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Math Help - proof please help

  1. #1
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    proof please help

    Sin2x + _sin2x = 2
    1-cosx 1+cosx

    i used sin^2x + cos^2 x = 1
    then got (sin^2x)/(1-cos^2x)=0
    and thats a diff of 2 squares
    so its
    (sin^2x)/(1-cosx)(1+cosx) = 0

    but where does the two come from
    what did i do wrong??
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  2. #2
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    Hello deej813
    Quote Originally Posted by deej813 View Post
    Sin2x + _sin2x = 2
    1-cosx 1+cosx

    i used sin^2x + cos^2 x = 1
    then got (sin^2x)/(1-cos^2x)=0
    and thats a diff of 2 squares
    so its
    (sin^2x)/(1-cosx)(1+cosx) = 0

    but where does the two come from
    what did i do wrong??
    I'm not clear what the question is here. My best guess is that you have to solve the equation
    \frac{\sin2x}{1-\cos x}+\frac{\sin2x}{1+\cos x}=2
    If that's it then you'd write:
    \Rightarrow \frac{\sin2x(1+\cos x + 1-\cos x)}{(1-\cos x)(1+\cos x)}=2

    \Rightarrow \frac{4\sin x \cos x}{1-\cos^2 x}=2, using \sin 2x = 2\sin x\cos x

    \Rightarrow \frac{2\sin x\cos x}{\sin^2x}=1

    \Rightarrow 2\cot x = 1

    \Rightarrow \tan x = 2

    \Rightarrow x = \arctan(2) + n\pi
    Grandad
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  3. #3
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    nah i've gotta prove that the LHS = RHS = 2
    so rearrange the LHS and get it to equal two
    but i keep getting 0
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  4. #4
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    Hello deej813

    Sorry, but the LHS just isn't identically equal to 2. If you followed my working you'd see that it's 4\cot x, and that's as far as you can go. So here it is once again:

    \frac{\sin2x}{1-\cos x}+\frac{\sin2x}{1+\cos x}
    =\frac{\sin2x(1+\cos x + 1-\cos x)}{(1-\cos x)(1+\cos x)}

    =\frac{2\sin2x}{1-\cos^2 x}
    = \frac{4\sin x\cos x}{\sin^2x}
    =4\cot x
    Grandad
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  5. #5
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    sorry its sin^2x on top of both fractions.
    does that make a difference??
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  6. #6
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    Hello deej813
    Quote Originally Posted by deej813 View Post
    sorry its sin^2x on top of both fractions.
    does that make a difference??
    It sure does! Instead of \sin2x on the top then, we get:
    \frac{\sin^2x}{1-\cos x}+\frac{\sin^2x}{1+\cos x}
    =\frac{\sin^2x(1+\cos x + 1-\cos x)}{(1-\cos x)(1+\cos x)}

    =\frac{2\sin^2x}{1-\cos^2 x}
    = \frac{2\sin^2 x}{\sin^2x}
    =2
    Grandad
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