Results 1 to 13 of 13

Math Help - Prove(9)

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Prove(9)

    Prove that \frac{1-cos\theta+sin\theta}{1+cos\theta+sin\theta}=tan\fr  ac{\theta}{2}.

    Hence or otherwise, find the values of \theta between 0 and \pi which satisfies the equation  1+sin\theta=2cos\theta. where cos<br />
\theta is not equals to zero
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by Punch View Post
    Prove that \frac{1-cos\theta+sin\theta}{1+cos\theta+sin\theta}=tan\fr  ac{\theta}{2}.

    Hence or otherwise, find the values of \theta between 0 and \pi which satisfies the equation  1+sin\theta=2cos\theta. where cos<br />
\theta is not equals to zero
    insert all the half angles stuff inside ..

    knowing these identities ,

    \sin \theta=2\sin \frac{\theta}{2}\cos \frac{\theta}{2}

    \cos \theta=1-2\sin^2 \frac{\theta}{2}

    so

    \frac{1-1+2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{1+1-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}

    \frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}

    \frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}+2\sin^2 \frac{\theta}{2}-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}

    bring out the common factor and do some necessary cancellings , then you will have your proof .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by mathaddict View Post
    insert all the half angles stuff inside ..

    knowing these identities ,

    \sin \theta=2\sin \frac{\theta}{2}\cos \frac{\theta}{2}

    \cos \theta=1-2\sin^2 \frac{\theta}{2}

    so

    \frac{1-1+2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{1+1-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}

    \frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}

    \frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}+2\sin^2 \frac{\theta}{2}-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}

    bring out the common factor and do some necessary cancellings , then you will have your proof .
    I don't see how the formulas used are related, what is the key idea to it?

    To convert all to sin or cos?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Punch View Post

    Prove that \frac{1-cos\theta+sin\theta}{1+cos\theta+sin\theta}=tan\fr  ac{\theta}{2}.
    Hi Punch,

    you can also start with Tan\frac{\theta}{2}=\frac{Sin(\frac{\theta}{2})}{C  os(\frac{\theta}{2})}

    then...

    \frac{1-Cos\theta+Sin\theta}{1+Cos\theta+Sin\theta}=\frac{  Sin(\frac{\theta}{2})}{Cos(\frac{\theta}{2})} ?

    giving...

    Cos\frac{\theta}{2}-Cos\theta\ Cos\frac{\theta}{2}+Sin\theta\ Cos\frac{\theta}{2}=Sin\frac{\theta}{2}+Cos\theta\ Sin\frac{\theta}{2}+Sin\theta\ Sin\frac{\theta}{2} ?

    Now we can use the CosACosB, SinASinB and CosASinB identities to get

    \ Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta-\frac{1}{2}Cos\frac{\theta}{2}+\frac{1}{2}Sin\frac  {3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2}

    =Sin\frac{\theta}{2}+\frac{1}{2}Sin\frac{3}{2}\the  ta-\frac{1}{2}Sin\frac{\theta}{2}+\frac{1}{2}Cos\frac  {\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta ?

    \frac{1}{2}Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta+\frac{1}{2}Sin\fra  c{3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2}

    =\frac{1}{2}Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta+\frac{1}{2}Sin\fra  c{3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2} ? yes, it's true
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by Archie Meade View Post
    Hi Punch,

    you can also start with Tan\frac{\theta}{2}=\frac{Sin(\frac{\theta}{2})}{C  os(\frac{\theta}{2})}

    then...

    \frac{1-Cos\theta+Sin\theta}{1+Cos\theta+Sin\theta}=\frac{  Sin(\frac{\theta}{2})}{Cos(\frac{\theta}{2})} ?

    giving...

    Cos\frac{\theta}{2}-Cos\theta\ Cos\frac{\theta}{2}+Sin\theta\ Cos\frac{\theta}{2}=Sin\frac{\theta}{2}+Cos\theta\ Sin\frac{\theta}{2}+Sin\theta\ Sin\frac{\theta}{2} ?

    Now we can use the CosACosB, SinASinB and CosASinB identities to get

    \ Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta-\frac{1}{2}Cos\frac{\theta}{2}+\frac{1}{2}Sin\frac  {3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2}

    =Sin\frac{\theta}{2}+\frac{1}{2}Sin\frac{3}{2}\the  ta-\frac{1}{2}Sin\frac{\theta}{2}+\frac{1}{2}Cos\frac  {\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta ?

