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    (Original post by Philip-flop)
    Expanded as in like?....
    Attachment 583512

    Where do I go from here though?
    From there I would probably multiply top and bottom by 1+tan(60)tan(45).

    Then use tan(45) = 1 and tan(60) = root(3) and then simplify.
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    Can anyone help me with this?...

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    (Original post by Philip-flop)
    Can anyone help me with this?...

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     sin(\frac{\pi}{2}) = \frac{1}{2} and  cos(\frac{\pi}{6})=\frac{\sqrt{3  }}{2}
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    (Original post by NotNotBatman)
     sin(\frac{\pi}{2}) = \frac{1}{2} and  cos(\frac{\pi}{6})=\frac{\sqrt{3  }}{2}
    I'm assuming you meant  sin (\frac{\pi}{6}) = \frac{1}{2}

    Thanks a lot for your help

    How did you realise to do that from that equation? I never would have spotted it if it wasn't for your help
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    Ffs, now I'm stuck on part (g).
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    Am I even doing it right for this question?...
    Attachment 583808583810

    EDIT: Don't worry, I managed to work this out in a very strange way
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    Guys do you know how to solve these questions. They are from C1

    1. Find vaalues of the constants A,B and C in identity
    4x2 - 18x +29 = A(x-4)2 - B (x-2) + C

    2. The equation x2 + 3px + p = 0, where p is a non-zero constant and has equal roots. Find the value of p.
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    (Original post by AB J)
    Guys do you know how to solve these questions. They are from C1

    1. Find vaalues of the constants A,B and C in identity
    4x2 - 18x +29 = A(x-4)2 - B (x-2) + C

    2. The equation x2 + 3px + p = 0, where p is a non-zero constant and has equal roots. Find the value of p.
    Expansion may help you can see what A is going to be anyway, because there's only one x^2 term on the RHS.

    For 2, if you are given that a quadratic has equal roots.. what calculation can you perform?
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    (Original post by AB J)
    Guys do you know how to solve these questions. They are from C1

    1. Find vaalues of the constants A,B and C in identity
    4x2 - 18x +29 = A(x-4)2 - B (x-2) + C

    2. The equation x2 + 3px + p = 0, where p is a non-zero constant and has equal roots. Find the value of p.
    Please start a new thread. This is not C3 trigonometry.
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    (Original post by Philip-flop)
    Ffs, now I'm stuck on part (g).
    Name:  C3 Exe 7 Q7e.png
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Size:  7.6 KB
    Am I even doing it right for this question?...
    Attachment 583808583810

    EDIT: Don't worry, I managed to work this out in a very strange way
    Note that \cos(90-u)= \sin u, then rewrite the second sin term as an appropriate cos.
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    I'm probably being stupid. But how do I even work out part (a) of this question? I'm assuming it's to do with the difference of two squares?...
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    (Original post by Philip-flop)
    I'm probably being stupid. But how do I even work out part (a) of this question? I'm assuming it's to do with the difference of two squares?...
    Name:  C3 Exe 7A Q12(a).png
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    I would change the sin^2(theta) into 1 - cos^2(theta) and then work from there
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    (Original post by asinghj)
    I would change the sin^2(theta) into 1 - cos^2(theta) and then work from there
    oh right so I would do?...

     cos^2 \theta - sin^2 \theta

     cos^2 \theta - (1-cos^2 \theta)

     cos^2 \theta - 1 + cos^2 \theta

     2cos^2 \theta -1.... But then where do I go from here?
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    (Original post by Philip-flop)
    oh right so I would do?...

     cos^2 \theta - sin^2 \theta

     cos^2 \theta - (1-cos^2 \theta)

     cos^2 \theta - 1 + cos^2 \theta

     2cos^2 \theta -1.... But then where do I go from here?
    what is the double angle formula:
     cos2 \theta  ?
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    (Original post by asinghj)
    what is the double angle formula:
     cos2 \theta  ?
    I'm not sure I know. The Edexcel C3 Modular Maths textbook hasn't even explained double angle formulas. But from what I've just found out...

     cos2 \theta

     = cos( \theta + \theta)

     = cos \theta cos \theta - sin \theta sin \theta ... and so on from there (what I showed in my last comment).
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    (Original post by Philip-flop)
    I'm not sure I know. The Edexcel C3 Modular Maths textbook hasn't even explained double angle formulas. But from what I've just found out...

     cos2 \theta

     = cos( \theta + \theta)

     = cos \theta cos \theta - sin \theta sin \theta ... and so on from there (what I showed in my last comment).
    oh ok.. so yes

     = cos \theta cos \theta - sin \theta sin \theta

    this can be simplified to

     cos^2 \theta - sin^2 \theta

    because  cos \theta cos \theta = cos \theta \times cos \theta
    and this made me realise you didn't have to change the  sin^2 \theta into anything...

    oh and watch this video
    http://www.examsolutions.net/tutoria...=C3&topic=1420
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    (Original post by asinghj)
    oh ok.. so yes

     = cos \theta cos \theta - sin \theta sin \theta

    this can be simplified to

     cos^2 \theta - sin^2 \theta

    because  cos \theta cos \theta = cos \theta \times cos \theta
    and this made me realise you didn't have to change the  sin^2 \theta into anything...

    oh and watch this video
    http://www.examsolutions.net/tutoria...=C3&topic=1420
    Oh right, I see. So for this question my answer would literally be...

     cos^2 \theta - sin^2 \theta

     =cos 2\theta
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    (Original post by Philip-flop)
    I'm probably being stupid. But how do I even work out part (a) of this question? I'm assuming it's to do with the difference of two squares?...
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    For part a), later on in the textbook you'll learn about the double angle formula and the expression will be a form you recognise.

    At this stage of the textbook, you should be looking to write the expression as the right-hand-side as one of the addition formulae.

    So \cos^2 \theta - \sin^2 \theta = \cos \theta \cos \theta - \sin \theta \sin \theta.

    Can you see what to do from here?
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    (Original post by notnek)
    For part a), later on in the textbook you'll learn about the double angle formula and the expression will be a form you recognise.

    At this stage of the textbook, you should be looking to write the expression as the right-hand-side as one of the addition formulae.

    So \cos^2 \theta - \sin^2 \theta = \cos \theta \cos \theta - \sin \theta \sin \theta.

    Can you see what to do from here?
    I don't think I know where to go from there.

    All I know is...

     cos \theta cos \theta - sin \theta sin \theta

     = cos( \theta + \theta)

     = cos2 \theta
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    (Original post by Philip-flop)
    I don't think I know where to go from there.

    All I know is...

     cos \theta cos \theta - sin \theta sin \theta

     = cos( \theta + \theta)

     = cos2 \theta
    You don't need to do anything else. \cos 2\theta is a single trig function so you're done.
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    (Original post by notnek)
    You don't need to do anything else. \cos 2\theta is a single trig function so you're done.
    Oh I see. Thank you so much notnek !!
 
 
 
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