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Atkinsoha

  • one year ago

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

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  1. Atkinsoha
    • one year ago
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    ^opps, show all steps too please.

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

  3. Atkinsoha
    • one year ago
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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.

  4. satellite73
    • one year ago
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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

  5. satellite73
    • one year ago
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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

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

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

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

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

  10. Atkinsoha
    • one year ago
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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..

  11. satellite73
    • one year ago
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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

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

  13. satellite73
    • one year ago
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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

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

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

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

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

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

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

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

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

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

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

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

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

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

  27. satellite73
    • one year ago
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    sure

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

  29. satellite73
    • one year ago
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    k

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

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

  33. Atkinsoha
    • one year ago
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    Yes.

  34. satellite73
    • one year ago
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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

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

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

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

  38. satellite73
    • one year ago
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    yes exactly

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

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

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

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

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

  44. satellite73
    • one year ago
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    adjacent over hypotenuse

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

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

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

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

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

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

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

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

  53. Atkinsoha
    • one year ago
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    thank you!

  54. Atkinsoha
    • one year ago
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    wait quick

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

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

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

  58. satellite73
    • one year ago
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    oh no

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

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

  61. satellite73
    • one year ago
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    yw

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