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anonymous

  • one year ago

Find the exact value of each expression: a. sin20degcos80de-cos20degsin80deg b. cos 5pi/12cos7pi/12-sin5pi/12sin7pi/12

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  1. anonymous
    • one year ago
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    @jim_thompson5910 does that make sense or should i write out again?

  2. jim_thompson5910
    • one year ago
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    you don't need to write "deg"

  3. jim_thompson5910
    • one year ago
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    hint: see page 2 of http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf specifically the "Sum and Difference Formulas" section

  4. anonymous
    • one year ago
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    so it would be the formula stating that sin(a +or-b)=sinacosb-cosasinb?

  5. jim_thompson5910
    • one year ago
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    correct

  6. jim_thompson5910
    • one year ago
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    you'll use the cosine version for part b

  7. anonymous
    • one year ago
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    oh okay so for the test tomorrow I should just memorize those formulas? but so now to solve it, it would be (20+80) or (20-80)? so the answer is 100 or -60?

  8. jim_thompson5910
    • one year ago
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    sin(20)cos(80)-cos(20)sin(80) = sin(20-80) = sin(-60) = ???

  9. anonymous
    • one year ago
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    oh wait oops

  10. anonymous
    • one year ago
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    would it be sin(60)?

  11. anonymous
    • one year ago
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    @jim_thompson5910

  12. jim_thompson5910
    • one year ago
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    sin(-60) is not equal to sin(60)

  13. anonymous
    • one year ago
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    oh .. i thought you put the question marks as in that it was wrong

  14. jim_thompson5910
    • one year ago
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    no it's my way of putting a blank

  15. anonymous
    • one year ago
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    it's pi/3

  16. jim_thompson5910
    • one year ago
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    so you'd fill in the blank

  17. anonymous
    • one year ago
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    ohh ok sorry

  18. jim_thompson5910
    • one year ago
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    sin(-60) = -sin(60) = _______________

  19. anonymous
    • one year ago
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    -pi/3

  20. anonymous
    • one year ago
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    @jim_thompson5910

  21. jim_thompson5910
    • one year ago
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    no not -pi/3

  22. jim_thompson5910
    • one year ago
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    you're thinking of -60 degrees = -pi/3 radians

  23. jim_thompson5910
    • one year ago
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    use the unit circle and determine -sin(60)

  24. anonymous
    • one year ago
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    okay i've been looking at the unit circle for 5 minutes and because you're determining -sin60 you would be looking in the 3 and fourth quadrants right?

  25. anonymous
    • one year ago
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    and looking for a distance of 60 degrees the opposite way? I'm just trying to remember what my teacher said.. He said something like that @jim_thompson5910

  26. jim_thompson5910
    • one year ago
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    |dw:1433398564125:dw|

  27. jim_thompson5910
    • one year ago
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    |dw:1433398585790:dw|

  28. jim_thompson5910
    • one year ago
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    |dw:1433398602906:dw|

  29. jim_thompson5910
    • one year ago
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    determine the y coordinate of that point, then make it negative that will be the value of -sin(60)

  30. anonymous
    • one year ago
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    root 3/2?

  31. jim_thompson5910
    • one year ago
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    \[\Large \sin(60) = \frac{\sqrt{3}}{2}\] \[\Large -\sin(60) = -\frac{\sqrt{3}}{2}\]

  32. anonymous
    • one year ago
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    okay so I just look for it on the unit circle and determine it through the coordinate?

  33. jim_thompson5910
    • one year ago
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    yes, x = cos(theta) and y = sin(theta)

  34. anonymous
    • one year ago
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    ok! got it!

  35. anonymous
    • one year ago
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    now for part b do you find pi/6 on the unit circle and look at x coordinate to determine the value?

  36. anonymous
    • one year ago
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    @jim_thompson5910

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