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Math Help - Solution and Deduction

  1. #1
    Super Member dhiab's Avatar
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    Solution and Deduction

    1 ) Solve in R this equation: ( let:tan x= t)
    2 )Deduct the values of :
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  2. #2
    Super Member malaygoel's Avatar
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    cos^2\frac{\pi}{8}=\frac{cos\frac{\pi}{4}+1}{2}

    Similarly, other values can be found.
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  3. #3
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    Quote Originally Posted by dhiab View Post
    1 ) Solve in R this equation: ( let:tan x= t)
    2 )Deduct the values of :
    2\sin{x}\cos{x} = \sin{(2x)}.


    So if \cos{(2x)} + 2\sin{x}\cos{x} = 0

    \cos{(2x)} + \sin{(2x)} = 0

    \cos{(2x)} = -\sin{(2x)}

    1 = -\frac{\sin{(2x)}}{\cos{(2x)}}

    -\tan{(2x)} = 1

    2x = \left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\} + 2\pi n, where n \in \mathbf{Z}

    x = \frac{1}{2}\left[\left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\} + 2\pi n\right]

    x = \left\{\frac{\pi}{8}, \frac{5\pi}{8}\right\} + \pi n, where n \in \mathbf{Z}.
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  4. #4
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    Blast you, Prove It! You got in just ahead of me!
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  5. #5
    Super Member dhiab's Avatar
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    Quote Originally Posted by Prove It View Post
    2\sin{x}\cos{x} = \sin{(2x)}.


    So if \cos{(2x)} + 2\sin{x}\cos{x} = 0

    \cos{(2x)} + \sin{(2x)} = 0

    \cos{(2x)} = -\sin{(2x)}

    1 = -\frac{\sin{(2x)}}{\cos{(2x)}}

    -\tan{(2x)} = 1

    2x = \left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\} + 2\pi n, where n \in \mathbf{Z}

    x = \frac{1}{2}\left[\left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\} + 2\pi n\right]

    x = \left\{\frac{\pi}{8}, \frac{5\pi}{8}\right\} + \pi n, where n \in \mathbf{Z}.
    Hello THANK YOU
    You have:

    Subst :
    Now solve the quadratic equation.....
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  6. #6
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    Quote Originally Posted by dhiab View Post
    Hello THANK YOU
    You have:

    Subst :
    Now solve the quadratic equation.....
    I've just realised I made a mistake...

    I got to the step

    -\tan{(2x)} = 1

    \tan{(2x)} = -1

    2x = \left\{\pi - \frac{\pi}{4}, 2\pi - \frac{\pi}{4}\right\} + 2\pi n, n \in \mathbf{Z}

    2x = \left\{\frac{3\pi}{4}, \frac{7\pi}{4}\right\} + 2\pi n

    x = \frac{1}{2}\left[\left\{\frac{3\pi}{4}, \frac{7\pi}{4}\right\} + 2\pi n\right]

    x = \left\{\frac{3\pi}{8}, \frac{7\pi}{8}\right\} + \pi n, n \in \mathbf{Z}.


    There was nothing wrong with my solution process, I just forgot about the negative sign.

    You don't need to use the other method...
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  7. #7
    Super Member dhiab's Avatar
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    Prove it said:










    .
    but :
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  8. #8
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    Quote Originally Posted by dhiab View Post
    Prove it said:










    .
    but :
    Wrong. Read the post properly.
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  9. #9
    Super Member dhiab's Avatar
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    Quote Originally Posted by Prove It View Post
    Wrong. Read the post properly.
    HELLO : they know that

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  10. #10
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    Quote Originally Posted by dhiab View Post
    HELLO : they know that

    That's right - I forgot the period was different for tan...

    It looks like you've got it now though.
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