Results 1 to 8 of 8

Math Help - Half-Angle Formulas

  1. #1
    Banned
    Joined
    Feb 2008
    Posts
    24

    Half-Angle Formulas

    Find the value of cos\;\left(\frac{\alpha}{2}\right) if tan\;\alpha = -0.2917 and (90<\alpha <180)

    I'm not exactly sure how to type the degrees sign using latex. And I'm also lost on how to tackle this problem.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by TI-84 View Post
    Find the value of cos\;\left(\frac{\alpha}{2}\right) if tan\;\alpha = -0.2917 and (90<\alpha <180)

    I'm not exactly sure how to type the degrees sign using latex. And I'm also lost on how to tackle this problem.
    Double angle formula: \cos(2A) = 2 \cos^2 A - 1 \Rightarrow \cos^2 A = \frac{\cos (2A) + 1}{2}.

    Let A = \frac{\alpha}{2}:

    \cos^2 \left( \frac{\alpha}{2}\right) = \frac{\cos \alpha + 1}{2} \Rightarrow \cos \left( \frac{\alpha}{2}\right) = \pm \sqrt{\frac{\cos \alpha + 1}{2}} .

    You know \tan\;\alpha = -0.2917 where 90<\alpha <180 ..... from this you can get \cos \alpha.

    Substitute the value of \cos \alpha into the above formula and use the fact that 90< \alpha <180 \Rightarrow 45 < \frac{\alpha}{2} < 90 to justify using the positive root. Capisce?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Feb 2008
    Posts
    24
    How exactly do I know what the value of cos\;\alpha is?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by TI-84 View Post
    How exactly do I know what the value of cos\;\alpha is?
    Substitute \tan\ \alpha = -0.2917 into 1 + \tan^2 \alpha = \sec^2 \alpha.

    Solve for \sec^2 \alpha and hence get \cos \alpha. Use the given fact that 90<\alpha <180 and hence \cos \alpha < 0 to justify keeping the negative value for \cos \alpha.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Feb 2008
    Posts
    24
    Quote Originally Posted by mr fantastic View Post
    Substitute \tan\ \alpha = -0.2917 into 1 + \tan^2 \alpha = \sec^2 \alpha.

    Solve for \sec^2 \alpha and hence get \cos \alpha. Use the given fact that 90<\alpha <180 and hence \cos \alpha < 0 to justify keeping the negative value for \cos \alpha.
    Alright so what I'm getting from what you said is 1 + (tan\;-.2917)^2 = sec^2\;\alpha Which in turn gives sec^2\;\alpha = 1.00002592\Longrightarrow sec\;\alpha = \sqrt{1.00002592}\Longrightarrow cos\;\alpha = \frac{1}{\sqrt{1.00002592}} I'm pretty sure I screwed up. Its some tricky stuff.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by TI-84 View Post
    Alright so what I'm getting from what you said is 1 + (tan\;-.2917)^2 = sec^2\;\alpha Which in turn gives sec^2\;\alpha = 1.00002592\Longrightarrow sec\;\alpha = \sqrt{1.00002592}\Longrightarrow cos\;\alpha = \frac{1}{\sqrt{1.00002592}} I'm pretty sure I screwed up. Its some tricky stuff.
    Since \tan(\alpha)=-0.2917 you have to plug in only the value (without the tan-function!). Your equation then becomes:

    1+(-0.2917)^2 = (\sec(\alpha))^2

    Now go on!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Feb 2008
    Posts
    24
    I got cos\;\alpha = .9597 Is that right?

    If it is do I follow that up by \cos \left( \frac{.9597}{2}\right) = \pm \sqrt{\frac{.9597 + 1}{2}}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by TI-84 View Post
    I got cos\;\alpha = .9597 Is that right?

    If it is do I follow that up by \cos \left( \frac{.9597}{2}\right) = \pm \sqrt{\frac{.9597 + 1}{2}}
    Quote Originally Posted by mr fantastic View Post
    Double angle formula: \cos(2A) = 2 \cos^2 A - 1 \Rightarrow \cos^2 A = \frac{\cos (2A) + 1}{2}.

    Let A = \frac{\alpha}{2}:

    \cos^2 \left( \frac{\alpha}{2}\right) = \frac{\cos \alpha + 1}{2} \Rightarrow \cos \left( \frac{\alpha}{2}\right) = \pm \sqrt{\frac{\cos \alpha + 1}{2}} .

    You know \tan\;\alpha = -0.2917 where 90<\alpha <180 ..... from this you can get \cos \alpha.

    Substitute the value of \cos \alpha into the above formula and use the fact that 90< \alpha <180 \Rightarrow 45 < \frac{\alpha}{2} < 90 to justify using the positive root. Capisce?
    Quote Originally Posted by mr fantastic View Post
    Substitute \tan \alpha = -0.2917 into 1 + \tan^2 \alpha = \sec^2 \alpha.

    Mr F says: So 1 + (-0.2917)^2 = \sec^2 \alpha.

    Solve for \sec^2 \alpha and hence get \cos \alpha.

    Mr F says: So \sec^2 \alpha = 1.0851 \Rightarrow \cos^2 \alpha = 0.9216 \Rightarrow \cos \alpha = \pm 0.9600.

    Use the given fact that 90<\alpha <180 and hence \cos \alpha < 0 to justify keeping the negative value for \cos \alpha.

    Mr F says: So \cos \alpha = - 0.9600.
    Therefore \cos \left( \frac{\alpha}{2}\right) = \pm \sqrt{\frac{-0.9600 + 1}{2}} = \pm \sqrt{\frac{0.04}{2}} = \pm \sqrt{0.02}.

    But 45 < \frac{\alpha}{2} < 90. Therefore \cos \left( \frac{\alpha}{2}\right) = \sqrt{0.02} = 0.1414.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Half-Angle formulas
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 2nd 2010, 09:57 AM
  2. Half Angle Formulas
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: March 17th 2010, 06:02 PM
  3. Replies: 0
    Last Post: March 1st 2010, 09:53 AM
  4. Help with double angle and half angle formulas
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 20th 2009, 05:30 PM
  5. double angle/half angle formulas
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 4th 2008, 05:19 PM

Search Tags


/mathhelpforum @mathhelpforum