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Math Help - Trigonometric Function

  1. #1
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    Trigonometric Function

    Ok I'm asked to find the other Trig. Functions given one value.
    Here's the question:

    sec = -2 tan > 0

    So sec = 1/x, so I get 1/(-1/2) = -2.
    If I'm understanding it correctly, x = cos, therefore x = -1 and r = 2?
    Also since tan > 0, that means y is also negative since tan = y/x and x is already -1?
    I just want to know if I'm understanding it correctly.
    Thanks.
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  2. #2
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    Quote Originally Posted by ktcrew View Post
    Ok I'm asked to find the other Trig. Functions given one value.
    Here's the question:

    sec = -2 tan > 0

    So sec = 1/x, so I get 1/(-1/2) = -2.
    If I'm understanding it correctly, x = cos, therefore x = -1 and r = 2?
    Also since tan > 0, that means y is also negative since tan = y/x and x is already -1?
    I just want to know if I'm understanding it correctly.
    Thanks.
    You really need to work on your notation.

    I would translate what you wrote into something more meaningful, but there are too many conceptual errors in there to make it worth the effort.

    We know that
    sec(q) = -2 and tan(q) > 0.

    Thus
    cos(q) = \frac{1}{sec(q)} = \frac{1}{-2} = -\frac{1}{2}.

    This means that q is in either quadrant II or quadrant III. Given that tan(q) > 0, we can reduce this to q being in quadrant II.

    So, to find sin(q), just use sin^2(q) + cos^2(q) = 1. The value of sin(q) will be positive since q is in QII.

    Now the rest is easy:
    tan(q) = \frac{sin(q)}{cos(q)}

    csc(q) = \frac{1}{sin(q)}

    cot(q) = \frac{1}{tan(q)}

    -Dan
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  3. #3
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    Sorry about the notation, I haven't really learned all the tags yet.
    But I don't quite understand tan(q) being in quadrant 2.
    Shouldn't sin(q) be negative since cos(q) is a negative value? And a negative numerator divided by a negative denominator would result in a positive number because the signs cancel each other, no?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ktcrew View Post
    Sorry about the notation, I haven't really learned all the tags yet.
    My comment wasn't about the lack of the fancy looking tags. It was about statements like
    "sec = -2 and tan > 0"

    You need to place an angle in that statement. I'll admit it's a little fussy, but it's just good practice at being clear in your writing.

    Quote Originally Posted by ktcrew View Post
    But I don't quite understand tan(q) being in quadrant 2.
    Shouldn't sin(q) be negative since cos(q) is a negative value? And a negative numerator divided by a negative denominator would result in a positive number because the signs cancel each other, no?
    Yes, you are right. I should have said that angle q is in QIII. The tangent function is positive in quadrants I and III, it is negative in QII. My apologies.

    -Dan
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  5. #5
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    Ok, thanks.
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