Question

please do this fast! I will ignore bustard answers##​

Answer 1

Step-by-step explanation:

We have,

$\cos( \frac{2\pi}{15} ) \cos( \frac{4\pi}{15} ) \cos( \frac{8\pi}{15} ) \cos( \frac{16\pi}{15} )$

$\cos( \frac{2\pi}{15} ) \cos(2. \frac{2\pi}{15} ) \cos( {2}^{2} . \frac{2\pi}{15} ) \cos( {2}^{3} .\frac{2\pi}{15} )$

Now,

We know that

$\cos( \alpha ) \cos(2 \alpha ) \cos( {2}^{2} \alpha ) \cos( {2}^{3} \alpha ) .... \cos( {2}^{n - 1} \alpha ) = \frac{ \sin( {2}^{n} \alpha ) }{ {2}^{n} \sin( \alpha ) }$

Using this, we have,

$= \frac{ \sin( \frac{32\pi}{15} ) }{16 \sin( \frac{2\pi}{15} ) }$

$= \frac{1}{16}$

$= \frac{ \sin(2\pi + \frac{2\pi}{15} ) }{16 \sin( \frac{2\pi}{15} ) }$

$= \frac{ \sin( \frac{2\pi}{15} ) }{16 \sin( \frac{2\pi}{15} ) }$

Answer 2

Answer:

Hope this helps uh......