展開する 3sin^2(2x-3x^2+1)cos(2x-3x^2+1)(-6x+2)

解

展開する

3 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) (- 6 x + 2)

- 18 x sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) + 6 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1)

解答ステップ

3 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) (- 6 x + 2)

分配法則を適用する:

a (b + c) = ab + a c

3 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) (- 6 x + 2) = 3 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) (- 6 x) + 3 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) \cdot 2

= 3 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) (- 6 x) + 3 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) \cdot 2

簡素化

3 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) (- 6 x) + 3 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) \cdot 2 : - 18 x sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) + 6 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1)

= - 18 x sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1) + 6 sin^{2} (2 x - 3 x^{2} + 1) cos (2 x - 3 x^{2} + 1)