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Thread: Trig identities

  1. #1
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    Question Trig identities

    Hi,

    I hope someone can help. I'm suppose to show that the following equation is true given the diagram shown:
    Trig identities-screen-shot-2017-05-29-5.06.57-pm.png

    The textbook solution is as follows:
    Trig identities-screen-shot-2017-05-29-5.08.13-pm.png

    I'm confused how we could possibly get the coordinates (-y, x), as explained above in the solution. It just does not make any sense to me. It would be great for someone to explain why this is the case, with the aid of a visualization.

    Sincerely,
    Olivia
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  2. #2
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    Re: Trig identities

    Take the triangle on the left,0 with one leg, on the x-axis, and the other parallel to the y-axis, and rotate it 90 degrees counter clock-wise. The x-axis, rotated 90 degrees, becomes the y-axis, and the leg parallel to the y-axis becomes parallel to the x-axis. So the point at the vertex of the right triangle on the circle, with coordinates (x, y) is rotated to the point with coordinates (-y, x).

    Since this is unit circle, that point has x= cos(\theta), y= sin(\theta). So the point after rotating has x= -sin(\theta), and y= cos(\theta). That is, cos(\theta+ \frac{\pi}{2})= -sin(\theta) and [tex]sin(\theta+ \frac{\pi}{2})= cos(\theta)
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  3. #3
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    Re: Trig identities

    the right triangles in quads I and II are congruent ...
    Attached Thumbnails Attached Thumbnails Trig identities-trig_identity.jpg  
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    Re: Trig identities

    ummm sorry, but what you said isn't clear to me at all. "Take the triangle on the left,0 with one leg, on the x-axis, and the other parallel to the y-axis, and rotate it 90 degrees counter clock-wise" - what does this look like?

    There are other things which didn't make sense to me, and I think that a drawing would be tremendously helpful. Please let me know if you can do this. I would really appreciate it.
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  5. #5
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    Re: Trig identities

    you posted ...

    I'm confused how we could possibly get the coordinates (-y, x), as explained above in the solution.
    then you said ...

    ... and I think that a drawing would be tremendously helpful.
    did you even look at the labeled diagram I attached to post #3 ? Maybe I need to post it again ...



    From Geometry, the right triangles in quadrants I and II are congruent by ASA and their corresponding parts are congruent.

    recall SOCAHTOA?

    in quadrant I ...

    $\sin{\theta} = \dfrac{opposite}{hypotenuse} = \dfrac{y}{r}$

    The right triangle in quadrant II is the reference triangle for angle $\left(\theta + \dfrac{\pi}{2}\right)$ that is in standard position.

    $\cos\left(\theta + \dfrac{\pi}{2}\right) = \dfrac{adjacent}{hypotenuse} = \dfrac{-y}{r}$

    therefore ... $\cos\left(\theta + \dfrac{\pi}{2}\right) = -\sin{\theta}$



    There are other things which didn't make sense to me ...
    then you need to state what doesn't make sense ...
    Last edited by skeeter; May 29th 2017 at 06:27 PM.
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    Re: Trig identities

    Okay that makes more sense now since you were more specific. Thank you!
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    Re: Trig identities

    Hi, I actually have one more question regarding this math problem! I understand why point P would have the coordinates (-y,x) and point Q would has the coordinates (x,y)... but I don't entirely understand how the triangle in quadrant II is the reference triangle for cos(θ+pi/2)? How was that found?
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  8. #8
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    Re: Trig identities

    If you draw a horizontal line from point P to the y-axis, you turn the triangle into a rectangle. Let's call the point of intersection R $(0,x)$. O is the standard label for the origin $(0,0)$. The triangle PRO is congruent to the triangle in quadrant I, just rotated $\dfrac{\pi}{2}$ radians about the origin. Do you see that? Then, triangle PRO is also congruent with the triangle you already had in quadrant II.
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