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Math Help - Analytic Trigonometry help...

  1. #1
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    Analytic Trigonometry help...

    Hi,

    Today we started working on what our book calls, Analytic Trigonometry. I am quite lost as to how I solve these problems. My college has a "Math Lab," but they were extremely busy when I went during my break between classes. I wasn't able to get any help there before I had to take off for my next class.

    The first is (I will post some more after I figure this out if I still need the help):

    2sin^2 theta -1 = 0

    So I add one to the RS and divide by two leaving me with...

    sin^2 theta = 1/2

    Then I take the sqrt of both sides and get:

    sin theta = 1/sqrt(2).

    That is where I am lost. On some of the problems I am able to look at the unit circle to find the answer I am looking for, but what is 1/sqrt(2) on the unit circle? Some are simple, like the sin of sqrt of 1/2...

    Then after I find that number I always seem to have issues getting the points. In this case I would draw the sin curve and put a line at 1/2.

    Thanks in advance for the help! It is greatly appreciated!


    -Tom
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  2. #2
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    You're looking for angle, theta, with the property that sin(theta)=1/sqrt{2}.

    This happens to be pi/4. You just have to be familiar with certain values of the 30-69-90 and 45-45-90 triangles. Lot's of times the angle can't be expressed in a closed form by hand.
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  3. #3
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    Quote Originally Posted by TomCat View Post
    Hi,

    Today we started working on what our book calls, Analytic Trigonometry. I am quite lost as to how I solve these problems. My college has a "Math Lab," but they were extremely busy when I went during my break between classes. I wasn't able to get any help there before I had to take off for my next class.

    The first is (I will post some more after I figure this out if I still need the help):

    2sin^2 theta -1 = 0

    So I add one to the RS and divide by two leaving me with...

    sin^2 theta = 1/2

    Then I take the sqrt of both sides and get:

    sin theta = 1/sqrt(2).

    That is where I am lost. On some of the problems I am able to look at the unit circle to find the answer I am looking for, but what is 1/sqrt(2) on the unit circle? Some are simple, like the sin of sqrt of 1/2...

    Then after I find that number I always seem to have issues getting the points. In this case I would draw the sin curve and put a line at 1/2.

    Thanks in advance for the help! It is greatly appreciated!


    -Tom
    If the answer is not obvious to you, use a calculator

    sin(theta) = 1/sqrt(2)
    theta = arcsin(1/sqrt(2)) = 45 degrees

    to get the answer and then draw the picture to gain familiarity.

    But are you really supposed to figure these out without a calculator?
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  4. #4
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    No. I believe that we shouldn't be using a calculator.

    I'm still stumped though. What I'm trying to do is find where the sin theta =1/sqrt(2) line intercepts the sin curve I believe.

    Here's what we did in class...

    tan^2 theta = 1/3

    tan theta = + or - 1/sqrt(3).

    Then we graph tan and draw the lines 1/sqrt(3) and - 1/sqrt(3). Where those two lines intercept the graph, from 0 to 2pi are what I'm looking for.

    So theta = pi/6. Then I have to add pi to that and get 7pi/6. That was for the positive...

    The negative is -pi/6, so add pi and get 5pi/6 and add pi to that and get 11pi/6.

    So the intercepts are {pi/6, 5pi/6, 7pi/6, 11pi/6} .

    That one is very simple to me.

    Would the intercepts for the problem I posted above be {pi/4, 5pi/5}. My prof did not do a good job explaining this at all...

    Thanks again!


    -Tom
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