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Prove that

$$\frac{2cos(2^n\theta)+1}{2cos\theta+1}= (2cos\theta -1)(2cos2\theta -1)(2cos2^2\theta -1)...(2cos2^{n-1}\theta -1)$$

How do i do this

thanks

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    Do you have anything that you have tried... – 2017-02-02
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    This looks like a good candidate for induction – 2017-02-02
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    @Riemann-bitcoin. can you show me some start? – 2017-02-02

2 Answers 2

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$$ (2cos\theta+1)(2cos\theta -1)(2cos2\theta -1)(2cos2^2\theta -1)...(2cos2^{n-1}\theta -1)= $$ $$ (2cos2\theta+1)(2cos2\theta -1)(2cos2^2\theta -1)...(2cos2^{n-1}\theta -1)= $$ $$ (2cos4\theta+1)(2cos4\theta -1)(2cos2^3\theta -1)...(2cos2^{n-1}\theta -1)= $$ $$\vdots$$ $$=(2cos(2^n\theta)+1)$$

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We check if $$\frac{2cos(2^n\theta)+1}{2cos\theta+1}= (2cos\theta -1)(2cos2\theta -1)(2cos2^2\theta -1)...(2cos2^{n-1}\theta -1)$$ holds for n=1. Which is $$ \frac{2cos(2\theta)+1}{2cos(\theta)+1}=2cos(\theta)-1 $$ we reformulate this as $$ 2cos(2\theta)+1=(2cos(\theta)+1)(2cos(\theta)-1) $$ or $$ 2cos(2\theta)+1=4cos^2(\theta)-1 $$ recall $2cos^2x-1=cos(2x)$ which gives us $2cos^2x=cos(2x)+1$.

so we have $$ 2cos(2\theta)+1=2(cos(2\theta)+1)-1 $$ so $$ 2cos(2\theta)+1=2cos(2\theta)+1 $$ So now assume this holds for some n=k, thus we assume $$ $$ $$\frac{2cos(2^k\theta)+1}{2cos\theta+1}= (2cos\theta -1)(2cos2\theta -1)(2cos2^2\theta -1)...(2cos2^{k-1}\theta -1)$$ holds; then show it holds for n=k+1.