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Visualize this type of Combined Trigonometric Functions

  1. Sep 7, 2016 #1
    1. The problem statement, all variables and given/known data

    Show that sin 600° . cos 330° + cos 120° . sin 150° = - 1

    2. Relevant equations

    I know that sinΘ = opposite/hypotenuse and cosΘ = adjacent/hypotenuse.

    3. The attempt at a solution

    I am equipped with knowledge about what sinΘ and cosΘ is from right angled triangle.
    I stand in front of a wall and visualize opposite side, adjacent side and hypotenuse but I am unable to clearly picture what sin 600° . cos 330° + cos 120° . sin 150° = - 1 mean.

    What is the meaning of the equation?
    What is the meaning of -1 on the right hand side of the equation?
     
  2. jcsd
  3. Sep 7, 2016 #2

    RUber

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    I would begin with the idea of sin z and cos z as the y and x coordinates of the unit circle at an angle of z. This makes the angles greater than 90 degrees make more sense.
    Secondly, since on a circle, 360 degrees is the same as 0 degrees (back to the starting point), these trig functions are periodic over 360 degrees. With that knowledge, you can change sin(600) into sin(240).
    Are there any rules or identities you have been learning lately that might help you change the equation around into something else?
     
  4. Sep 7, 2016 #3

    PeroK

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    ##\sin## and ##\cos## are periodic with period 360°. You should also understand how the values of ##\sin## and ##\cos## relate to the four "quadrants". Does that mean anything to you?
     
  5. Sep 7, 2016 #4

    fresh_42

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    What do you know about the symmetries of ##\sin## and ##\cos##, i.e. e.g. how are ##\sin x## and ##\sin -x## related? Or ##\sin x## and ##\cos(x + 90°)##?
     
  6. Sep 7, 2016 #5

    RUber

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    Once you draw it out, you are dealing with 30-60-90 triangles. You should be able to find the sine and cosine of the angles indicated.
    The final equation will make sense once you put the values in place of the trig functions. I assume this was what was implied in the direction to "show" that the equation is true.
     
  7. Sep 7, 2016 #6
    sin 600° . cos 330° + cos 120° . sin 150°
    From the above, it can imply
    sin (360° + Θ) . cos (360° - Θ) + cos (360° - Θ) . sin (360° - Θ)
    I am not trying to solve to obtain the result as -1, but instead I am trying to see what phenomenon the equation is trying to portray.
    And I have no idea what -1 on the right hand side is trying to tell me.
    Please put some light on the equation and let me know what the equation is all about.
     
  8. Sep 7, 2016 #7

    RUber

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    This is simply an equation like 1/2 + 1/2 = 1. Instead of 1/2, you have some functions to evaluate first.
     
  9. Sep 7, 2016 #8

    PeroK

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    Actually solving the equation might shed some light on what it means.
     
  10. Sep 7, 2016 #9
    To me it appears that a person 'x' has instantaneously come to express such trigonometry by looking at some occurrence.
    I want to assume that I have a robotic hand that is governed by expression:
    sin 600° . cos 330° + cos 120° . sin 150°
    If such robotic hand is operated upon by such equation, how will the robotic hand behave?
     
  11. Sep 7, 2016 #10

    PeroK

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    I have no idea what that means. To give you a hand, your first step might be:

    ##\sin(600) \cos(330) + \cos(120) \sin(150) = \sin(240) \cos(-30) + (-\cos(60)) \sin(30)##

    You might also want to look at this:

    https://www.liverpool.ac.uk/~maryrees/homepagemath191/trigid.pdf
     
  12. Sep 7, 2016 #11

    RUber

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    Remember that once you put an angle into your sine and cosine functions, they are just numbers.
    For example,
    ## \sin(90) = 1,\quad \cos(45)=\frac{\sqrt{2}}{2}, \quad \cos(60) = \frac12.##
     
  13. Mar 18, 2017 #12
    Please may I know how you would visualize the above statement?
     
  14. Mar 18, 2017 #13

    PeroK

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    Using the graphs of the sine and cosine functions.
     
  15. Mar 18, 2017 #14
    I have a confusion related to the above statement.
    When I consider the right-hand side of the equation ##= \sin(240) \cos(-30) + (-\cos(60)) \sin(30)## there are two trigonometric functions separated by plus sign. Now, if I draw a graph, how do I visualize the product of ##= \sin(240) \cos(-30)## and ##+ (-\cos(60)) \sin(30)## by viewing them as hands of wall clock? Is it possible to understand the combined function in terms of hands of a wall clock?
     
  16. Mar 18, 2017 #15

    PeroK

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    You don't try to visualise the products, just the individuals sine and cosine graphs:

    https://www.mathsisfun.com/algebra/trig-sin-cos-tan-graphs.html
     
  17. Mar 18, 2017 #16
    Please let me know why not to try to visualize the products.
    ##= \sin(240) \cos(-30)## and ##+ (-\cos(60)) \sin(30)##
    I am unable to figure out the following:
    If I resolve the above statement then sine is opposite/hypotenuse and cosine is adjacent/hypotenuse so how is this knowledge going to fit in the right-hand side of the equation?
     
  18. Mar 18, 2017 #17

    PeroK

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    I suggest you study the link I gave you.
     
  19. Mar 18, 2017 #18

    Ray Vickson

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    You are wasting your time.
    (1) There is probably no good "visualization" of the equation, and looking for one is very likely futile.
    (2) You can find (numerical) values for the sines and cosines involved, and can just go ahead and substitute those values into the left-hand-side of your equation, to see if--after algebraic (not geometric!) simplification--you get -1.
    (3) Don't bother trying to find an "interpretation" of (-1) in geometric terms, for example; -1 is just a number.
     
  20. Mar 18, 2017 #19
    sin 600° . cos 330° + cos 120° . sin 150° = - 1
    sin 600° will resolve to a value.
    cos 330° will resolve to a value.
    cos 120° will resolve to a value.
    sin150° will resolve to a value.
    And the answer is -1.
    I think of sin 600° as the point at distance of opposite side and hypotenuse is rotated until the measurement reaches 600° from 0°. But the sine is multiplied by cosine. So, when sine gets multiplied by cosine this is where I am loosing the flow.
    Does my explanation make sense?
     
    Last edited: Mar 18, 2017
  21. Mar 18, 2017 #20

    Mark44

    Staff: Mentor

    Not really.
    600° is located on the unit circle at the same point as 240°. The reference triangle for 240° is the same as that for 60°, but both the sine and cosine of this angle are negative. So sin(240°) = ##-\frac {\sqrt 3} 2## while sin(60°) = ##+\frac{\sqrt 3} 2##.
    330° is located on the unit circle at the same point as -30°, so the sines of 30° and -30° are opposite in sign, while the cosines of these angles are equal (##\frac{\sqrt 3} 2##).
    The reference triangle for 120° is the same as for 60°, with the sines of both angles being equal and the cosines being opposite in sign.
    The reference triangle for 150° is the same as for 30°, again with the sines of both angles being equal and the cosines being opposite in sign.

    So sin 600° * cos 330° + cos 120° * sin 150° = sin 240° * cos(-30°) - cos 60° * sin 30° = -sin 60° * cos(-30°) - cos 60° * sin 30° = ##-\frac {\sqrt 3}2 * \frac{\sqrt 3} 2 - \frac 1 2 * \frac 1 2 = -\frac 3 4 - \frac 1 4 = -1##.

    If you know the trig functions of a few reference triangle in the unit circle, the rest is arithmetic.
     
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