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Precalculus trig question?

  1. Jun 9, 2005 #1
    Can anyone tell me how the text arrived at the following answer?

    cos 7pi/6 = -sqrt3/2

    I just don't understand how to draw the triangle inside the unit circle and to visualize the cosine ratio in order to arrive at the answer(what are the lengths of the adjacent and hypotenuse?)..why is it negative too?

    baby steps please?

    thank you to anyone that can help.
  2. jcsd
  3. Jun 9, 2005 #2


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    Well, 7 pi/6 is "the same as" pi/6 but in the 3rd quadrant. If you think of a cosine graph, then when you get to pi, the graph is -1. pi is equal to 6pi/6, so 7pi/6 is just that plus pi/6. So you evaluate cos pi/6 but make it negative since the function is negative here.

    That's not a very rigorous argument, but it might help...

    You could also evaluate

    [tex]cos( \pi + \frac{\pi}{6})[/tex] by expanding it thus
    [tex]cos( \pi + \frac{\pi}{6}) = \cos (\pi)\cos(\pi/6) - \sin(\pi)\sin(\pi/6)[/tex]
    which equals -(sqrt3)/2.

    I'm not quite sure what the unit circle is about; I'll have to think about that, though it sounds like you use constraints on a triangle to calculate cos as a ratio. (Which you probably knew already!)
  4. Jun 9, 2005 #3
    Okay, now you have run into something I hate! The way books get to their answers sometimes makes us feel like a trick has been played.

    They happened to know the lengths for a different (yet highly popular) 30-60-90 triangle -- one for which they themselves just memorized the lengths of the sides and then used for "their" answer. That is, we are on a unit circle with a triangle of H=1, but they left the circle, went to a different triangle, and came back with their answer (which is correct, but is not plainly visible in our unit-circle triangle) -- let me explain.

    A) First, I'll back up and confirm where we are. We already established that 7pi/6 is really just a 30 degree triangle below the x axis.

    B) The ratio of Cosine is A/H;

    C) It turns out that a 90 degree triangle with a hypotenuse of 2, leg length of 1 and length of other leg being sqrt(3) is a 30-60-90 triangle. That’s their trick they are not being clear about. I had mentioned in my last post that you break out a protractor, draw a 30-60-90 triangle and then measure the ratio of A/H. That would force you to think of Cosine as a simple exercise in measuring lines and finding ratios. But what our books do (inadvertently) is hide the fact that we can derive the right answer; this is done by using this special 30-60-90 triangle where the H=2.

    D) Instead of working with our triangle, where H=1, we are going to walk across the room and work with a triangle with an H=2. Then, when we get the ratio of A/H, we’ll go back to our H=1 triangle and apply the ratio to it. We can do this because triangle ratios stay the same as long as they have all the same angles – i.e. even if the hypotenuses are of different length! So, let’s work with H=2 instead of H=1.

    E) 30-60-90 degree triangles have been studied for a long time, and are rather special. I am sorry to say this, I just automatically think “1-2-sqrt(3)” when I hear 30-60-90 (where the 2 is the length of H). There is a good page that talks about these fantastic 30-60-90 triangles: http://www.themathpage.com/aTrig/30-60-90-triangle.htm.

    F) As a side note: Starting with a 90 degree triangle where, H=2 we can confirm O=1, and A=sqrt(3) in respect to the Pythagoreans theorem. But that’s a side not…

    G) So, A=sqrt(3), H=2 in our 30-60-90 triangle. All we have to do is find A/H. That is sqrt(3)/2. And, because we know that we are in a negative A direction on the unit circle, the answer is negative. It’s all very mechanical in one sense.

    H) I have given you the steps to the answer, now I’d like to mention one other thing that drives some people nuts. Teachers don’t like to have square root signs in their denominators. So, if they asked you for the Tangent of 7pi/6, you would expect the answer to be O/A which is 1/sqrt(3). And it is. But they multiply top and bottom by sqrt(3) -- which is the same as multiplying by one! -- to get the radical out of the denominator. And that gives the funny looking answer: sqrt(3)/3. If you don’t know that they switched triangles on you (going from H=1 to H=2), and that they don’t like radicals in denominators, then you’d pull your hair out wondering how they ever made up such an answer.

    Steve Rives
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