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haleyking345

  • one year ago

Find six trig values if sec(x)=-3/2 and tan(x)>0 I got cos(x)=-2/3 but that is all I could get.

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  1. alanli123
    • one year ago
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    if sec x = - 3/2 then cos x = -2/3 therefore sin x = 1 - (-2/3)^2 = 1- 4/9 = 5/9 therefore, csc x = 9/5 furthermore tan x = sin x / cos x = 5/9 / -2/3 = -5/6. however, since tan x > 0 we have that angle x must be in quadrant 1 or 3. since cos x is negative, the angle x must be in quadrant 3 and thus tan x = 5/6 (tan is positive in quadrant 3) now, finally cot x = 6/5

  2. NoelGreco
    • one year ago
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    You're first step should be to draw the angle on the coordinate plane.

  3. terenzreignz
    • one year ago
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    @alanli123 Fatal Error :P It's sin² x = 1 - cos² x You used sin x = 1 - cos² x

  4. terenzreignz
    • one year ago
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    Hey, @haleyking345 Don't start writing down alanli's answers yet, they were all based on the conclusion that sin x = 5/9 which is WRONG Now relax and sit back... let's draw it on the coordinate plane, as @NoelGreco suggested. So, you know cosine is negative, and in which of the quadrants is cosine negative?

  5. haleyking345
    • one year ago
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    Quadrant 2

  6. terenzreignz
    • one year ago
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    And...? There's another.

  7. haleyking345
    • one year ago
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    oh quadrant 3

  8. terenzreignz
    • one year ago
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    Yep, Quadrant 2 or Quadrant 3. But it ALSO says that tangent is positive, and between quadrant 2 and quadrant 3, in only one of them is tangent positive, which is it?

  9. haleyking345
    • one year ago
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    Quadrant 3 right?

  10. terenzreignz
    • one year ago
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    Correct. So let's draw our triangle in quadrant 3... |dw:1361494367065:dw|

  11. terenzreignz
    • one year ago
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    Now, the cosine of the angle is -2/3... how do you figure that into your triangle?

  12. terenzreignz
    • one year ago
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    Remember, cosine is the adjacent side over the hypotenuse... or x/r, where x is the length of the horizontal leg, and r is the radius (hypotenuse)

  13. haleyking345
    • one year ago
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    put -2 on x-axis line and 3 on the longest side

  14. terenzreignz
    • one year ago
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    Very good. Hopefully, you figured out to put -2 on the x-axis line because it's to the left of the origin (0, 0) Anyway, |dw:1361494618728:dw| Now what's the length of the missing side? (HINT: Pythagorean Theorem)

  15. haleyking345
    • one year ago
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    is it square root of 5?

  16. terenzreignz
    • one year ago
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    Almost... but look, this is the y-value, the vertical leg of the triangle, right? And it's in the third Quadrant (QIII) And in the third quadrant, both x and y should be negative, no? :)

  17. haleyking345
    • one year ago
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    should it be - square root of 5?

  18. terenzreignz
    • one year ago
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    Yes. it should :P So far, is everything clear to you? :) |dw:1361494965439:dw|

  19. haleyking345
    • one year ago
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    yes it is

  20. terenzreignz
    • one year ago
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    Okay, so now, you can easily determine the sine of the angle :) sine is the opposite side over the hypotenuse, or x/r What's the sine?

  21. haleyking345
    • one year ago
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    - sqrt 5/3

  22. terenzreignz
    • one year ago
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    I saw the sine...and it opened up my LOL Yes, correct! And with the sine, you can easily get the cosecant :D What is it?

  23. haleyking345
    • one year ago
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    3/ -sqrt 5

  24. terenzreignz
    • one year ago
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    Yeah... but some teachers can be picky and make you rationalise the denominator... fine... \[\huge -\frac{3}{\sqrt{5}}=-\frac{3\sqrt{5}}{5} \] May as well put it up like that...

  25. terenzreignz
    • one year ago
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    Now, back to the triangle. You can also now easily determine the tangent of the triangle :) Tangent is the opposite side, over the adjacent side or y/x What's the tangent?

  26. terenzreignz
    • one year ago
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    tangent of the *angle sorry

  27. haleyking345
    • one year ago
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    sqrt 5/2

  28. terenzreignz
    • one year ago
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    Yep :) As you can see, the tangent is positive. Now with the tangent, you can easily get the cotangent. What is it? Also, rationalise it.

  29. haleyking345
    • one year ago
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    2/sqrt 5 or 2sqrt 5/5

  30. terenzreignz
    • one year ago
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    And there you have it, your six trigonometric functions :D I'd like to point out that had alanli used the correct identity for sine, that is sin²x = 1 - cos²x it would have given... sin²x = 1 - (-2/3)² = 1 - (4/9) = 5/9 But you still have to get the square root of this sin x = (+/-) sqrt(5)/sqrt(9) = (+/-)sqrt(5)/3

  31. haleyking345
    • one year ago
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    ok that makes sense. thank you very much!!!

  32. terenzreignz
    • one year ago
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    No problem :)

  33. terenzreignz
    • one year ago
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    Need to go now, anyway :D ---------------------- Terence out

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