# Can you simplify  cos(x)cos(2x)cos(4x)cos(8x)cos(16x) ... cos(2^n x)  ?,

Feb 8, 2018

$\sin \frac{{2}^{n + 1 x}}{{2}^{n + 1} \sin \left(x\right)}$

#### Explanation:

We want to simplify

$\cos \left(x\right) \cos \left(2 x\right) \cos \left(4 x\right) \cos \left(8 x\right) \ldots \cos \left({2}^{n} x\right)$

The trick is to use the double-angle identity repeatedly

• $\sin \left(2 x\right) = 2 \cos \left(x\right) \sin \left(x\right)$

Consider the following example

$\sin \left(16 x\right) = 2 \cos \left(8 x\right) \sin \left(8 x\right)$

$= {2}^{2} \cos \left(8 x\right) \cos \left(4 x\right) \sin \left(4 x\right)$

$= {2}^{3} \cos \left(8 x\right) \cos \left(4 x\right) \cos \left(2 x\right) \sin \left(2 x\right)$

$= {2}^{4} \cos \left(8 x\right) \cos \left(4 x\right) \cos \left(2 x\right) \cos \left(x\right) \sin \left(x\right)$

$\implies \cos \left(x\right) \cos \left(2 x\right) \cos \left(4 x\right) \cos \left(8 x\right) = \sin \frac{16 x}{{2}^{4} \sin \left(x\right)}$

If we generalize it

$\sin \left({2}^{n + 1} x\right) = 2 \cos \left({2}^{n} x\right) \sin \left({2}^{n} x\right)$

$= {2}^{2} \cos \left({2}^{n} x\right) \cos \left({2}^{n - 1} x\right) \sin \left({2}^{n - 1} x\right)$

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$= {2}^{n} \cos \left({2}^{n} x\right) \cos \left({2}^{n - 1} x\right) \ldots \cos \left(4 x\right) \cos \left(2 x\right) \sin \left(2 x\right)$

$= {2}^{n + 1} \cos \left({2}^{n} x\right) \cos \left({2}^{n - 1} x\right) \ldots \cos \left(2 x\right) \cos \left(x\right) \sin \left(x\right)$

$\implies \cos \left(x\right) \cos \left(2 x\right) \cos \left(4 x\right) \cos \left(8 x\right) \ldots \cos \left({2}^{n} x\right) = \sin \frac{{2}^{n + 1} x}{{2}^{n + 1} \sin \left(x\right)}$