展開する (3sin(x)+3cos(x))^2

解

展開する

(3 sin (x) + 3 cos (x))^{2}

9 sin (2 x) + 9

解答ステップ

(3 sin (x) + 3 cos (x))^{2}

完全平方式を適用する:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

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

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

簡素化

(3 sin (x))^{2} + 2 \cdot 3 sin (x) \cdot 3 cos (x) + (3 cos (x))^{2} : 9 (sin (2 x) + 1)

= 9 (sin (2 x) + 1)

分配法則を適用する:

a (b + c) = ab + a c

9 (sin (2 x) + 1) = 9 sin (2 x) + 9 \cdot 1

= 9 sin (2 x) + 9 \cdot 1

数を乗じる:

9 \cdot 1 = 9

= 9 sin (2 x) + 9