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Math Help - To divide by zero in a trigonometric identity proof

  1. #1
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    To divide by zero in a trigonometric identity proof

    Alright, so I did this test at school, and I was asked to prove this identity:

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

    So I went along my merry way:

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

    and prepared to make it into

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

    But being the clueless person I am, I included the intermediary step

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

    which is wrong and loses marks even if it still reaches the answer.

    The correct step in the given solution was to make a common denominator through:

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

    And the thing is, I don't understand what the difference between the two methods is. As far as I can see, the given correct solution multiplies both sides of the fraction by

    cos(A)cos(B)

    and I'm doing the same thing except multiplying by each cos(A) cos(B) one at a time instead of both at the same time. And as far as I can see, multiplying by two given numbers m and n is the same as multiplying by mn, right?

    I asked my teacher about this and she said something about not accounting for cos(A) = 0. She justified the given correct solution with something like "the existence of tan(A) in the original solution automatically rules out cos(A) = 0 and that's why we can multiply both sides of it, but you made it into an addition and needed an extra condition 'for cos(A) = 0' in that line" (I don't remember exactly)

    The proof concludes:

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

    = \frac{sin(A+B)}{sin(A-B)} (double angle formula)

    = RHS

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

    Could someone shed some light on this? Thanks.
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  2. #2
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    Re: To divide by zero in a trigonometric identity proof

    Quote Originally Posted by icedtrees View Post
    Alright, so I did this test at school, and I was asked to prove this identity:

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

    So I went along my merry way:

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

    and prepared to make it into

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

    But being the clueless person I am, I included the intermediary step

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

    which is wrong and loses marks even if it still reaches the answer.

    The correct step in the given solution was to make a common denominator through:

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

    And the thing is, I don't understand what the difference between the two methods is. As far as I can see, the given correct solution multiplies both sides of the fraction by

    cos(A)cos(B)

    and I'm doing the same thing except multiplying by each cos(A) cos(B) one at a time instead of both at the same time. And as far as I can see, multiplying by two given numbers m and n is the same as multiplying by mn, right?

    Yes that's correct.

    I asked my teacher about this and she said something about not accounting for cos(A) = 0. She justified the given correct solution with something like "the existence of tan(A) in the original solution automatically rules out cos(A) = 0 and that's why we can multiply both sides of it, but you made it into an addition and needed an extra condition 'for cos(A) = 0' in that line" (I don't remember exactly)

    The proof concludes:

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

    = \frac{sin(A+B)}{sin(A-B)} (double angle formula)

    = RHS

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

    Could someone shed some light on this? Thanks.
    Hi icedtrees,

    Since \tan A=\frac{\sin A}{\cos A} for \tan A to have any meaningful value \cos A\neq 0. This is what your teacher had said. So in this case \cos A\neq 0\mbox{ and }\cos B\neq 0. Otherwise \tan A \mbox{ and }\tan B would not exist. Is that clear to you?
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  3. #3
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    Re: To divide by zero in a trigonometric identity proof

    Thankyou Sudharaka, I understand what is meant by that now. But to clarify, proving it with



    and then multiplying through by cos(A) should not be incorrect?
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    Re: To divide by zero in a trigonometric identity proof

    Quote Originally Posted by icedtrees View Post
    Thankyou Sudharaka, I understand what is meant by that now. But to clarify, proving it with



    and then multiplying through by cos(A) should not be incorrect?
    Whether you multiply by cos(A)cos(B) directly or by cos(A) and cos(B) in two steps there is no difference. Although the former is a little easier and reduces the number of steps.
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  5. #5
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    Re: To divide by zero in a trigonometric identity proof

    The identity has certain constraints on the domains of both sides.

    On the right, we see that A and B must differ, due to the denominator.
    More specifically, A-B must not be any multiple of 180 degrees,
    otherwise the right side is undefined.

    On the left, we see again that A and B must differ.
    However, tanA and tanB are both undefined for any odd multiple of 90 degrees.

    If you place A=90 degrees and B=30 degrees for example into the right side,
    you get a solution, but try placing those angle values into the left!

    Hence, tanA and tanB are both undefined when cosA and cosB are zero.

    Therefore, multiplying by cosA/cosA=cosB/cosB=(cosAcosB)/(cosAcosB)=1
    is not an issue,
    since cosA and cosB cannot be zero for the allowable domain of this identity.
    If you are aware of this, then your method is fine.
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