Question

prove that ........

(cos24)(cos48)(cos96)(cos192)=1/16


pls help​

Answer 1

Step-by-step explanation:

Given, cos(24°)•cos(48°)•cos(96°)•cos(192°)

Now, we know that,

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

so, given expression reduced to

=cos(24°)•cos(2•(24°))•cos(4•(24°))•cos(8•(24°))

=cos(24°)•cos(2•(24°))•cos(2²•(24°))•cos(2³•(24°))

$= \frac{ \sin({2}^{4}(24^{0} )) }{ {2}^{4} \sin(24^{0} ) }$

$= \frac{ \sin(384)^{0} }{16 \sin(24)^{0} }$

$= \frac{ \sin(360^{0} + {24}^{0} ) }{16 \sin( {24}^{0} ) }$

We know that,

sin(2π+θ)=sin(θ), so,

$= \frac{ \sin( {24}^{0} ) }{16 \sin( {24}^{0} ) }$

$= \frac{1}{16}$

Hence, proved

Answer 2

Answer:

Hi have a great day ahead