Question

-cosπ/7*cos3π/7*cos2π/7помогите пожалуйста

Answer 1

-cosπ/7*cos3π/7*cos2π/7
$-cos \frac{ \pi }{7} *cos \frac{2 \pi }{7} *cos \frac{3 \pi }{7} = \\ \\ =- \frac{2sin \frac{ \pi }{7}*cos \frac{ \pi }{7} }{2sin \frac{ \pi }{7} } *cos \frac{2 \pi }{7} *cos \frac{3 \pi }{7}= \\ \\ =- \frac{sin \frac{2 \pi }{7} }{2sin \frac{ \pi }{7} } *cos \frac{2 \pi }{7} *cos \frac{3 \pi }{7} = \\ \\ =- \frac{2sin \frac{2 \pi }{7}*cos \frac{2 \pi }{7} }{4sin \frac{ \pi }{7} } *cos \frac{3 \pi }{7} =$
$=- \frac{sin \frac{4 \pi }{7} }{4sin \frac{ \pi }{7} } *cos( \pi - \frac{ 4\pi }{7} )= \frac{sin \frac{4 \pi }{7} }{4sin \frac{ \pi }{7} } *cos \frac{ 4\pi }{7} = \\ \\ = \frac{2sin \frac{4 \pi }{7} *cos \frac{ 4\pi }{7}}{8sin \frac{ \pi }{7} } = \frac{sin \frac{8 \pi }{7} }{8sin \frac{ \pi }{7} } = \frac{sin( \pi + \frac{ \pi }{7} )}{8sin \frac{ \pi }{7} } = \\ \\ =- \frac{sin \frac{ \pi }{7} }{8sin \frac{ \pi }{7} } =- \frac{1}{8} =-0,125$

Answer 2

Для решения данного выражения воспользуемся формулой тригонометрии:

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)

Применим данную формулу несколько раз:

cos(π/7) = cos(π/4 - π/28) = cos(π/4)cos(π/28) + sin(π/4)sin(π/28) = (sqrt(2)/2)cos(π/28) + (sqrt(2)/2)sin(π/28) = sqrt(2)/2 * (cos(π/28) + sin(π/28))

cos(2π/7) = cos(π/7 + π/7) = cos(π/7)cos(π/7) - sin(π/7)sin(π/7) = cos²(π/7) - sin²(π/7)

cos(3π/7) = cos(π/7 + 2π/7) = cos(π/7)cos(2π/7) - sin(π/7)sin(2π/7)

cos(2π/7) = cos(π/4 - π/28) = cos(π/4)cos(π/28) + sin(π/4)sin(π/28) = (sqrt(2)/2)cos(π/28) + (sqrt(2)/2)sin(π/28) = sqrt(2)/2 * (cos(π/28) + sin(π/28))

Теперь мы можем заменить cos(π/7), cos(2π/7), и cos(3π/7) на выражения, полученные выше, и получить:

cos(π/7)cos(2π/7)cos(3π/7) = (sqrt(2)/2 * (cos(π/28) + sin(π/28))) * (cos²(π/7) - sin²(π/7)) * (cos(π/7)cos(2π/7) - sin(π/7)sin(2π/7))

Заметим, что cos(π/7)cos(2π/7) - sin(π/7)sin(2π/7) = cos(π/7 + 2π/7) = cos(3π/7), поэтому мы можем заменить это выражение на cos(3π/7) и далее упростить:

cos(π/7)cos(2π/7)cos(3π/7) = (sqrt(2)/2 * (cos(π/28) + sin(π/28))) * (cos²(π/7) - sin²(π/7)) * cos(3π/7)

= (sqrt(2)/2 * (cos(π/28) + sin(π/28))) * (2cos²(π/7) - 1) * cos(3π/7)

= sqrt(2)/4 * (cos(π/28) + sin(π/28)) * (2cos²(π/7) - 1) * (cos(π - 4π/7))

= sqrt(2)/4 * (cos(π/28) + sin(π/28)) * (2cos²(π