Question #7207e

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Answer 1 · 2017-03-04T14:36:31

See below.

From the identity:

$sin (2^n x) = 2 sin (2^(n-1) x) \ cos (2^(n-1) x)$

Splitting out that first sine term again:

$= 2 * [2 sin (2^(n-2) x) \ cos (2^(n-2) x) ]\ cos (2^(n-1) x)$

And again:

$= 2 * 2 * [2 sin (2^(n-3) x) \ cos (2^(n-3) x)] \ cos (2^(n-2) x) \ cos (2^(n-1) x)$

I am stopping there :) but you see the pattern