6cos80°-3√3/2sin140° = 6cos80° - 3√3/2sin(180° - 40°)

Сначала посчитаем sin(180°-40°) = sin140° = sin(180°-40°) = sin40° = √3/2

Теперь посчитаем cos80° = cos(2 40°) = cos²40 - sin²40 = ((1 + cos80°) - sin80°)/2 - sin²40 = ((1 + cos80°) - sin80°)/2 - (1 - cos40°)/2 = (1 + cos80° - sin80° - 1 + cos40°)/2 = (cos80° + cos40° - sin80°)/2 = (cos80° + cos(2 20°) - sin(2 * 40°))/2 = (cos80° + cos20° - 2sin40°cos40°)/2 = (cos80° + cos20° - 2sin40°√(1 - sin²40°))/2 = (cos80° + cos20° - 2sin40°√(1 - (√3/2)²))/2 = (cos80° + cos20° - 2sin40°√(1 - 3/4))/2 = (cos80° + cos20° - 2sin40°√(1/4))/2 = (cos80° + cos20° - sin40°)/2 = (cos80° + cos20° - √3/2)/2 = (cos80° + cos20° - √3/2)/2

Теперь заменим в исходном уравнении:

6cos80°-3√3/2sin140° = 6(cos80° + cos20° - √3/2)/2 - 3(√3/2)(√3/2)/2 = 6cos80° + 6cos20° - 3/2 - 9/4 = 3 + 6cos20° + 6cos80° - 9/4 = 3 + 6(cos20° + cos80°) - 9/4 = 3 + 6*cos100° - 9/4 = 3 - 6/2 - 9/4 = 3 - 3/2 - 9/4 = 6/4 - 3/4 - 9/4 = -6/4 = -3

Итак, выражение равно -3.