展開する-(2cos(x)+1)(2cos(x)-1)sin^2(x)

解

展開する

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

- 4 sin^{2} (x) cos^{2} (x) + sin^{2} (x)

解答ステップ

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

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

2つの正方形の差

(2 cos (x) + 1) (2 cos (x) - 1) を適用します : 4 cos^{2} (x) - 1

= - sin^{2} (x) (4 cos^{2} (x) - 1)

分配法則を適用する:

a (b - c) = ab - a c

- sin^{2} (x) (4 cos^{2} (x) - 1) = - sin^{2} (x) \cdot 4 cos^{2} (x) - (- sin^{2} (x)) \cdot 1

= - sin^{2} (x) \cdot 4 cos^{2} (x) - (- sin^{2} (x)) \cdot 1

簡素化

- sin^{2} (x) \cdot 4 cos^{2} (x) - (- sin^{2} (x)) \cdot 1 : - 4 sin^{2} (x) cos^{2} (x) + sin^{2} (x)

= - 4 sin^{2} (x) cos^{2} (x) + sin^{2} (x)