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Math Help - New to trigonometry

  1. #1
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    New to trigonometry

    Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.

    There are two values that fulfil sin\theta=\frac{\sqrt{3}}{2}

    they are \theta=\:\:,

    second part of the question:

    for sin2\theta=\frac{\sqrt{3}}{2} there are 4 values,

    if \theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the  ta_4 then

    \theta_3-\theta_1=\theta_4-\theta_2=

    and \theta_2-\theta_1=\theta_4-\theta_3=

    thanks for any help.
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  2. #2
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    Quote Originally Posted by davidman View Post
    Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.

    There are two values that fulfil sin\theta=\frac{\sqrt{3}}{2}

    they are \theta=\:\:,

    second part of the question:

    for sin2\theta=\frac{\sqrt{3}}{2} there are 4 values,

    if \theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the  ta_4 then

    \theta_3-\theta_1=\theta_4-\theta_2=

    and \theta_2-\theta_1=\theta_4-\theta_3=

    thanks for any help.
    First, draw yourself an equilateral triangle of side length 2 units.

    Then draw a straight line perpendicular to the base and going through the "top" angle.

    You will now have two 30^\circ, 60^\circ, 90^\circ triangles.

    Their dimensions are 1, \sqrt{3}, 2 (you can check this using Pythagoras).


    Remember that \sin{\theta} = \frac{\textrm{Opposite}}{\textrm{Hypotenuse}}.

    Which angle (out of 30^\circ and 60^\circ) has the Opposite side with length \sqrt{3} units and the Hypotenuse as 2 units?
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by davidman View Post
    Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.

    There are two values that fulfil sin\theta=\frac{\sqrt{3}}{2}

    they are \theta=\:\:,

    second part of the question:

    for sin2\theta=\frac{\sqrt{3}}{2} there are 4 values,

    if \theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the  ta_4 then

    \theta_3-\theta_1=\theta_4-\theta_2=

    and \theta_2-\theta_1=\theta_4-\theta_3=

    thanks for any help.
    I think that you mean to say that \sin\theta=\frac{\sqrt{3}}{2} has two solutions for 0\leq\theta\leq2\pi ? And you wish to find the solutions?

    Well, \sqrt3/2 is given by evaluating \sin\theta at what is commonly referred to as one of the "special angles". Unfortunately, I can't draw too well on this forum, but here's something for you to look at:

    Math Concepts Explained: Special Angles in Trigonometry
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  4. #4
    Super Member bigwave's Avatar
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    Q1,q2

    Quote Originally Posted by davidman View Post
    Ok, I've been looking around on wikipedia etc. but I really don't know how to find the answer to this question at all... would be great with some input.

    There are two values that fulfil sin\theta=\frac{\sqrt{3}}{2}

    they are \theta=\:\:,

    second part of the question:

    for sin2\theta=\frac{\sqrt{3}}{2} there are 4 values,

    if \theta=\theta_1,\:\:\theta_2,\:\:\theta_3,\:\:\the  ta_4 then

    \theta_3-\theta_1=\theta_4-\theta_2=

    and \theta_2-\theta_1=\theta_4-\theta_3=

    thanks for any help.
    sin\theta=\frac{\sqrt{3}}{2}

    \frac{\pi}{3},\frac{2\pi}{3}

    these are quadrants 1 and 2 respectfully
    Last edited by bigwave; January 23rd 2010 at 07:55 PM. Reason: latex error
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  5. #5
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    Quote Originally Posted by Prove It
    Which angle (out of 30^\circ and 60^\circ) has the Opposite side with length \sqrt{3} units and the Hypotenuse as 2 units?
    sin60=\frac{\sqrt{3}}{2}\:\:sin30=\frac{1}{2}\:\:s  in90=1

    Thanks!

    Quote Originally Posted by VonNemo19
    I think that you mean to say that \sin\theta=\frac{\sqrt{3}}{2} has two solutions for 0\leq\theta\leq2\pi ? And you wish to find the solutions?
    Exactly that, yes.

