Results 1 to 6 of 6

Thread: proof please help

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    96

    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??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    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
    $\displaystyle \frac{\sin2x}{1-\cos x}+\frac{\sin2x}{1+\cos x}=2$
    If that's it then you'd write:
    $\displaystyle \Rightarrow \frac{\sin2x(1+\cos x + 1-\cos x)}{(1-\cos x)(1+\cos x)}=2$

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

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

    $\displaystyle \Rightarrow 2\cot x = 1$

    $\displaystyle \Rightarrow \tan x = 2$

    $\displaystyle \Rightarrow x = \arctan(2) + n\pi$
    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2009
    Posts
    96
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello deej813

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

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

    $\displaystyle =\frac{2\sin2x}{1-\cos^2 x}$
    $\displaystyle = \frac{4\sin x\cos x}{\sin^2x}$
    $\displaystyle =4\cot x$
    Grandad
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2009
    Posts
    96
    sorry its sin^2x on top of both fractions.
    does that make a difference??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    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 $\displaystyle \sin2x$ on the top then, we get:
    $\displaystyle \frac{\sin^2x}{1-\cos x}+\frac{\sin^2x}{1+\cos x}$
    $\displaystyle =\frac{\sin^2x(1+\cos x + 1-\cos x)}{(1-\cos x)(1+\cos x)}$

    $\displaystyle =\frac{2\sin^2x}{1-\cos^2 x}$
    $\displaystyle = \frac{2\sin^2 x}{\sin^2x}$
    $\displaystyle =2$
    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Oct 19th 2010, 10:50 AM
  2. Replies: 0
    Last Post: Jun 29th 2010, 08:48 AM
  3. [SOLVED] direct proof and proof by contradiction
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Feb 27th 2010, 10:07 PM
  4. Proof with algebra, and proof by induction (problems)
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: Jun 8th 2008, 01:20 PM
  5. proof that the proof that .999_ = 1 is not a proof (version)
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: Apr 14th 2008, 04:07 PM

Search Tags


/mathhelpforum @mathhelpforum