Results 1 to 5 of 5

Math Help - Interesting identity

  1. #1
    Newbie
    Joined
    Feb 2009
    From
    South Africa
    Posts
    21

    Interesting identity

    Just got this in an exam:

    If tan(A+B)= \frac{Sin(A+B)}{cos(A+B)}

    Prove that tanA.tanB.tanC = tanA + tanB + tanC

    I think there's an error though cause I find that they are not equal when I assign values to A, B, and C. Any help? I'm probably wrong
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87
    I think we're missing some information. I believe the identity is true only if A, B, and C are angles of the same triangle (A + B + C = 180).

    If so, break up the tangents into sines and cosines.

    Then try using the product to sum formulas:

     \sin(\theta)\sin(\phi) = \frac{\cos(\theta - \phi) - \cos(\theta + \phi)}{2}

     \cos(\theta)\cos(\phi) = \frac{\cos(\theta - \phi) + \cos(\theta + \phi)}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2009
    From
    South Africa
    Posts
    21
    Ok I have checked the information and I left out the first line and you were correct, it only applies to a triangle.

    On the level that we do trigenometry, we can do compound angles sin(A+B) = sinA.CosB + CosA.SinB

    Then changing from Cos2A = 2Cos^2A -1 = 1 -2Sin^2A = Cos^2A - Sin^2A

    And from Sin2A = 2SinA.CosA

    Do you perhaps know how i would go about proving this identity using these rules?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Jan 2008
    Posts
    588
    Thanks
    87
    The product to sum formulas can be derived using those equations. For example:

     \cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B) ---- 1
     \cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B) ---- 2

    1 + 2 gives:

     \cos(A+B) + \cos(A-B) = 2\cos(A)(cos(B) which is also

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

    The product to sum formula for two sines is also derived in a similar way, adding  \sin(A+B) with  \sin(A-B)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello MarcoMP
    Quote Originally Posted by MarcoMP View Post
    Ok I have checked the information and I left out the first line and you were correct, it only applies to a triangle.

    On the level that we do trigenometry, we can do compound angles sin(A+B) = sinA.CosB + CosA.SinB

    Then changing from Cos2A = 2Cos^2A -1 = 1 -2Sin^2A = Cos^2A - Sin^2A

    And from Sin2A = 2SinA.CosA

    Do you perhaps know how i would go about proving this identity using these rules?
    Starting with the formula you initially quoted, then:

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

    =\frac{\tan A + \tan B}{1-\tan A \tan B}, by dividing top-and-bottom by \cos A \cos B
    \Rightarrow \tan(A+B) - \tan(A+B)\tan A\tan B=\tan A +\tan B

    But if A+B= 180 - C, then \tan(A+B) = -\tan C.

    \Rightarrow -\tan C +\tan A\tan B\tan C=\tan A + \tan B

    \Rightarrow \tan A\tan B\tan C =\tan A+\tan B+\tan C

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. interesting question
    Posted in the Business Math Forum
    Replies: 1
    Last Post: September 30th 2010, 11:26 PM
  2. Looks interesting
    Posted in the Math Forum
    Replies: 0
    Last Post: April 23rd 2010, 07:46 AM
  3. interesting combinatorial identity proof
    Posted in the Math Challenge Problems Forum
    Replies: 1
    Last Post: November 29th 2009, 04:34 PM
  4. Interesting one!
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 13th 2008, 05:46 AM
  5. Interesting X
    Posted in the Geometry Forum
    Replies: 1
    Last Post: June 16th 2008, 03:44 PM

Search Tags


/mathhelpforum @mathhelpforum