Question

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

Answer 1

L.H.S $=(2\cos \theta -1)(2\cos 2\theta -1)(2\cos 2^2\theta -1)\cdots (2\cos 2^{n-1}\theta -1)$

multiplying and dividing by $2cos\theta +1$

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

Now, $(2\cos \theta +1)(2\cos \theta -1)=4{\cos }^2\theta -1=2(1+\cos 2\theta )-1=2\cos 2\theta +1$

similarly, $(2\cos 2\theta +1)(2\cos 2\theta -1)=4{\cos }^{2}2\theta -1=2(1+\cos 2^2\theta )-1=2\cos 2^2\theta +1$
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$(2\cos 2^{n-1}\theta +1)(2\cos 2^{n-1}\theta -1)=4{\cos }^2{2}^{n-1}\theta -1=2(1+\cos 2^n\theta )-1=2\cos 2^n\theta +1$

Thus L.H.S
$=\frac{2\cos 2^n\theta +1}{2\cos \theta +1}=$ R.H.S