sin6x как разложить

Ответ от Алексей Стадников[гуру]
sin(6x) = sin(2·3x) = 2sin(3x)cos(3x)
sin(3x) = sin(2x + x) = sin(2x)cos(x) + cos(2x)sin(x)
cos(3x) = cos(2x + x) = cos(2x)cos(x) - sin(2x)cos(x)
sin(2x) = 2sin(x)cos(x)
cos(2x) = cos²(x) - sin²(x)
so, we get
2sin(3x)cos(3x)
=2(sin(2x)cos(x) + cos(2x)sin(x))(cos(2x)cos(x) - sin(2x)cos(x))
=2(2sin(x)cos²(x)+cos²(x)sin(x) - sin³(x))(cos³(x) -sin(x)cos²(x) - 2sin(x)cos²(x))
=2(3sin(x)cos²(x) - sin³(x))(cos³(x) - 3sin(x)cos²(x))
=2[3sin(x)cos^5(x) -9sin²(x)cos^5(x) -3sin^4(x)cos²(x) + 3sin^4(x)cos(x)]
=6sin(x)cos(x)[cos^4(x) - 3sin(x)cos^4(x) - sin³(x)cos(x) + sin³(x)]