • one year ago
  4. satellite73
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    you can look in your text for a proof of that result, but that is not what you are being asked to do. you are being asked to multiply out, the point of the exercise it to use that identity, not to derive it

    • one year ago
  5. satellite73
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    in other words, you are being asked a question to see if you know the identity multiplying out is now a matter of algebra

    • one year ago
  6. Atkinsoha
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    I just don't understand how you went from sin(x+y)sec(x)sec(y) to that..

    • one year ago
  7. satellite73
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    oh i have confused you that is not the answer at all that is the identity \[\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)\]

    • one year ago
  8. Atkinsoha
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    Yes, you have confused me..

    • one year ago
  9. satellite73
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    now your job is to multiply \[\left(\sin(x)\cos(y)+\cos(x)\sin(y)\right)\times \sec(x)\sec(y)\]\]

    • one year ago
  10. Atkinsoha
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    Maybe I should tell you this.. I've been gone from school for a few weeks now (death in the family) I have no idea how to do any of this..

    • one year ago
  11. satellite73
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    you get \[\sin(x)\cos(y)\sec(x)\sec(y)+\cos(x)\sin(y)\sec(x)\sec(y)\] as a first step

    • one year ago
  12. Atkinsoha
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    okay.. then were the heck do you go?

    • one year ago
  13. satellite73
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    that was multiplying out using the distributive law then \(\sec(x)=\frac{1}{\cos(x)}\) so you can cancel some stuff

    • one year ago
  14. satellite73
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    lets look at the first term \[\sin(x)\cos(y)\sec(x)\sec(y)\] that is equal to \[\frac{\sin(x)\cos(y)}{\cos(x)\cos(y)}\]

    • one year ago
  15. satellite73
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    cancel the \(\cos(y)\) top and bottom and get \[\frac{\sin(x)}{\cos(x)}\] which you can rewrite as \[\tan(x)\]

    • one year ago
  16. Atkinsoha
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    okay. that makes some sense..

    • one year ago
  17. Atkinsoha
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    basically the same thing for the other side too then?

    • one year ago
  18. satellite73
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    did you know \(\sec(x)=\frac{1}{\cos(x)}\)? this is really algebra at this step

    • one year ago
  19. satellite73
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    yes, same thing for the second term

    • one year ago
  20. satellite73
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    only this time you will get \(\tan(y)\)

    • one year ago
  21. Atkinsoha
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    Yes I did. Just didn't understand the problem.. so you get down to tan x=tanx?

    • one year ago
  22. Atkinsoha
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    ohh, yeah. tan x + tan y

    • one year ago
  23. satellite73
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    that is it, yes

    • one year ago
  24. Atkinsoha
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    So is that your answer then? tan x + tan y?

    • one year ago
  25. satellite73
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    that is my answer, yes

    • one year ago
  26. Atkinsoha
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    Okay, that makes sense. Do you have time to help me with one more problem?

    • one year ago
  27. satellite73
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    sure

    • one year ago
  28. Atkinsoha
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    okay, let me take a screen shot.

    • one year ago
  29. satellite73
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    k

    • one year ago
  30. Atkinsoha
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    • one year ago
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  31. Atkinsoha
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    I kind of understand it.. but not really :/

    • one year ago
  32. satellite73
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    ok we can do this, but it is going to take a minute ready?

    • one year ago
  33. Atkinsoha
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    Yes.

    • one year ago
  34. satellite73
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    we are going to use the same formula that we used above \[\sin(u+v)=\sin(u)\cos(v)+\cos(u)\sin(v)\] at the moment, we know only two of those four numbers

    • one year ago
  35. Atkinsoha
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    yes, we know sin u and cos v

    • one year ago
  36. satellite73
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    we need \(cos(u)\) do you know how to find it?

    • one year ago
  37. Atkinsoha
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    Don't you have to make a triangle or something?

    • one year ago
  38. satellite73
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    yes exactly

    • one year ago
  39. Atkinsoha
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    Okay, I'm not sure exactly how to come about finding it on a triangle

    • one year ago
  40. satellite73
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    |dw:1362154017344:dw|

    • one year ago
  41. satellite73
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    third side via pythagoras \[\sqrt{61^2-11^2}=60\]

    • one year ago
  42. Atkinsoha
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    Wouldn't it be 50?

    • one year ago
  43. satellite73
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    no \[\cos(u)=\frac{60}{61}\]

    • one year ago
  44. satellite73
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    adjacent over hypotenuse

    • one year ago
  45. Atkinsoha
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    Okay. so we still ned to find the sin one right?

    • one year ago
  46. satellite73
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    right but actually i made a mistake \[\cos(u)=-\frac{60}{61}\] because you are in quadrant 2

    • one year ago
  47. satellite73
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    now we need \(\sin(v)\) and it is the same idea as before |dw:1362154367573:dw|

    • one year ago
  48. satellite73
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    the third side is \[\sqrt{41^2-40^2}=9\]so \(\sin(v)=\frac{9}{41}\)

    • one year ago
  49. satellite73
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    now plug the numbers directly in to the formula

    • one year ago
  50. Atkinsoha
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    okay, then you just solve it right? multiplying and adding fractions basically?

    • one year ago
  51. satellite73
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    \[\frac{11}{61}\times (-\frac{40}{41})+\frac{9}{61}\times(- \frac{60}{61})\]

    • one year ago
  52. satellite73
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    yeah it is arithmetic from here on in gotta run, good luck

    • one year ago
  53. Atkinsoha
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    thank you!

    • one year ago
  54. Atkinsoha
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    wait quick

    • one year ago
  55. satellite73
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    oh i made a typo!!

    • one year ago
  56. Atkinsoha
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    for multiplying you flip the second fraction?

    • one year ago
  57. satellite73
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    \[\frac{11}{61}\times (-\frac{40}{41})+\frac{9}{41}\times(- \frac{60}{61})\]

    • one year ago
  58. satellite73
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    oh no

    • one year ago
  59. satellite73
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    flip nothing multiply straight across

    • one year ago
  60. Atkinsoha
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    okay. ill give it a try... thanks!

    • one year ago
  61. satellite73
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    yw

    • one year ago
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