    \frac{1}{2}Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta+\frac{1}{2}Sin\fra  c{3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2}

    =\frac{1}{2}Cos\frac{\theta}{2}-\frac{1}{2}Cos\frac{3}{2}\theta+\frac{1}{2}Sin\fra  c{3}{2}\theta+\frac{1}{2}Sin\frac{\theta}{2} ? yes, it's true
    My teacher once said that to solve a proving question, we have to prove from left to right or from right to left. And that we can't solve it by manipulating both sides and eventually getting A=A
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by mathaddict View Post
    insert all the half angles stuff inside ..

    knowing these identities ,

    \sin \theta=2\sin \frac{\theta}{2}\cos \frac{\theta}{2}

    \cos \theta=1-2\sin^2 \frac{\theta}{2}

    so

    \frac{1-1+2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{1+1-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}

    \frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}

    \frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}+2\sin^2 \frac{\theta}{2}-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}

    bring out the common factor and do some necessary cancellings , then you will have your proof .
    I think I need more help, I didn't solved it using this..
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    The calculations I showed prove that both sides are equal, Punch.

    If \frac{1-Cos\theta+Sin\theta}{1+Cos\theta+Sin\theta}=\frac{  Sin\frac{\theta}{2}}{Cos\frac{\theta}{2}}

    Then Cos\frac{\theta}{2}(1-Cos\theta+Sin\theta) must equal Sin\frac{\theta}{2}(1+Cos\theta+Sin\theta)

    The remaining calculations examine whether or not this is true,
    but i know what you mean.

    Using the double-angle formulae for CosA and SinA

    \frac{1-Cos\theta+Sin\theta}{1+Cos\theta+Sin\theta}=\frac{  1-\frac{1-tan^2\frac{\theta}{2}}{1+tan^2{\frac{\theta}{2}}}+  \frac{2tan\frac{\theta}{2}}{1+tan^2{\theta}{2}}}{1  +\frac{1-tan^2\frac{\theta}{2}}{1+tan^2\frac{\theta}{2}}+\f  rac{2tan\frac{\theta}{2}}{1+tan^2\frac{\theta}{2}}  }

    =\frac{\frac{1+tan^2\frac{\theta}{2}-(1-tan^2\frac{\theta}{2})+2tan\frac{\theta}{2}}{1+tan  ^2\frac{\theta}{2}}}{\frac{1+tan^2\frac{\theta}{2}  +1-tan^2\frac{\theta}{2}+2tan\frac{\theta}{2}}{1+tan^  2\frac{\theta}{2}}}

    =\frac{\frac{1+tan^2\frac{\theta}{2}-1+tan^2\frac{\theta}{2}+2tan\frac{\theta}{2}}{1+ta  n^2\frac{\theta}{2}}}{\frac{1+tan^2\frac{\theta}{2  }+1-tan^2\frac{\theta}{2}+2tan\frac{\theta}{2}}{1+tan^  2\frac{\theta}{2}}}

    =\frac{2tan^2\frac{\theta}{2}+2tan\frac{\theta}{2}  }{2+2tan\frac{\theta}{2}}

    =\frac{2tan\frac{\theta}{2}(1+tan\frac{\theta}{2})  }{2(1+tan\frac{\theta}{2})}=tan\frac{\theta}{2}
    Last edited by Archie Meade; January 28th 2010 at 04:55 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by mathaddict View Post
    insert all the half angles stuff inside ..

    knowing these identities ,

    \sin \theta=2\sin \frac{\theta}{2}\cos \frac{\theta}{2}

    \cos \theta=1-2\sin^2 \frac{\theta}{2}

    so

    \frac{1-1+2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{1+1-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}} -----1

    \frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}} ---------2

    \frac{2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}+2\sin^2 \frac{\theta}{2}-2\sin^2 \frac{\theta}{2}+2\sin \frac{\theta}{2}\cos \frac{\theta}{2}} --------3

    bring out the common factor and do some necessary cancellings , then you will have your proof .
    Which step don u understand ? 1,2 or 3 , or the steps after that ?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Nov 2009
    Posts
    202
    u can also use the following identities:
    1)1-cos(theta)=2sin^2(theta/2)
    2)1+cos(theta)=2cos^2(theta/2)
    3)sin(theta)=2sin(theta/2)cos(theta/2) which has already been given by mathaddict.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Dec 2009
    Posts
    755
    After looking at this now, I still do not understand the workings provided.

    Sorry but I guess I need a complete and detailed working.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Hi Punch,

    if you get stuck at any of the steps 1 to 7, let me know.

    \frac{1-Cos\theta+Sin\theta}{1+Cos\theta+Sin\theta}=Tan\fr  ac{\theta}{2}

    The evaluation of the left fraction is Tan\frac{\theta}{2}

    This is "tangent of the half-angle",
    therefore we may express the Cos\theta and Sin\theta from the left using "half-angle formulae".

    The "double angle formulae" for Cos\theta and Sin\theta are

    Cos2\theta=\frac{1-Tan^2\theta}{1+Tan^2\theta},\ Sin2\theta=\frac{2Tan\theta}{1+Tan^2\theta} (1)

    If we replace \theta with \frac{\theta}{2}, we will have the "half-angle" formulae.

