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    On a past paper for my maths exam board (ccea northern ireland) i got this question:

    Use induction to prove
    2^(n + 1) sin x cos x cos (2x) cos (4x) … cos [2^(n)x] = sin [2^(n + 1)x]
    where n is a non-negative integer.

    i literally have no clue how to even start this, i think it might involve de moivre's theorem somehow. I looked at my mark scheme and it says to use n=0...
    And please, baby steps. i have the general way of solving this already on a mark scheme, i just need it explained. For example, how do you know where one term stops and the next starts, and why do you use n=0 instead of n=1?

    (the paper i got this off was CCEA, FP2, May/June 2010, Question 5)
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    hey, I've moved your thread to the maths forum, you'll get a better response here
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    Thanks! how did you do that? I'm new here so there's some things i'm just getting used to.
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    (Original post by mathboy64)
    On a past paper for my maths exam board (ccea northern ireland) i got this question:

    Use induction to prove
    2^(n + 1) sin x cos x cos (2x) cos (4x) … cos [2^(n)x] = sin [2^(n + 1)x]
    where n is a non-negative integer.

    i literally have no clue how to even start this, i think it might involve de moivre's theorem somehow. I looked at my mark scheme and it says to use n=0...
    And please, baby steps. i have the general way of solving this already on a mark scheme, i just need it explained. For example, how do you know where one term stops and the next starts, and why do you use n=0 instead of n=1?

    (the paper i got this off was CCEA, FP2, May/June 2010, Question 5)


    Intro to mathematical induction:
    http://nrich.maths.org/4718

    May help
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    (Original post by mathboy64)
    On a past paper for my maths exam board (ccea northern ireland) i got this question:

    Use induction to prove
    2^(n + 1) sin x cos x cos (2x) cos (4x) … cos [2^(n)x] = sin [2^(n + 1)x]
    where n is a non-negative integer.

    i literally have no clue how to even start this, i think it might involve de moivre's theorem somehow. I looked at my mark scheme and it says to use n=0...
    And please, baby steps. i have the general way of solving this already on a mark scheme, i just need it explained. For example, how do you know where one term stops and the next starts, and why do you use n=0 instead of n=1?

    (the paper i got this off was CCEA, FP2, May/June 2010, Question 5)
    You don't need de moivre's, it's a lot simpler than that. The reason why they use n=0 is because the principle of induction works wherever you start it. For example, if you started at n=5, assumed true for n=k and thus true for n=k+1 then the statement is true for all n >= 5 then you could do the cases n=0,1,2,3,4 separately. For some things, for example,  (1- \dfrac{1}{2})(1- \dfrac{1}{3})(1- \dfrac{1}{4})...(1- \dfrac{1}{n}) = \dfrac{1}{n} this is obviously not valid for n=0 or n=1 so we start at n=2. It's possible here and makes most sense to start from n=0 because it should be quite common knowledge that 2sinxcosx=sin2x.
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    So,  2^{n + 1} sin x cos x cos (2x) cos (4x) … cos (2^nx) \equiv sin (2^{n + 1}x)

For n=0, cos (2^0x) \equiv sin (2^{0 + 1}x)?

\Rightarrow cosx \equiv sin2x

    I'm geting the feeling this is wrong, and i must do:

    

For  n=0, 2^{0 + 1} sin x cos (2^0x) \equiv sin (2^{0 + 1}x)

\Rightarrow 2 sin x cos x \equiv sin 2x

    But how would i know to do this? Would i just have to recognise the "...cos(2^nx)," notice that if n=0, cos(2^nx)=cosx, and so only count the term on the L.H.S. until i see the cosx?
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    (Original post by mathboy64)
    So,  2^{n + 1} sin x cos x cos (2x) cos (4x) … cos (2^nx) \equiv sin (2^{n + 1}x)

For n=0, cos (2^0x) \equiv sin (2^{0 + 1}x)?

\Rightarrow cosx \equiv sin2x

    I'm geting the feeling this is wrong, and i must do:

    

For  n=0, 2^{0 + 1} sin x cos (2^0x) \equiv sin (2^{0 + 1}x)

\Rightarrow 2 sin x cos x \equiv sin 2x

    But how would i know to do this? Would i just have to recognise the "...cos(2^nx)," notice that if n=0, cos(2^nx)=cosx, and so only count the term on the L.H.S. until i see the cosx?
    It is the latter, if you look at it, the only thing that changes on the LHS is cos(2^nx) so you keep 2^(n+1)sinx and as n increases, multiply by the required cos(x) then cos(x)cos(2x) then cos(x)cos(2x)cos(4x) etc.
 
 
 
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