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Math Help - Trig identities with special triangles

  1. #1
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    Trig identities with special triangles

    the answer in the back of the book is 2-root(3) the answer i keep getting is 2+root(3)

    Find the exact value of each by choosing the values for angle a and b and applying an appropriate sum or difference identity:

    tan 15

    tan 15 = tan (pi/4 + pi/6)

    tan (A+B)= sin( A+B) / cos (A+B)

    so:

    sin (A+B) = sin pi/4 cos pi/6 + cos pi/4 sin pi/6
    (1/root(2))(root(3)/2) + (1/root(2))(1/2)
    = (root(3) +1)/(2root(2))

    cos (A+B) = cos pi/4 cos pi/6 - sin pi/4 sin pi/6
    = (1/root(2))(root(3)/2) - (1/root(2))(1/2))
    = (root(3)-1)/(2root(2))

    tan (a+b) = ((root(3)+1)/(2root(2))/((root(3)-1)/(2root(2))
    = (root(3)+1)/(2root(2) times (2root(2))/(root(3)-1)
    (cancel out the 2root(2))
    =(root(3)+1)/(root(3)-1) <---- multiply both sides by conjugate (root(3)+1)
    =(3+2root(3) +1)/ 3-1
    =(4+2root(3))/2
    simplify
    =2+root(3)
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  2. #2
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    Quote Originally Posted by tmas View Post
    the answer in the back of the book is 2-root(3) the answer i keep getting is 2+root(3)

    Find the exact value of each by choosing the values for angle a and b and applying an appropriate sum or difference identity:

    tan 15

    tan 15 = tan (pi/4 + pi/6)

    tan (A+B)= sin( A+B) / cos (A+B)

    so:

    sin (A+B) = sin pi/4 cos pi/6 + cos pi/4 sin pi/6
    (1/root(2))(root(3)/2) + (1/root(2))(1/2)
    = (root(3) +1)/(2root(2))

    cos (A+B) = cos pi/4 cos pi/6 - sin pi/4 sin pi/6
    = (1/root(2))(root(3)/2) - (1/root(2))(1/2))
    = (root(3)-1)/(2root(2))

    tan (a+b) = ((root(3)+1)/(2root(2))/((root(3)-1)/(2root(2))
    = (root(3)+1)/(2root(2) times (2root(2))/(root(3)-1)
    (cancel out the 2root(2))
    =(root(3)+1)/(root(3)-1) <---- multiply both sides by conjugate (root(3)+1)
    =(3+2root(3) +1)/ 3-1
    =(4+2root(3))/2
    simplify
    =2+root(3)
    15 = 45 - 30 NOT 45 + 30 (and certainly not for that matter pi/4 + pi/6).
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  3. #3
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    Another alternative is the half angle identity

    \displaystyle \tan{\frac{\theta}{2}} = \frac{1 - \cos{\theta}}{\sin{\theta}}.

    Here you would make \displaystyle \theta = 30^{\circ} or \displaystyle \theta = \frac{\pi}{6}^C (like Mr F said, don't switch between degrees and radians - make a choice and stick with it).
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