    Therefore the half-angle formulae for Cos\theta and Sin\theta are

    Cos\theta=\frac{1-Tan^2\left(\frac{\theta}{2}\right)}{1+Tan^2\left(\  frac{\theta}{2}\right)}

    Sin\theta=\frac{2Tan\left(\frac{\theta}{2}\right)}  {1+Tan^2\left(\frac{\theta}{2}\right)} (2)

    Then, we can write the ratio on the left in terms of Tan\frac{\theta}{2}

    \left[\frac{1-\frac{1-Tan^2\left(\frac{\theta}{2}\right)}{1+Tan^2\left(\  frac{\theta}{2}\right)}+\frac{2Tan\left(\frac{\the  ta}{2}\right)}{1+Tan^2\left(\frac{\theta}{2}\right  )}}{1+\frac{1-Tan^2\left(\frac{\theta}{2}\right)}{1+Tan^2\left(\  frac{\theta}{2}\right)}+\frac{2Tan\left(\frac{\the  ta}{2}\right)}{1+Tan^2\left(\frac{\theta}{2}\right  )}}\right] (3)

    Writing 1 as \frac{1+Tan^2\left(\frac{\theta}{2}\right)}{1+Tan^  2\left(\frac{\theta}{2}\right)}


    \frac{\frac{1+Tan^2\left(\frac{\theta}{2}\right)-\left[1-Tan^2\left(\frac{\theta}{2}\right)\right]+2Tan\left(\frac{\theta}{2}\right)}{1+Tan^2\left(\  frac{\theta}{2}\right)}}{\frac{1+Tan^2\left(\frac{  \theta}{2}\right)+1-Tan^2\left(\frac{\theta}{2}\right)+2Tan\left(\frac  {\theta}{2}\right)}{1+Tan^2\left(\frac{\theta}{2}\  right)}} (4)

    gives \frac{1+Tan^2\left(\frac{\theta}{2}\right)-\left[1-Tan^2\left(\frac{\theta}{2}\right)\right]+2Tan\left(\frac{\theta}{2}\right)}{1+Tan^2\left(\  frac{\theta}{2}\right)+1-Tan^2\left(\frac{\theta}{2}\right)+2Tan\left(\frac  {\theta}{2}\right)} (5)

    =\frac{2Tan^2\left(\frac{\theta}{2}\right)+2Tan\le  ft(\frac{\theta}{2}\right)}{2+2Tan\left(\frac{\the  ta}{2}\right)} (6)

    \frac{2\left[Tan\left(\frac{\theta}{2}\right)\right]\left[Tan\left(\frac{\theta}{2}\right)+1\right]}{2\left[Tan\left(\frac{\theta}{2}\right)+1\right]}=Tan\left(\frac{\theta}{2}\right) (7)
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Okay. I haven't seen what you have mentioned in (1) which refers to the formulas used. The other part are pretty much about simplifying which are rather okay with me.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Hi Punch,

    Those identities are listed as standard, though they can be developed from

    Cos(A+B)=CosACosB-SinASinB\ \Rightarrow\ Cos(2\theta)=Cos(\theta+\theta)=Cos^2\theta-Sin^2\theta

    =\frac{Cos^2\theta-Sin^2\theta}{1}

    and since Cos^\theta+Sin^2\theta=1

    this is \frac{Cos^2\theta-Sin^2\theta}{Cos^2\theta+Sin^2\theta}=\frac{\frac{  1}{Cos^2\theta}}{\frac{1}{Cos^2\theta}}\frac{Cos^2  \theta-Sin^2\theta}{Cos^2\theta+Sin^2\theta} =\frac{\frac{Cos^2\theta}{Cos^2\theta}-\frac{Sin^2\theta}{Cos^2\theta}}{\frac{Cos^2\theta  }{Cos^2\theta}+\frac{Sin^2\theta}{Cos^2\theta}}

    =\frac{1-Tan^2\theta}{1+Tan^2\theta}


    Sin(A+B)=SinACosB+CosASinB

    Sin(\theta+\theta)=Sin(2\theta)=2Sin\theta\ Cos\theta

    =\frac{\frac{1}{Cos^2\theta}}{\frac{1}{Cos^2\theta  }}2Sin\theta\ Cos\theta=\frac{2\frac{Sin\theta}{Cos\theta}}{\fra  c{Cos^2\theta+Sin^2\theta}{Cos^2\theta}}

    =\frac{2Tan\theta}{1+Tan^2\theta}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. can anybody prove the g'(x) = g (x)? please
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 6th 2011, 03:59 AM
  2. If A^2 = -I then prove...
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 3rd 2011, 11:35 PM
  3. Prove(5)
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 22nd 2010, 04:37 AM
  4. Prove the Following
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 17th 2010, 04:11 PM
  5. Replies: 2
    Last Post: August 28th 2009, 02:59 AM

Search Tags


/mathhelpforum @mathhelpforum