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    have attempted this question loads now and can't get anywhere:

    by using 2cos(ntheta)=e^(intheta)+e^(-intheta) for suitable values of n, show that:
    32cos^6(theta)=cos(6theta)+6cos( 4theta)+15cos(2theta)+10

    have tried binomial expansion but I couldn't get anywhere, is that the right way to start?
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      Yes, it is. Picking n = 1, 2 \cos \theta = e^{i \theta} + e^{- i \theta}, then raise both sides to the six and expand. Try grouping terms of the expansion of the form e^{ik \theta} and e^{-ik \theta}, and use what you've been given for cos(n theta).
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      Big thanks
     
     
     
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