Results 1 to 5 of 5

Math Help - Double and half angle identites help

  1. #1
    Newbie
    Joined
    Jul 2012
    From
    unknown
    Posts
    21

    Double and half angle identites help

    Hi.I am having trouble solving the following quesions:
    Q1
    \frac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}=sec2\alpha-tan2\alpha
    Q2
    \frac{sin2\alpha}{1+cos2\alpha}=tan2\alpha
    Q3
    \sqrt{\frac{1+sin\alpha}{1-sin\alpha}} = \frac{sin\frac{a}{2}+cos\frac{a}{2}}{sin\frac{a}{2  }-cos\frac{a}{2}}

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,835
    Thanks
    1604

    Re: Double and half angle identites help

    Quote Originally Posted by hacker804 View Post
    Hi.I am having trouble solving the following quesions:
    Q1
    \frac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}=sec2\alpha-tan2\alpha
    Q2
    \frac{sin2\alpha}{1+cos2\alpha}=tan2\alpha
    Q3
    \sqrt{\frac{1+sin\alpha}{1-sin\alpha}} = \frac{sin\frac{a}{2}+cos\frac{a}{2}}{sin\frac{a}{2  }-cos\frac{a}{2}}

    Thanks
    Q1

    \displaystyle \begin{align*} \sec{2\alpha} - \tan{2\alpha} &\equiv \frac{1}{\cos{2\alpha}} - \frac{\sin{2\alpha}}{\cos{2\alpha}} \\ &\equiv \frac{1 - \sin{2\alpha}}{\cos{2\alpha}} \\ &\equiv \frac{1 - 2\sin{\alpha}\cos{\alpha}}{\cos^2{\alpha} - \sin^2{\alpha}} \\ &\equiv \frac{\cos^2{\alpha} - 2\sin{\alpha}\cos{\alpha} + \sin^2{\alpha}}{\cos^2{\alpha} - \sin^2{\alpha}} \\ &\equiv \frac{\left(\cos{\alpha} - \sin{\alpha}\right)^2 }{\left(\cos{\alpha} - \sin{\alpha}\right)\left(\cos{\alpha} + \sin{\alpha}\right)} \\ &\equiv \frac{\cos{\alpha} - \sin{\alpha}}{\cos{\alpha} + \sin{\alpha}} \end{align*}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,835
    Thanks
    1604

    Re: Double and half angle identites help

    Quote Originally Posted by hacker804 View Post
    Hi.I am having trouble solving the following quesions:
    Q1
    \frac{cos\alpha-sin\alpha}{cos\alpha+sin\alpha}=sec2\alpha-tan2\alpha
    Q2
    \frac{sin2\alpha}{1+cos2\alpha}=tan2\alpha
    Q3
    \sqrt{\frac{1+sin\alpha}{1-sin\alpha}} = \frac{sin\frac{a}{2}+cos\frac{a}{2}}{sin\frac{a}{2  }-cos\frac{a}{2}}

    Thanks
    Q2 is wrong, it should be obvious that \displaystyle \begin{align*} \tan{2\alpha} \equiv \frac{\sin{2\alpha}}{\cos{2\alpha}} \end{align*}, NOT \displaystyle \begin{align*} \frac{\sin{2\alpha}}{1 + \cos{2\alpha}} \end{align*}.

    Q3

    \displaystyle \begin{align*} \sqrt{\frac{1 + \sin{\alpha}}{1 - \sin{\alpha}}} &\equiv \sqrt{\frac{1 + \sin{\left[2\left(\frac{\alpha}{2}\right)\right]}}{1 - \sin{\left[2\left(\frac{\alpha}{2}\right)\right]}}} \\ &\equiv \sqrt{ \frac{ 1 + 2\sin{ \left( \frac{\alpha}{2} \right) } \cos{ \left( \frac{\alpha}{2} \right) } }{ 1 - 2\sin{ \left( \frac{\alpha}{2} \right) }\cos{ \left( \frac{\alpha}{2} \right) } } } \\ &\equiv \sqrt{ \frac{ \sin^2{ \left( \frac{\alpha}{2} \right) } + 2\sin{ \left( \frac{\alpha}{2} \right) } \cos{ \left( \frac{\alpha}{2} \right) } + \cos^2{ \left( \frac{\alpha}{2} \right) }  }{ \sin^2{ \left( \frac{\alpha}{2} \right) } - 2\sin{ \left( \frac{\alpha}{2} \right) }\cos{ \left(\frac{\alpha}{2} \right)} + \cos^2{ \left(\frac{\alpha}{2}\right) } } } \end{align*}

    \displaystyle \begin{align*} &\equiv \sqrt{ \frac{ \left[ \sin{\left(\frac{\alpha}{2}\right)} + \cos{\left(\frac{\alpha}{2}\right)} \right]^2 }{ \left[ \sin{\left(\frac{\alpha}{2}\right)} - \cos{\left(\frac{\alpha}{2}\right)} \right]^2 } } \\ &\equiv \frac{ \sin{\left(\frac{\alpha}{2}\right)} + \cos{\left(\frac{\alpha}{2}\right)} }{ \sin{\left(\frac{\alpha}{2}\right)} - \cos{\left(\frac{\alpha}{2}\right)} } \end{align*}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2012
    From
    unknown
    Posts
    21

    Re: Double and half angle identites help

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

    Explain this please.How did you complete the square?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,835
    Thanks
    1604

    Re: Double and half angle identites help

    Quote Originally Posted by hacker804 View Post
    \frac{cos^{2}\alpha-2sin\alpha cos\alpha+sin^{2}\alpha}{cos^{2}\alpha-sin^{2}\alpha}

    Explain this please.How did you complete the square?
    Surely you know that \displaystyle \begin{align*} \cos^2{\alpha} + \sin^2{\alpha} \equiv 1 \end{align*}...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simplify the Expression (Double-Angle or Half-Angle)
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 3rd 2009, 04:38 PM
  2. More double angle and Half angle problems
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: July 21st 2009, 07:54 PM
  3. Half angle or double angle Identities
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: July 21st 2009, 06:20 PM
  4. Help with double angle and half angle formulas
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 20th 2009, 06:30 PM
  5. double angle/half angle formulas
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 4th 2008, 06:19 PM

Search Tags


/mathhelpforum @mathhelpforum