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anonymous

  • 4 years ago

How do you express trigonometric functions of degrees in terms of a reference angle? Please use the following as examples: 4a. cot 285º b. sec -105º c. csc 600º d. tan 3º 6a tan 160º b. csc 115º c. sec 235º d. cot 5º

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  1. Mertsj
    • 4 years ago
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    The reference angle is the angle the terminal side of the angle makes with the x axis.

  2. Mertsj
    • 4 years ago
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    So let's begin. It helps to draw them

  3. Mertsj
    • 4 years ago
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    |dw:1328759040338:dw|

  4. anonymous
    • 4 years ago
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    you mean 285º?

  5. Mertsj
    • 4 years ago
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    |dw:1328759118643:dw|

  6. Mertsj
    • 4 years ago
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    No. I meant 270. that's why there was 15 to go.

  7. anonymous
    • 4 years ago
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    i see. thanks! what is the terminal side of the angle?

  8. Mertsj
    • 4 years ago
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    |dw:1328759258167:dw|

  9. Mertsj
    • 4 years ago
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    Ready for the next one?

  10. anonymous
    • 4 years ago
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    OH. thank you so much! yes.

  11. Mertsj
    • 4 years ago
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    |dw:1328759377288:dw|

  12. Mertsj
    • 4 years ago
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    Same reference angle as the first one.

  13. anonymous
    • 4 years ago
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    ok.

  14. Mertsj
    • 4 years ago
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    Now that we can draw them, we need to pay attention to the directions and the signs.

  15. Mertsj
    • 4 years ago
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    285 degrees. That angle is in quadrant 4. The reference angle is 75 degrees. So the cotangent of 285 is the same as the cotangent of 75 except for the sign.

  16. Mertsj
    • 4 years ago
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    All functions are positive in the first quadrant and all reference angles will be first quadrant angles. However, the cotangent is negative in the fourth quadrant so we have to write: \[\cot (285)=-\cot(75)\]

  17. Mertsj
    • 4 years ago
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    Are you with me?

  18. anonymous
    • 4 years ago
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    yes.

  19. anonymous
    • 4 years ago
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    i guess i'm a little confused with the quadrant thing...

  20. Mertsj
    • 4 years ago
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    Do you know that the cos is like x and the sin is like y?

  21. anonymous
    • 4 years ago
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    I didn't...

  22. anonymous
    • 4 years ago
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    what else should I know?

  23. Mertsj
    • 4 years ago
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    that tan = y/x cot=x/y sec=1/x csc=1/y

  24. Mertsj
    • 4 years ago
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    Knowing that will help you find the sign of the functions in the various quadrants.

  25. anonymous
    • 4 years ago
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    ok, thank you... i don't think I had the hang of that either.

  26. Mertsj
    • 4 years ago
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    For example: since the sec is the reciprocal of the cos, it will be positive in quadrants 1 and 4 because that is where x is positive.

  27. anonymous
    • 4 years ago
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    I see.

  28. Mertsj
    • 4 years ago
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    So let's do the next one sec(-105)=

  29. Mertsj
    • 4 years ago
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    We have seen that -105 is a third quadrant angle. Will the secant be positive or negative in the third quadrant? Remember the sec agrees with the cos that the cos is like x

  30. anonymous
    • 4 years ago
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    the secant will be negative right?

  31. Mertsj
    • 4 years ago
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    yes. So we would write: \[\sec(-105)=-\sec(75)\]

  32. anonymous
    • 4 years ago
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    could you write it as sec(-75) or would that be wrong?

  33. anonymous
    • 4 years ago
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    also, i just rechecked the book and 4d is actually in radians (tan=3) and 6d is also in radians (cot=5) how do we do those? I understand the degrees now... but i don't understand the radians.

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