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C3 Trig Watch

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    So basically I've been trig homework all night because it's so hard haha!
    Just wondering if anyone can find a solution to:

    Cotx - Cosx / (1 - sinx) = cotx

    If you could explain your workings out itd be really helpful.

    Thanks
    James


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    (Original post by james0902)
    So basically I've been trig homework all night because it's so hard haha!
    Just wondering if anyone can find a solution to:

    Cotx - Cosx / (1 - sinx) = cotx

    If you could explain your workings out itd be really helpful.

    Thanks
    James


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    Subtract \cot(x) from both sides then it's just when \cos(x)=0
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    Multiply by sinx/sinx.
    Is this question basically proving the identity \displaystyle \frac{\cot x -\cos x}{1-\sin x} ?
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    I mean you have to simplify the left to find the right. The two sides cannot interact with one another.


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    (Original post by B_9710)
    Multiply by sinx/sinx.
    Is this question basically proving the identity \displaystyle \frac{\cot x -\cos x}{1-\sin x} ?
    Yes it's proving the identity.


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    (Original post by james0902)
    Yes it's proving the identity.


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    Multiply both numerator and denominator of LHS by sinx.
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    (Original post by B_9710)
    Multiply both numerator and denominator of LHS by sinx.
    I'm really struggling so what would that give me? Would that make

    sinx (cotx - cosx)

    Is that right?


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    (Original post by james0902)
    I'm really struggling so what would that give me? Would that make

    sinx (cotx - cosx)

    Is that right?


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    Think about what cotx is in terms of cosx and sinx. What is sinxcotx?
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    (Original post by james0902)
    Yes it's proving the identity.


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    Personally I'd multiply top and bottom by 1+\sin(x)

    This gets a nice difference of two square on the denominator which is equal to \cos^2(x) then the numerator just takes a bit messing around with to get get into a fraction in terms of sine/cosine, by which point loads of things cancel and you're soon left with \cot(x). But I'm not sure how simple you'd find messing around with a fraction as a numerator, should be straight forward enough if you know what's happening.
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    (Original post by james0902)
    So basically I've been trig homework all night because it's so hard haha!
    Just wondering if anyone can find a solution to:

    Cotx - Cosx / (1 - sinx) = cotx

    If you could explain your workings out itd be really helpful.

    Thanks
    James




    Easy solution!!

    1. We start off by multiplying both sides of your expression by (1-Sinx)
    2. Divide both sides by Cotx.You should get (Cotx-Cosx)/Cotx=1-Sinx
    3. Now realise that you can split the fraction on the left so the LHS becomes (Cotx/Cotx)-(Cosx/Cotx)=1-Sinx
    4. Hence it simplifies to 1-(Cosx/Cotx).
    5. Cotx is 1/tanx therefore it now becomes 1-Cosxtanx
    6. Tanx=Sinx/Cosx hence the two Cosx's cancel and you are left with 1-Sinx on LHS which is equal to the RHS.

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    (Original post by BJA98)
    Easy solution!!

    1. We start off by multiplying both sides of your expression by (1-Sinx)
    2. Divide both sides by Cotx.You should get (Cotx-Cosx)/Cotx=1-Sinx
    3. Now realise that you can split the fraction on the left so the LHS becomes (Cotx/Cotx)-(Cosx/Cotx)=1-Sinx
    4. Hence it simplifies to 1-(Cosx/Cotx).
    5. Cotx is 1/tanx therefore it now becomes 1-Cosxtanx
    6. Tanx=Sinx/Cosx hence the two Cosx's cancel and you are left with 1-Sinx on LHS which is equal to the RHS.

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    wat
 
 
 
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