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anonymous

  • one year ago

given the sin a=-2/3 and a is in quadrant what are the exact values of the following trigonometric functions of a? cos a= tan a= csc a=

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  1. anonymous
    • one year ago
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    what quadrant is a in?

  2. anonymous
    • one year ago
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    3

  3. anonymous
    • one year ago
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    Sin=y/r Cos=x/r Tan=y/x Csc=r/r

  4. anonymous
    • one year ago
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    |dw:1433564957553:dw|

  5. anonymous
    • one year ago
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    And remember that in quadrant 3 only tang is positive

  6. anonymous
    • one year ago
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    Sinx is negative in the 3rd quadrant

  7. anonymous
    • one year ago
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    and also in the fourth quadrant

  8. anonymous
    • one year ago
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    sin=-2/3

  9. anonymous
    • one year ago
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    Do you know how to approach this question ?

  10. anonymous
    • one year ago
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    has anyone suggested drawing a triangle yet? it would really make things easy

  11. anonymous
    • one year ago
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    no not really

  12. anonymous
    • one year ago
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    the do it label the "opposite side " 2, the hypotenuse 3, and use pythagoras to find the length of the adjacent side

  13. anonymous
    • one year ago
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    |dw:1433565255841:dw|

  14. anonymous
    • one year ago
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    5

  15. anonymous
    • one year ago
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    oh no!

  16. anonymous
    • one year ago
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    \[2^2+?^2=3^2\\ ?=\sqrt{3^2-2^2}\]

  17. anonymous
    • one year ago
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    1^2

  18. anonymous
    • one year ago
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    is it right @satellite73

  19. anonymous
    • one year ago
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    no not at all

  20. anonymous
    • one year ago
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    what is \(3^2\)?

  21. anonymous
    • one year ago
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    9

  22. anonymous
    • one year ago
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    ok and \(2^2\)?

  23. anonymous
    • one year ago
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    4

  24. anonymous
    • one year ago
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    and \[9-4\]?

  25. anonymous
    • one year ago
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    5

  26. anonymous
    • one year ago
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    so what is \[\sqrt{3^2-2^2}\]?

  27. anonymous
    • one year ago
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    5^2?

  28. anonymous
    • one year ago
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    no dear not \(5^2\) but rather \(\sqrt 5\)

  29. anonymous
    • one year ago
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    my bad

  30. anonymous
    • one year ago
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    then what next

  31. anonymous
    • one year ago
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    Then you use the sides to find the rest of the trig functions

  32. anonymous
    • one year ago
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    |dw:1433566350237:dw|

  33. anonymous
    • one year ago
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    now you can take any trig ratio you want, just remember what quadrant you are in to make sure to know if it is positive or negative

  34. anonymous
    • one year ago
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    so for cos a

  35. anonymous
    • one year ago
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    so for cos a its 5/3 @satellite73

  36. anonymous
    • one year ago
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    and for tan a 2/sqrt5

  37. anonymous
    • one year ago
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    You can't leave a radical in the denominator. You would have to multiply the 2/sqrt5 by sqrt5/sqrt5

  38. anonymous
    • one year ago
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    so like 3/2sqrt5

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