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Math Help - cosine

  1. #1
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    Exclamation cosine

    It's one of those trigonometric ones that always make my brain go huh?

    If cos x= (-4/5) and tan x>0, find the value of Sin x.

    I figured that since Cos was negative, it was in either quadrants II or III. Since Tan is positive, it must be in III. After that....I get lost.

    ~Wynne
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    Quote Originally Posted by Blodwynne View Post
    It's one of those trigonometric ones that always make my brain go huh?

    If cos x= (-4/5) and tan x>0, find the value of Sin x.

    I figured that since Cos was negative, it was in either quadrants II or III. Since Tan is positive, it must be in III. After that....I get lost.

    ~Wynne
    Think of a right triangle. The cosine of an angle is the ratio of the adjacent side to the hypotenuse. So, ignoring the negative for now:

    Code:
            /|
      5   /  |
        /    | b
      /      |
    / θ      |
    ----------
        4
    By the Pythagorean theorem, we have

    b = \sqrt{25 - 16} = \sqrt9 = 3 (this is a Pythagorean triple: all sides are integers).

    Now, \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, so we have

    \sin\theta = \frac b5 = \frac35.

    But, our \theta is in quadrant I. Adding 180^\circ to get into quadrant III will make the sine negative:

    \sin x = -\frac35
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by Blodwynne View Post
    It's one of those trigonometric ones that always make my brain go huh?

    If cos x= (-4/5) and tan x>0, find the value of Sin x.

    I figured that since Cos was negative, it was in either quadrants II or III. Since Tan is positive, it must be in III. After that....I get lost.

    ~Wynne
    Draw a diagram

    cosine-trig.jpg

    \sin(x)=\frac{opposite}{hypotenuse}=\frac{-3}{5}

    Good luck

    I'm too slow
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    Quote Originally Posted by TheEmptySet View Post
    I'm too slow
    Yes, but your graph with the triangle oriented the proper way is probably easier to understand than my explanation. Nice work.
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    Needless to say, that was far off from what I got. It turns out I did what I always do: I overthunk the problem. >.<

    Sorry, but another trig problem is giving me issues. it is:

    47. sin(cot^-1 5/12)

    What I know so far:

    Cotangent is the inverse of tangent.
    This seems like another ratio of sides.

    ~Wynne
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    Quote Originally Posted by Blodwynne View Post
    Needless to say, that was far off from what I got. It turns out I did what I always do: I overthunk the problem. >.<

    Sorry, but another trig problem is giving me issues. it is:

    47. sin(cot^-1 5/12)

    What I know so far:

    Cotangent is the inverse of tangent.
    This seems like another ratio of sides.

    ~Wynne
    Indeed it is. Set it up the same way:

    x = {\rm arccot}\frac5{12}\Rightarrow\cot x = \frac5{12}

    You know that \cot\theta = \frac{\text{adjacent}}{\text{opposite}}, so draw your triangle.

    And since the cotangent is positive, and since the range of arccot is [0,\;\pi] (or sometimes it is defined as \left[-\frac\pi2,\;\frac\pi2\right]), the angle you are dealing with ( x) is in quadrant I.

    Edit: I thought I should add: Be careful when you say that cotangent is the "inverse" of tangent. The word inverse usually implies an inverse function--e.g., arcsin is the inverse of sin. The proper word in your case is "reciprocal": the cotangent is the reciprocal of the tangent (reciprocal meaning a number's multiplicative inverse).
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  7. #7
    Behold, the power of SARDINES!
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    Quote Originally Posted by Blodwynne View Post
    Needless to say, that was far off from what I got. It turns out I did what I always do: I overthunk the problem. >.<

    Sorry, but another trig problem is giving me issues. it is:

    47. sin(cot^-1 5/12)

    What I know so far:

    Cotangent is the inverse of tangent.
    This seems like another ratio of sides.

    ~Wynne
    Here is another diagram. I always draw them it helps.

    We know that it has to be in quadrant I because the range of the arccot function is [0,\pi] and the argument is positive

    cosine-trig.jpg

    \sin(\cos^{-1}\left( \frac{5}{12}\right))=\frac{12}{13}

    Edit: Too slow again. Twice in the same post
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