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anonymous

  • 5 years ago

0=arcsin(-1/2) find cos(x), tan(x), cot(x), sec(x), csc(x)

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  1. anonymous
    • 5 years ago
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    something is missing from the 0 = arcsin(-1/2)...because arcsin(-1/2) cant be 0

  2. anonymous
    • 5 years ago
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    \[\sin^{-1}(-1/2) =0 \] or \[\sin^{-1} (x)=0\]???

  3. anonymous
    • 5 years ago
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    because arcsine(-1/2) can't be zero as sklee said

  4. anonymous
    • 5 years ago
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    I meant arcsin(-1/2)=x :)

  5. anonymous
    • 5 years ago
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    I wrote it wrong, the 0 is supposed to be an x. It was originally a theta, but the keyboard does not have the theta symbol so i replaced it with an x. x=arcsin(-1/2) find cos(x), tan(x), cot(x), sec(x), csc(x)

  6. anonymous
    • 5 years ago
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    ok first you find the x which is clearly either in the 3rd or 4th quadrant because it's negative , that's: \[x=210^o, x=330^o\]

  7. anonymous
    • 5 years ago
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    Now just substitute for each value of x in each function.

  8. anonymous
    • 5 years ago
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    if you're wondering how I got the values of x, it's because sinx has a value of 1/2 when x=30, then I took the correspondent to 30 in both 3rd and 4th quadrant in order to get the negative value., to do that to get the value in the 3rd quadrant just add 180 to 30 to get 210. and to get the angle in the 4th quadrant just subtract 30 from 360 to get 330

  9. anonymous
    • 5 years ago
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    jajuan, how old are you?

  10. anonymous
    • 5 years ago
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    he looks like a baby in the picture:P.. just kidding

  11. anonymous
    • 5 years ago
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    I understand all of the degrees and radians, but for some reason it didn't run through my head to apply 330 to the other functions lol and I am 20

  12. anonymous
    • 5 years ago
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    well don't take 330.. take -30, they are the same

  13. anonymous
    • 5 years ago
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    cool thanks

  14. anonymous
    • 5 years ago
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    you're welcome.. I just hope the answer makes sense to you, if it does, you can fan me.. I don't mind at all ;)

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