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Math Help - tough one to solve

  1. #1
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    tough one to solve

    I need help to verify if this is correct and I can't find the mistake can you?

    csc^2(2x)sin(x)tan(x)+2csc(x)cos(x)=sec(x)csc^2(2x )
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  2. #2
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    Hello, talor!

    I need help to verify if this is correct and I can't find the mistake, can you?
    . .
    What "mistake" are you looking for?

    \csc^2(2x)\sin(x)\tan(x)+2\csc(x)\cos(x)\;=\;\sec(  x)\csc^2(2x)

    Well, it's not an identity . . . Try x = \frac{\pi}{4}


    Left side: . \underbrace{\csc^2\!\left(\frac{\pi}{2}\right)}\un  derbrace{\sin\left(\frac{\pi}{4}\right)}\underbrac  e{\tan\left(\frac{\pi}{4}\right)} + \underbrace{2\cdot\csc\left(\frac{\pi}{4}\right)}\  underbrace{\cos\left(\frac{\pi}{4}\right)}

    . . . . . . = \quad\;\; 1^2\quad\cdot\quad \frac{1}{\sqrt{2}}\quad\cdot\quad 1 \quad+\quad 2\cdot \sqrt{2} \quad\cdot\quad \frac{1}{\sqrt{2}} \quad = \quad \boxed{\frac{1}{\sqrt{2}} + 2}


    Right side: . \underbrace{\sec\left(\frac{\pi}{4}\right)}\underb  race{\csc^2\left(\frac{\pi}{2}\right)}

    . . . . . . = \qquad \sqrt{2} \quad\cdot\quad 1^2 \quad = \quad\boxed{ \sqrt{2}}


    . . . See?

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  3. #3
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    clarification on hard one to solve

    I need to prove if there are any errors and if so identify them and explain why they are errors. If there are no errors explain why. I need help get gets to confusing to follow for me. thanks erin

    Csc^2(2x)sin(x)tan(x)+2 csc(x)cos(x)=sec(x)csc^2(2x)
    Csc(x)[csc(x)sin(x)tan(x)+2 cos(x)]=sec(x)csc^2(2x)
    [csc(x)sin(x)tan(x)+ 2 cos(x)] sec(x)csc(2x)
    csc(x)sin(x)(cos(x)/sin(x)) + 2 cos(x) =sec(x)csc(x)
    cos(x)[csc(x+2)] = sec(x)csc(2x)
    cos(x)csc(2x)=sec(x)csc(2x)
    cos(x)=sec(x)
    sec(x)cos(x) = 1
    (1/cos(x))cos(x) = 1
    1=1
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by talor View Post
    I need to prove if there are any errors and if so identify them and explain why they are errors. If there are no errors explain why. I need help get gets to confusing to follow for me. thanks erin

    Csc^2(2x)sin(x)tan(x)+2 csc(x)cos(x)=sec(x)csc^2(2x)
    Csc(x)[csc(x)sin(x)tan(x)+2 cos(x)]=sec(x)csc^2(2x)
    [csc(x)sin(x)tan(x)+ 2 cos(x)] = sec(x)csc(2x)
    csc(x)sin(x)(cos(x)/sin(x)) + 2 cos(x) =sec(x)csc(x)
    cos(x)[csc(x+2)] = sec(x)csc(2x)
    cos(x)csc(2x)=sec(x)csc(2x)
    cos(x)=sec(x)
    sec(x)cos(x) = 1
    (1/cos(x))cos(x) = 1
    1=1
    i assume that there should be an = sign there.
    take a look on that red text. going to the next line, \csc(x) and one \csc(2x) were gone (an intention of cancellation).. however, you cannot cancel them.. (why?)
    so there is the error..
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  5. #5
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    Hello, talor!

    The error is that the statement is not true!

    And sorry, I don't understand your steps.
    . . And did you notice the {\color{red}2x} ?


    \csc^2({\color{red}2x})\sin(x)\tan(x) + 2\csc(x)\cos(x)\;=\;\sec(x)\csc^2({\color{red}2x})

    The left side is: . \frac{1}{\sin^2(2x)}\cdot\sin(x)\cdot\frac{\sin(x)  }{\cos(x)} + 2\cdot\frac{1}{\sin(x)}\cdot\cos(x)

    . . =\;\frac{\sin^2(x)}{\sin^2(2x)\cos(x)} + \frac{2\cos(x)}{\sin(x)} \;=\;\frac{\sin^2(x)}{[2\sin(x)\cos(x)]^2\cos(x)} + \frac{2\cos(x)}{\sin(x)}

    . . = \;\frac{\sin^2(x)}{4\sin^2(x)\cos^3(x)} + \frac{2\cos(x)}{\sin(x)} \;=\;\frac{1}{4\cos^3(x)} + \frac{2\cos(x)}{\sin(x)}

    . . = \;\boxed{\frac{\sin(x) + 8\cos^4(x)}{4\sin(x)\cos^3(x)}}\;\;{\color{blue}[1]}


    The right side is: . \frac{1}{\cos(x)}\cdot\frac{1}{\sin^2(2x)} \;=\;\frac{1}{\cos(x)}\cdot\frac{1}{[2\sin(x)\cos(x)]^2}

    . . = \;\frac{1}{\cos(x)}\cdot\frac{1}{4\sin^2(x)\cos^2(  x)} \;=\;\boxed{\frac{1}{4\sin^2(x)\cos^3(x)}}\;\;{\co  lor{blue}[2]}


    And [1] and [2] are definitely not equal . . .

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  6. #6
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    making sure

    I saw that and thought that maybe they were dividing the whole equation by 2 or squaring the problem but that didn't look right But I didn't know how to explain in words I knew it was wrong but don't know the right way to explain other than you can not do it. thanks for helping me. validate I knew the mistake. Erin
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  7. #7
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    soroban

    this was a preassigned question were the steps were already done and we had to find the mistake if any and explain why it was a mistake. Thank you for all your help and guidance, this has been a long learning process for me. Erin
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  8. #8
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by talor View Post
    I saw that and thought that maybe they were dividing the whole equation by 2 or squaring the problem but that didn't look right But I didn't know how to explain in words I knew it was wrong but don't know the right way to explain other than you can not do it. thanks for helping me. validate I knew the mistake. Erin
    there are a lot of x's for which \csc (x) \not= \csc (2x) so you cannot just cancel it..

    in fact, suppose, they are equal.
    \csc (x) = \csc (2x) can also be written as \frac{1}{\sin x} = \frac{1}{\sin(2x)} or \sin x = \sin (2x)

    then, \sin x = 2\sin (x)\cos (x) \Longleftrightarrow \sin x - 2\sin (x)\cos (x) = 0 \Longleftrightarrow \sin x(1 - 2\cos (x)) = 0

    so, either \sin x = 0 or 1 - 2\cos (x) = 0 \Leftrightarrow \cos x = \frac{1}{2}

    \sin x = 0 cannot happen otherwise \csc x is undefined.

    it means that \cos x = \frac{1}{2} should be true. hence x=\pm \frac{\pi}{3} + 2n\pi for any integer n.

    this only means that \csc (x) = \csc (2x) is true only when x=\pm \frac{\pi}{3} + 2n\pi for any integer n..
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