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OscarGon1234

  • 2 years ago

sinθ if cscθ= -2

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  1. lonnie455rich
    • 2 years ago
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    well what do you know about those 2 functions in relation to each other

  2. OscarGon1234
    • 2 years ago
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    That's all it says on my homework paper, 'Use the appropriate reciprocal identity to find each function value.'

  3. lonnie455rich
    • 2 years ago
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    ya, but you didn't answer my question. you have soh cah toa. and the three inverse functions. which one is the inverse of sin

  4. lonnie455rich
    • 2 years ago
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    or maybe call it reciprocal? im not good at my language

  5. OscarGon1234
    • 2 years ago
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    As far as I know it's reciprocal is csc=1/sin

  6. lonnie455rich
    • 2 years ago
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    alright then. what do you think the answer is

  7. OscarGon1234
    • 2 years ago
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    something like -1/2.

  8. OscarGon1234
    • 2 years ago
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    I just have different ones like tanθ if cotθ=sqrt6/2 and I don't really understand

  9. lonnie455rich
    • 2 years ago
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    so your correct on this problem. do you know about radians yet for the other problems?

  10. OscarGon1234
    • 2 years ago
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    WHat do you mean radians?

  11. lonnie455rich
    • 2 years ago
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    and cot and tan are reciprocals as well. radians-degrees. have you now done the unit circle yet?

  12. OscarGon1234
    • 2 years ago
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    I don't recall so I am going to have to say no.

  13. lonnie455rich
    • 2 years ago
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    well. hmm.. you need to learn about radians and degrees for trigonometry. but just to avoid that. remember that cot and tan are reciprocal identities as well. but you cant have a radical in the denominator

  14. OscarGon1234
    • 2 years ago
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    Yea, so how would I go on in doing that? I am in Pre-cal if that makes a difference.

  15. lonnie455rich
    • 2 years ago
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    http://tutorial.math.lamar.edu/cheat_table.aspx print that off. it will do wonders for your trig identities. do you know how to rationalize the denominator

  16. lonnie455rich
    • 2 years ago
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    tell me where you are at on the second problem

  17. OscarGon1234
    • 2 years ago
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    All I can really understand is that tanθ= 1/sqrt6/2, but I don't know how to make it like without the sqrt in the denominator.

  18. lonnie455rich
    • 2 years ago
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    hm. what is the reciprocal of \[\frac{ \sqrt{6} }{ 2 }\]

  19. OscarGon1234
    • 2 years ago
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    Umm tbh I'm not really sure :/

  20. napalmgrenade
    • 2 years ago
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    You multiply the numerator and denominator by your square rooted number to get rid of the root in the bottom, this is known as rationalizing it. Here is an example |dw:1385354774715:dw|

  21. OscarGon1234
    • 2 years ago
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    Oh ok I see so now I have to get the reciprocal.

  22. OscarGon1234
    • 2 years ago
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    but it would be 6/2sqrt2

  23. lonnie455rich
    • 2 years ago
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    reciprocal would be flipping your fraction like if you were dividing by fractions \[\frac{ 2 }{ \sqrt{6} }\] this is the reciprocal of my above fraction. and when you multiply 2 square roots like that above you would have 2 not 1

  24. lonnie455rich
    • 2 years ago
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    so if you multiply this \[\frac{ 2(\sqrt{6)} }{ \sqrt{6}*\sqrt{6} }\] that is how you rationalize the denominator

  25. OscarGon1234
    • 2 years ago
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    oh I see now, so then the answer is 2/√6, I don't have to rationalize anything I believe.

  26. lonnie455rich
    • 2 years ago
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    keeping in mind \[\sqrt{6}*\sqrt{6}=\sqrt{36}\]

  27. OscarGon1234
    • 2 years ago
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    Ok, thanks a lot! really appreciate the help!

  28. lonnie455rich
    • 2 years ago
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    you cant have square roots in the denominator. you have to rationalize it like I did in the above step

  29. lonnie455rich
    • 2 years ago
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    see how I multiplied the top and bottom by \[\sqrt{6}\]

  30. OscarGon1234
    • 2 years ago
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    so 2*sqrt6/6

  31. napalmgrenade
    • 2 years ago
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    Some teacher dont require rationalizing, but it is generally required, so if it wasn't required you would be right. From your answer above, look to simplify it !

  32. lonnie455rich
    • 2 years ago
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    looks like you can simplify it a little more think of it as \[\frac{ 2 }{ 6 }*\sqrt{6}\]

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