    Quote Originally Posted by bigwave
    sin\theta=\frac{\sqrt{3}}{2}

    \frac{\pi}{3},\frac{2\pi}{3}

    these are quadrants 1 and 2 respectfully
    this brings me to my next question... how do you get two answers when I only get one from evaluating triangles with "special angles"? And how to go from degrees to radian... ?
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  6. #6
    Super Member bigwave's Avatar
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    you may be looking at a triangle in just one quadrant

    see the attached very familiar demonstration of "special angles"

    (\cos\theta,\sin\theta)
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  7. #7
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    OK, now that you have found the "focus angle", we now go to a unit circle approach.

    A unit circle is a circle of radius 1.

    Have a look at the picture below:



    For any angle made with the positive x axis (call it \theta), the vertical distance is \sin{\theta} and the horizontal distance is \cos{\theta}.

    If your angle is 60^\circ = \frac{\pi}{3}^C, can you see another angle which has the same vertical distance?
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  8. #8
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    Ok, I see how there would be two solutions. Thinking in degrees, 180-\theta=\theta_2

    and I googled converting radians/degrees and I must admit that was pretty straightforward...

    however, about the second half of the original question,

    sin2\theta=\frac{\sqrt{3}}{2}

    considering the first part of the question, 2\theta=60\:\:, \theta=30 makes sense the way I see it. But how are there four solutions and not two?
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  9. #9
    Super Member bigwave's Avatar
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    radians x \left(\frac{180^o}{\pi}\right) = degrees


    degrees x \left(\frac{\pi}{180^o}\right)= radians



    Last edited by bigwave; January 23rd 2010 at 10:10 PM. Reason: anwered below
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  10. #10
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    Quote Originally Posted by davidman View Post
    Ok, I see how there would be two solutions. Thinking in degrees, 180-\theta=\theta_2

    and I googled converting radians/degrees and I must admit that was pretty straightforward...

    however, about the second half of the original question,

    sin2\theta=\frac{\sqrt{3}}{2}

    considering the first part of the question, 2\theta=60\:\:, \theta=30 makes sense the way I see it. But how are there four solutions and not two?
    OK, for any given domain, the number of solutions depends on the period of the function.

    Can you see that if you go around the circle (i.e. if you travel 360^\circ) you get to the same angle?

    If you travel another 360^\circ, you get back to the same angle again.

    And so on, and so on...


    So if you wanted to show all solutions to

    \sin{x} = \frac{\sqrt{3}}{2}

    then x = \left\{ 60^\circ, 120^\circ \right \} + 360^\circ n, where n is some integer representing the number of times you have gone around the unit circle.


    Now if x = 2\theta then

    2\theta = \left \{ 60^\circ, 120^\circ \right\} + 360^\circ n.


    Now, to solve for \theta we have to divide both sides of the equation by 2. This includes the 360^\circ...


    So \theta = \left \{ 30^\circ, 60^\circ \right\} + 180^\circ n.


    So letting n = 0 we have

    \theta = \left \{ 30^\circ, 60^\circ \right \}.


    If we let n = 1 we have

    \theta = \left \{ 210^\circ, 240^\circ \right \} .


    If we let n = 2 we have

    \theta = \left \{ 390^\circ, 420^\circ \right \}. These solutions are out of the original 360^\circ.


    So the four solutions are

    \theta = \left \{ 30^\circ, 60^\circ, 210^\circ, 240^\circ \right \}.
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  11. #11
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    Quote Originally Posted by Prove It View Post
    Awesome explanation.
    Ooooooooohhhhhhhhhh, I get it now!

    \theta_1=\frac{\pi}{6}\:,\:\theta_2=\frac{\pi}{3}\  :,\:\theta_3=\frac{7}{6}\pi\:,\:\theta_4=\frac{12}  {9}\pi
    Last edited by davidman; January 23rd 2010 at 09:57 PM.
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