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Math Help - Proving cofunction identity using subtraction formula.

  1. #1
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    Proving cofunction identity using subtraction formula.

    Got this question in textbook assignments:

    Prove the cofunction identity using the addition and subtraction formulas.

    20. cot (pi/2 - u) = tan u

    so to my knowledge, I turn it into:

    1 / tan (pi/2 - u) and try to apply subtraction formula but then tan pi/2 isn't defined is it? because 1/0 isn't defined. Is this the case?



    [EDIT] I've also run into this now and don't want to spam the forums so I will post here, prove:

    sin(pi/2 - x) = sin(pi/2 + x)

    I start with RHS = sin(pi/2) cos x + cos (pi/2) sin x
    = (1) cos x + (0) sin x
    = cos x

    Haha this is probly also simple but alas it eludes me...
    Last edited by DannyMath; March 27th 2010 at 01:10 PM.
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  2. #2
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    Hello, DannyMath!

    Your reasoning is correct ... We can't work it that way.


    Prove the cofunction identity using the addition and subtraction formulas.

    . . 20.\;\;\cot\left(\tfrac{\pi}{2}- u\right) \:=\: \tan u
    If you don't know the addition/subtraction formula for cotangent, we can re-invent it.

    \cot(A \pm B) \;=\;\frac{1}{\tan(A+B)} \;=\;\frac{1\mp\tan A\tan B}{\tan A \pm \tan B} \;=\;  \frac{1 \mp \dfrac{1}{\cot A}\,\dfrac{1}{\cot B}}{\dfrac{1}{\cot A} \pm\dfrac{1}{\cot B}}


    Multiply by \frac{\cot A\cot B}{\cot A\cot B}\!:\quad\boxed{\cot(A \pm B) \;=\;\frac{\cot A\cot A \mp 1}{\cot B \pm \cot A}}



    We have: . \cot\left(\tfrac{\pi}{2} - u\right) \;=\;\frac{\cot\frac{\pi}{2}\cot u + 1}{\cot u - \cot\frac{\pi}{2}} \;=\;\frac{0\cdot\cot u + 1}{\cot u - 0} \;=\;\frac{1}{\cot u} \;=\;\tan u

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  3. #3
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  4. #4
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    Ok I hesitate to ask but I understand everything until after

    <br /> <br />
\frac{1 \mp \dfrac{1}{\cot A}\,\dfrac{1}{\cot B}}{\dfrac{1}{\cot A} \pm\dfrac{1}{\cot B}}<br />


    The steps between the above and

    <br /> <br />
\boxed{\cot(A \pm B) \;=\;\frac{\cot A\cot B \mp 1}{\cot B \pm \cot A}}<br />

    have me stumped. I tried for an hour to work it out on paper and googled it. Does it have something to do with multiplying by its reciprocal?
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  5. #5
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    Quote Originally Posted by DannyMath View Post
    Ok I hesitate to ask but I understand everything until after

    <br /> <br />
\frac{1 \mp \dfrac{1}{\cot A}\,\dfrac{1}{\cot B}}{\dfrac{1}{\cot A} \pm\dfrac{1}{\cot B}}<br />
    Multiply both numerator and denominator by cot(A) cot(B) to get
    \frac{cot(A)cot(B)\mp 1}{cot(B)+ cot(A)}.


    The steps between the above and

    <br /> <br />
\boxed{\cot(A \pm B) \;=\;\frac{\cot A\cot B \mp 1}{\cot B \pm \cot A}}<br />

    have me stumped. I tried for an hour to work it out on paper and googled it. Does it have something to do with multiplying by its reciprocal?
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  6. #6
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    Ok I think I understand now. I was getting confused cause I know that when you multiply something by its reciprocal you get 1, but I'm starting to realize that's not what we're doing in this case haha. Thanks guys!
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