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    sin 3θ=sin (2θ+θ )=sin 2θ cosθ+cos 2θ sinθ

    how did the mark scheme get that answer?
    i know by expanding the brackets you get sin2θ+ sinθ
    and with the sin2θ you use a double angle formula.
    but not so sure how the MS got sin2θ and the cos 2θ
    thanks!
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    Use this identity

     sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

    with

     A = 2\theta

     B = \theta

    then it follows that

     sin(2\theta + \theta) = sin(\theta)cos(2\theta) + cos(\theta)sin(2\theta)
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    I've been stuck on this trig question for so long its just :cry2:

    Solve in the 0≤theta≤360

     sin^2(theta)/2 = 2sin(theta)

    Any help would be really appreciated!
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    use identity

     sin(2x) = 2sin(x)cos(x)

    with

     2x = \theta

    to get

     sin(\theta) = 2sin(\frac{\theta}{2})cos(\frac{  \theta}{2})

    See if that is easier to solve!
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    (Original post by spread_logic_not_hate)
    use identity

     sin(2x) = 2sin(x)cos(x)

    with

     2x = \theta

    to get

     sin(\theta) = 2sin(\frac{\theta}{2})cos(\frac{  \theta}{2})

    See if that is easier to solve!
    I've tried that.. but i'm stuck again. oh boy.

    I honestly don't understand the  2x = \theta . :no:
    Could you explain it please?
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    No problem.

    Ok so we have this

     sin(\theta) = 2sin(\frac{\theta}{2})cos(\frac{  \theta}{2})

    Gonna substitute in for  2sin\theta in the original equation

     sin^2\frac{\theta}{2} = 4sin(\frac{\theta}{2})cos(\frac{  \theta}{2})

    Bring everything over to the LHS

     sin^2\frac{\theta}{2} -4sin(\frac{\theta}{2})cos(\frac{  \theta}{2}) = 0

    Factorise out  sin\frac{\theta}{2}

     sin\frac{\theta}{2}[sin\frac{\theta}{2} -4cos(\frac{\theta}{2})] = 0

    One solution is obtained from here as  sin\frac{\theta}{2} = 0 which implies  \theta = 0, 180, 360

    Now solve

     sin\frac{\theta}{2} -4cos(\frac{\theta}{2}) = 0

    Bring cos to RHS

     sin\frac{\theta}{2} =4cos(\frac{\theta}{2})

    Divide by cos and use that  \frac{sin\frac{\theta}{2}}{cos(\  frac{\theta}{2})} = tan\frac{\theta}{2} to get

     tan\frac{\theta}{2} =4

    This gives  \theta = 2arctan(4) = 151.9 degrees
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    thanks spread_logic_not_hate (:

    i apologise for spamming this thread.
    i wish i was more gifted with numbers
    :}

    can someone help me with this question?

    c) given that (x-2)(x^3 + ax^2 + bx + c)

    find values of the constants a b and c

    thanks so much!
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    ^Factorise just by long division to find the constants.

    And can anyone help me on this one? If sinx = p/q

    Write cosec2x in terms of p and q.

    Thank you in advance!
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    By writing that \sin3 \theta as sin(2theta+theta)

    show that sin3theta=3\sin \theta-4\sin^3 \theta

    lol sorry for the poor latex, I'm still learning to get the hang of it, but anyways can someone please answer this question step by step thank you.
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    ^ For this question you need to apply the sin(A+B) formula but instead with sin(2A+1A) and expand!

    Tell me if you get stuck again!
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    (Original post by Apple24)
    ^ For this question you need to apply the sin(A+B) formula but instead with sin(2A+1A) and expand!

    Tell me if you get stuck again!
    haha thnx for that i already figured it out, I had to use sin(a+b) and double angle formulae's.
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    Hi,

    Does anybody know where I could download EDEXCEL C3 and C4 that was sat in Jan 2010.

    Many Thanks
 
 
 
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Updated: February 21, 2010
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