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Math Help - Proof of a trig relation using induction

  1. #1
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    Proof of a trig relation using induction

    The number x and the real number  \theta are such that  x + \frac{1}{x}= 2 cos\theta . Use
    mathematical induction to show that  x^n+ \frac{1}{x^n}= 2 cosn \theta for all positive integers n > 1.

    Thank you
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  2. #2
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    Hello differentiate
    Quote Originally Posted by differentiate View Post
    The number x and the real number  \theta are such that  x + \frac{1}{x}= 2 cos\theta . Use
    mathematical induction to show that  x^n+ \frac{1}{x^n}= 2 cosn \theta for all positive integers n > 1.

    Thank you
    Hint:

    Start with:
    \left(x^k+\dfrac1{x^k}\right)\left(x+\dfrac1x\righ  t)=x^{k+1}+\dfrac{1}{x^{k+1}}+x^{k-1}+\dfrac{1}{x^{k-1}}

    \Rightarrow x^{k+1}+\dfrac{1}{x^{k+1}}=\left(x^k+\dfrac1{x^k}\  right)\left(x+\dfrac1x\right)-\left(x^{k-1}+\dfrac{1}{x^{k-1}\right)}
    Then assume that the result is true for all values up to n = k, and the RHS becomes 2(2\cos n\theta\cos\theta - \cos(n-1)\theta), which simplifies to 2\cos(n+1)\theta.

    Can you complete the proof?

    Grandad
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by differentiate View Post
    The number x and the real number  \theta are such that  x + \frac{1}{x}= 2 cos\theta . Use
    mathematical induction to show that  x^n+ \frac{1}{x^n}= 2 cosn \theta for all positive integers n > 1.

    Thank you
    There is also a solution without using induction...

    Start of solution...

    Let z=rcis{\alpha}, a complex number.

    We will prove:

    If z + \frac{1}{z}= 2 cos\theta

    then: x^n+ \frac{1}{z^n}= 2 cosn \theta

    Using de Moivre's formula, and z=rcis{\alpha} on z + \frac{1}{z}= 2 cos\theta.

    The result quickly proved...
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