Results 1 to 10 of 10

Math Help - Can't start

  1. #1
    Super Member
    Joined
    Dec 2009
    Posts
    755

    Can't start

    Given that sin(A+B) = 2sin(A-B), express tan A in terms of tan B
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Dear Punch,

    Sin(A-B)=SinACosB-SinBCosA

    Sin(A+B)=SinACosB+SinBCosA

    Given the above relations I think you should be able to express TanA interms of TanB.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,567
    Thanks
    1426
    Quote Originally Posted by Punch View Post
    Given that sin(A+B) = 2sin(A-B), express tan A in terms of tan B
    \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \mp \cos{\alpha}\sin{\beta}.


    Given that \sin{(A + B)} = 2\sin{(A - B)}

    \sin{A}\cos{B} - \cos{A}\sin{B} = 2(\sin{A}\cos{B} + \cos{A}\sin{B})

    \sin{A}\cos{B} - \cos{A}\sin{B} = 2\sin{A}\cos{B} + 2\cos{A}\sin{B}

    -3\cos{A}\sin{B} = \sin{A}\cos{B}

    \frac{-3\sin{B}}{\cos{B}} = \frac{\sin{A}}{\cos{A}}

    -3\tan{B} = \tan{A}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Thanks Prove it.. i feel so grateful to you

    I have another question,

    Given that cos x = p, express sin4x in terms of p
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Dear Punch,

    Sin4x=2Sin2xCos2x

    Sin4x=2(2SinxCosx)Cos2x

    Sin4x=4\sqrt{(1-Cos^2x)}Cosx(2Cos^2x-1)

    By substituting Cosx=p and further simplification will give you the answer.

    Hope this helps.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2009
    Posts
    755
    Quote Originally Posted by Sudharaka View Post
    Dear Punch,



    Sin4x=2Sin2xCos2x



    3. Sin4x=2(2SinxCosx)Cos2x



    4. Sin4x=4\sqrt{(1-Cos^2x)}Cosx(2Cos^2x-1)



    By substituting Cosx=p and further simplification will give you the answer.



    Hope this helps.



    I don't get how you get from the 3 to the 4
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,567
    Thanks
    1426
    Quote Originally Posted by Punch View Post
    Thanks Prove it.. i feel so grateful to you

    I have another question,

    Given that cos x = p, express sin4x in terms of p
    \sin{4x} = 2\sin{2x}\cos{2x}

     = 2\cdot 2\sin{x}\cos{x}(\cos^2{x} - \sin^2{x})

     = 4\cos{x}\sqrt{1 - \cos^2{x}}[\cos^2{x} - (1 - \cos^2{x})]

     = 4\cos{x}(2\cos^2{x} - 1)\sqrt{1 - \cos^2{x}}

     = 4p(2p^2 - 1)\sqrt{1 - p^2}.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Dear Punch,

    Do you know that,

    Sin2A=2SinACosA

    SinA=\sqrt{1-Cos^2A}

    These are the two equations I used. In case you do not know them I think you should refer List of trigonometric identities - Wikipedia, the free encyclopedia When going from 3 to 4 I used the second identity.

    Hope this helps.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Dec 2009
    Posts
    755
    got it! thanks!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello everyone

    If I might make a slight correction:
    Quote Originally Posted by Prove It View Post
    \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \mp \cos{\alpha}\sin{\beta}...
    should, of course, be:
    \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}
    So the final result will be:
    3\tan B = \tan A
    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Please help me start this
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 8th 2009, 05:40 PM
  2. not sure where to start
    Posted in the Statistics Forum
    Replies: 4
    Last Post: October 5th 2009, 02:03 PM
  3. Don't know where to start...
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: February 15th 2009, 01:57 AM
  4. can someone help me start this?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 5th 2008, 12:21 AM
  5. How do I start
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 17th 2005, 09:32 PM

Search Tags


/mathhelpforum @mathhelpforum