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Math Help - Trig. Ratios for Angles Greater than 90 degrees.

  1. #1
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    Trig. Ratios for Angles Greater than 90 degrees.

    1. State all the angles between 0 degrees and 360 degrees that make each equation true.
    d) tan135 = -tan______________ (there's a blank space here)

    So, I understand that you have to use the appropriate equation, but the textbook gives two answers in the book, while I only got one. Also, I don't really understand what this process is doing. Could someone explain what the equations mean for each quadrant?

    I attached a photo from the textbook so you can see what I'm confused about. The textbook doesn't really explain well, and I can't find anything similar online. So, I'd really appreciate an explanation of this! Thanks.

    And this was my solution (for one of the two possible answers according to the textbook):
    tan(180-135) = -tanX
    45 = X
    Attached Thumbnails Attached Thumbnails Trig. Ratios for Angles Greater than 90 degrees.-20140510_135345.jpg   Trig. Ratios for Angles Greater than 90 degrees.-20140510_135333.jpg  
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  2. #2
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    Re: Trig. Ratios for Angles Greater than 90 degrees.

    Quote Originally Posted by eleventhhour View Post
    1. State all the angles between 0 degrees and 360 degrees that make each equation true.
    d) tan135 = -tan______________ (there's a blank space here)

    So, I understand that you have to use the appropriate equation, but the textbook gives two answers in the book, while I only got one. Also, I don't really understand what this process is doing. Could someone explain what the equations mean for each quadrant?

    I attached a photo from the textbook so you can see what I'm confused about. The textbook doesn't really explain well, and I can't find anything similar online. So, I'd really appreciate an explanation of this! Thanks.

    And this was my solution (for one of the two possible answers according to the textbook):
    tan(180-135) = -tanX
    45 = X
    The best way to understand all this I think is to be able to visualize the geometry.

    Imagine a circle with radius 1, centered at (0,0).

    Now draw a line segment from (0,0) to somewhere on that circle. Note the angle that segment makes with the x-axis.

    That point on the circle will have coordinates (x,y).

    Let the angle the segment makes with the axis be $\theta$. Then

    $\cos(\theta) = x$

    $\sin(\theta) = y$

    $\tan(\theta) = \dfrac y x$

    and this is true for whatever angle/point on the circle you choose.

    So for a given $\theta$ there are going to be 2 points on the circle that have a given value for $\cos(\theta)$ and $\sin(\theta)$ and $\tan(\theta)$

    Suppose you have a point (x,y) in the first quadrant that makes angle $\theta$ with the x-axis.

    From above you can see that the points

    (x,y) and (x,-y) will have the same cosine

    (x,y) and (-x,y) will have the same sine

    (x, y) and (-x,y) will have cosines that are the negative of one another

    (x,y) and (x,-y) will have sines that are the negative of one another

    so the cosine is preserved when reflecting about the y axis and the sine is preserved when reflecting about the x axis.

    the cosine is negated when reflecting about the x axis, and the sine is negated when reflecting about the y axis.

    The tangent is thus preserved when reflecting about the origin, i.e. (x,y) -> (-x,-y) and is negated when reflecting about a single axis.

    Now with a bit of thought you can turn these reflections into operations on $\theta$

    for example a reflection about the y axis is $\theta \to (\pi - \theta)$

    see if you can come up with the angular operation for reflection about the x-axis and about the origin.
    Thanks from eleventhhour
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  3. #3
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    Re: Trig. Ratios for Angles Greater than 90 degrees.

    Ugh, terrible mistake in here.

    cosine is preserved reflecting about the x axis and negated reflecting about the y axis.

    sine is preserved reflecting about the y axis and negated reflecting about the x axis.
    Thanks from JeffM
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