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Atkinsoha

  • 2 years 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
    • 2 years ago
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    ^opps, show all steps too please.

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

  3. Atkinsoha
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Yes, you have confused me..

  9. satellite73
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    okay.. then were the heck do you go?

  13. satellite73
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    okay. that makes some sense..

  17. Atkinsoha
    • 2 years ago
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    basically the same thing for the other side too then?

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

  19. satellite73
    • 2 years ago
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    yes, same thing for the second term

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

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

  22. Atkinsoha
    • 2 years ago
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    ohh, yeah. tan x + tan y

  23. satellite73
    • 2 years ago
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    that is it, yes

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

  25. satellite73
    • 2 years ago
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    that is my answer, yes

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

  27. satellite73
    • 2 years ago
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    sure

  28. Atkinsoha
    • 2 years ago
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    okay, let me take a screen shot.

  29. satellite73
    • 2 years ago
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    k

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

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

  33. Atkinsoha
    • 2 years ago
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    Yes.

  34. satellite73
    • 2 years 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
    • 2 years ago
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    yes, we know sin u and cos v

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

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

  38. satellite73
    • 2 years ago
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    yes exactly

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

  40. satellite73
    • 2 years ago
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    |dw:1362154017344:dw|

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

  42. Atkinsoha
    • 2 years ago
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    Wouldn't it be 50?

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

  44. satellite73
    • 2 years ago
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    adjacent over hypotenuse

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

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

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

  49. satellite73
    • 2 years ago
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    now plug the numbers directly in to the formula

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

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

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

  53. Atkinsoha
    • 2 years ago
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    thank you!

  54. Atkinsoha
    • 2 years ago
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    wait quick

  55. satellite73
    • 2 years ago
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    oh i made a typo!!

  56. Atkinsoha
    • 2 years ago
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    for multiplying you flip the second fraction?

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

  58. satellite73
    • 2 years ago
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    oh no

  59. satellite73
    • 2 years ago
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    flip nothing multiply straight across

  60. Atkinsoha
    • 2 years ago
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    okay. ill give it a try... thanks!

  61. satellite73
    • 2 years ago
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    yw

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