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Math Help - Solve Trigonometric Equations

  1. #1
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    Solve Trigonometric Equations

    So I just have a question that I am not sure about.

    Why is the maximum possible value for sinx 1 and then there are no solutions?

    could someone please try their best to explain this concept to me and the reason why this is true?

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    So I just have a question that I am not sure about.

    Why is the maximum possible value for sinx 1 and then there are no solutions?

    could someone please try their best to explain this concept to me and the reason why this is true?

    Thanks in advance!
    no solutions to what equation ?
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  3. #3
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    Definition of sine and cosine

    Hello skeske1234
    Why is the maximum possible value for sinx 1
    See the attached diagram.

    Imagine a point P
    that starts at the point (1, 0) on the x-axis, and moves anticlockwise around a circle, centre (0, 0), radius 1 unit.

    As it does so, the line OP
    rotates through an angle \theta, where \theta = 0 initially, and \theta = 360^o = 2\pi radians when OP has made one complete rotation.

    Then the definition of \sin\theta
    is the y-coordinate of P at any moment: PT in my diagram. (And \cos\theta is the x-coordinate of P = OT.)

    Since P is restricted to lying on the circle, its x-
    and y-coordinates always lie between -1 and +1. Hence the maximum and minimum values of \sin\theta and \cos\theta are +1 and -1 respectively.

    and then there are no solutions?
    Do you mean no solutions to the equation \sin x = 1? Because if you do, that's not true. As you'll see from my diagram, \sin 90^o = 1 (and so is \sin 450^o, \sin 810^o, ... and so on.)

    Grandad
    Attached Thumbnails Attached Thumbnails Solve Trigonometric Equations-untitled.jpg  
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  4. #4
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    Talking

    Quote Originally Posted by skeske1234 View Post
    Why is the maximum possible value for sinx [the value of] 1 and then there are no solutions?
    I don't know what you mean by "and then there are no solutions", but I think you mean to ask why sine never takes on values that are larger than 1.

    Think back to the definition of the sine. For a right triangle, isn't the sine ratio equal to "(opposite) over (hypotenue)"? For a right triangle, isn't the hypotenuse always the longest of the sides?

    Once you get to viewing the sine as a function, rather than a ratio of sides of a right triangle, you can let the angle be ninety degrees (which obviously wouldn't work in a right triangle), and get a value of "1" for the sine.

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