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    how do i solve sin(theta)cos(theta)-sin(theta)+cos(theta) = 1?
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    (Original post by alesha98)
    how do i solve sin(theta)cos(theta)-sin(theta)+cos(theta) = 1?
    If you write \sin \theta = x and \cos \theta = y it becomes xy - x + y - 1= 0 which factorises neatly as (x+1)(y-1) = (\sin \theta +1)(\cos \theta - 1) = 0 i.e:

    \displaystyle

\begin{equation*}\sin \theta \cos \theta - \sin \theta + \cos \theta - 1 \equiv (\sin \theta +1)(\cos \theta - 1)\end{equation*}

    So that you get two family of solutions:

    \sin \theta + 1 = 0 or \cos \theta - 1= 0.
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    (Original post by alesha98)
    how do i solve sin(theta)cos(theta)-sin(theta)+cos(theta) = 1?
    Bring the 1 over to the other side and see if you can put the LHS into two brackets.
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    sinϴcosϴ - sinϴ + cosϴ - 1 = (sinϴ + ?)(cosϴ - ?)
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    Express sin(theta) as a function of cos(theta) then you should be able to obtain a constant term from the function of cos(theta). This gives you the sin(theta) term but I didn't notice what Zacken had put. Don't listen to me.
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    (Original post by alesha98)
    how do i solve sin(theta)cos(theta)-sin(theta)+cos(theta) = 1?
    If you're not entirely sure why it factorises the way it does in my post, then consider this:

    \displaystyle 

\begin{equation*}\sin \theta \cos \theta - \sin \theta + \cos \theta - 1 \equiv \sin \theta (\cos \theta -1) + (\cos \theta - 1) \equiv (\cos \theta - 1)(\sin \theta +1)\end{equation*}

    By pulling out the common factor out (\cos \theta - 1).
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    (Original post by Zacken)
    If you're not entirely sure why it factorises the way it does in my post, then consider this:

    \displaystyle 

\begin{equation*}\sin \theta \cos \theta - \sin \theta + \cos \theta - 1 \equiv \sin \theta (\cos \theta -1) + (\cos \theta - 1) \equiv (\cos \theta - 1)(\sin \theta +1)\end{equation*}

    By pulling out the common factor out (\cos \theta - 1).
    Thankyou so much
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    (Original post by alesha98)
    Thankyou so much
    You're very welcome!
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    (Original post by Zacken)
    If you write \sin \theta = x and \cos \theta = y it becomes xy - x + y - 1= 0 which factorises neatly as (x+1)(y-1) = (\sin \theta +1)(\cos \theta - 1) = 0 i.e:

    \displaystyle

\begin{equation*}\sin \theta \cos \theta - \sin \theta + \cos \theta - 1 \equiv (\sin \theta +1)(\cos \theta - 1)\end{equation*}

    So that you get two family of solutions:

    \sin \theta + 1 = 0 or \cos \theta - 1= 0.
    so what if i take out  \sin \theta as a common factor? do i lose some answers?
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    (Original post by thefatone)
    so what if i take out  \sin \theta as a common factor? do i lose some answers?
    If you divide an equation by something which could be zero then you will lose solutions, for example x^2 - 2x = 0; dividing by x gives x - 2 = 0 -> x = 2, which misses the solution x = 0.
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    (Original post by thefatone)
    so what if i take out  \sin \theta as a common factor? do i lose some answers?
    Well, depends on what you mean. But I don't see how it would be useful.
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    (Original post by Zacken)
    Well, depends on what you mean. But I don't see how it would be useful.
    i did this

    \sin \theta \left(\cos \theta -1+ \cos \theta \right)=1
    which gave me
    \sin \theta \left(2\cos \theta -1 \right)=1
    then i would solve from there but that doesn't seem right...
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    (Original post by thefatone)
    i did this

    \sin \theta \left(\cos \theta -1+ \cos \theta \right)=1
    which gave me
    \sin \theta \left(2\cos \theta -1 \right)=1
    then i would solve from there but that doesn't seem right...
    It is right, it's just massively useless. How would you proceed from here?
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    (Original post by Zacken)
    It is right, it's just massively useless. How would you proceed from here?
    lol i know TeeEm would kill me for this

    but i'd continue to solve for values of theta

    so

    \sin\theta=1
    and
    2\cos\theta-1=1
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    (Original post by thefatone)
    ...
    That's not how it works. You can only do equate each factor to zero if your equation is of the form f(x) = 0.

    Otherwise, do you think we can do this: (x-1)^2 = x^2 - 2x + 1 = 0 \Rightarrow x(x-2) = 1 so x=1 and x-2=1 \Rightarrow x = 3? Is this valid.

    Nopes, it is not the case that ab = c then a=c or b=0. This is only true if c=0.
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    (Original post by Zacken)
    That's not how it works. You can only do equate each factor to zero if your equation is of the form f(x) = 0.

    Otherwise, do you think we can do this: (x-1)^2 = x^2 - 2x + 1 = 0 \Rightarrow x(x-2) = 1 so x=1 and x-2=1 \Rightarrow x = 3? Is this valid.

    Nopes, it is not the case that ab = c then a=c or b=0. This is only true if c=0.
    there we go that's what i was looking for
    thanks
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    (Original post by thefatone)
    there we go that's what i was looking for
    thanks
    You're welcome.
 
 
 
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