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Math Help - nid Help please... sum and difference

  1. #1
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    nid Help please... sum and difference

    Use the cosine , sine and tangent sum and difference identities to simplify each of the following:
    1.) cos (A + B) + cos (A - B)
    2.) cos (A + B) - cos (A - B)
    3.) sin (A + B) + sin (A - B)
    4.) sin (A + B) - sin (A - B)
    5.) tan20degrees + tan32degrees / (1- tan 20degrees X tan 32 degrees)
    6.) tan 35 degrees - tan 12 degrees / ( 1 + tan 35 degrees X tan 12 degrees)
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  2. #2
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    Do you know the identities? (I would like to learn them)
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  3. #3
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    here it is:
    cos ( A + B ) = cos A cos B + sin A sin B
    cos ( A + B ) = cos A cos B - sin A sin B
    sin ( A + B ) = sin A cos B + cos A sin B
    sin ( A - B) = sin A cos B - cos A sin B
    tan ( A - B ) = tan A - tan B / ( 1 + tan A tan B)
    tan ( A + B) = tan A + tan B / (1 - tan A tan B)

    im not sure w/ my answers... but it says simplify

    my answer in number 1 is 2cosAcosB am i correct?
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  4. #4
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    my answer in number 1 is 2cosAcosB am i correct?
    assuming the conversions in your post are correct, yes.

    2)2sinAsinB

    3)2sinAcosB

    4)2cosAsinB


    tan ( A - B ) = tan A - tan B / ( 1 + tan A tan B)
    tan ( A + B) = tan A + tan B / (1 - tan A tan B)
    do you mean: \tan(A-B)=\tan A-\frac{\tan B}{1+\tan A\tan B}

    or: \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}
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  5. #5
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    ^
    ^
    the 2nd one of course....
    btw how old are u?
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  6. #6
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    Quote Originally Posted by ^_^Engineer_Adam^_^
    btw how old are u?
    I'm 14 (15 in september)

    5)tan(20+32)=tan50

    6)tan(35-12)=tan23
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  7. #7
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    Hello, Adam!

    Do you need to see the steps?
    I'll do a few of them . . .


    Use the cosine, sine and tangent sum/difference identities to simplify:

    1)\;\cos(A + B) + \cos(A - B)

    We have: . . \cos(A + B) \qquad\quad\;\;+ \qquad\quad\cos(A - B)

    Then: . \overbrace{(\cos A\cos B - \sin A\sin B)} + \overbrace{(\cos A\cos B + \sin A\sin B)}

    . . . . . . . . . . . . . . . . = \;\;\;2\cos A\cos B



    3)\;\sin(A + B) +\sin (A - B)

    We have: . . \sin(A + B) \qquad\quad\;\;+ \qquad\quad\sin(A - B)

    Then: . \overbrace{(\sin A\cos B + \sin B\cos A)} + \overbrace{(\sin A\cos B - \sin B\cos A)}

    . . . . . . . . . . . . . . . . = \;\;\;2\sin A\cos B


    5)\;\frac{\tan20^o + \tan32^o}{1- \tan 20^o\cdot\tan32^o}

    We're exprected to recognize the form: . \frac{\tan A + \tan B}{1 - \tan A\!\cdot\tan B}\:=\:\tan(A + B)

    Therefore: . \frac{\tan20^o + \tan32^o}{1 - \tan20^o\!\cdot\tan32^o} \:=\:\tan(20^o + 32^o) \:=\:\tan52^o


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