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Math Help - trignometric problem

  1. #1
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    trignometric problem

    1. LHS=2cos 2^a+1/2cos a+1= RHS=(2cos a-1) (2cos 2a-1) (2cos 2^2a-1).....(2cos 2^n-1 a-1)

    here a is the angle n is number

    we have to prove LHS=RHS
    Last edited by prasum; April 15th 2010 at 04:05 AM.
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  2. #2
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    More information please

    Hello prasum
    Quote Originally Posted by prasum View Post
    1.2cos 2^a+1/2cos a+1=(2cos a-1) (2cos 2a-1) (2cos 2^2a-1).....(2cos 2^n-1 a-1)

    please help to solve specially n is making problem
    I think the reason you have not had a reply is that no-one understands what you have written. You appear to have written:
    1.2\cos2^a+\tfrac12\cos a+1=(2\cos a-1) (2\cos 2a-1) (2\cos 2^2a-1).....(2\cos 2^{n-1} a-1)
    So, two questions:
    What did you mean to write?

    So what? Do we have to prove that the two sides are equal?
    Grandad
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  3. #3
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    Thats not 1/2 thats whole first part is divided
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Hi Prasum

    Use Latex or if you dont know it and than try to use your brain your question is still not at all clear...

    Here is another try is this correct
    <br />
\frac{2\cos(2^{a})+1}{2\cos(a)+1}=(2\cos a-1) (2\cos 2a-1) (2\cos 2^2a-1).....(2\cos 2^{n-1} a-1)
    Last edited by ADARSH; April 15th 2010 at 04:31 AM. Reason: Heavy edition
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  5. #5
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    it is 2cos 2^n a+1 what yu have written is correct
    Last edited by prasum; April 15th 2010 at 06:32 PM.
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  6. #6
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    Hello prasum

    If you're serious about wanting help on this Forum - and, indeed, serious about improving your mathematical skills - you must learn to be more accurate in writing down mathematical expressions. This will save everyone a good deal of time, and help you to acquire the skills you need to solve problems like this.


    I think the question is this:
    Prove, for any positive integer n, and any value of a, that
    \frac{2\cos2^na+1}{2\cos a +1}=(2\cos a-1)(2\cos 2a-1)(2\cos 2^2a-1)...(2\cos 2^{n-1}a-1)
    This is done by Mathematical Induction, as follows.

    Let P(n) be the propositional function:
    (2\cos a-1)(2\cos 2a-1)(2\cos  2^2a-1)...(2\cos 2^{n-1}a-1)=\frac{2\cos2^na+1}{2\cos a +1}
    (For convenience, I've swapped the sides of the equation.)

    Then, multiplying both sides of the equation by (2\cos2^na-1):
    P(n)\Rightarrow (2\cos a-1)(2\cos 2a-1)(2\cos  2^2a-1)...(2\cos 2^{n-1}a-1)(2\cos2^na-1) =\frac{(2\cos2^na+1)(2\cos2^na-1)}{2\cos a +1}
    =\frac{4\cos^22^na-1}{2\cos a + 1}, using (a+b)(a-b) = a^2-b^2

    =\frac{2(2\cos^22^na-1)+1}{2\cos a + 1}


    =\frac{2\cos2^{n+1}a+1}{2\cos a + 1}, using \cos 2\theta = 2\cos^2\theta-1

    \Rightarrow P(n+1)
    Now P(1) states:
    2\cos a -1=\frac{2\cos2a+1}{2\cos a +1}
    The RHS is:
    \frac{2(2\cos^2a-1)+1}{2\cos a +1}
    =\frac{4\cos^2a-1}{2\cos a +1}

    =\frac{(2\cos a+1)(2\cos a -1)}{2\cos a +1}


    =2\cos a -1

    So P(1) is true. Therefore, by Induction, P(n) is true for all positive integers, n.

    Grandad
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  7. #7
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    thanks man
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