簡約 2i(cos(x)+isin(x))(2-cos(x)+isin(x))^2

解

簡約

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

- 6 sin (x) + (8 cos (x) + 2 cos (- x) - 8 cos^{2} (x) - 8 sin^{2} (x)) i

解答ステップ

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

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

分配法則を適用する:

a (b + c) = ab + a c

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

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

簡素化

2 i (2 - cos (x) + i sin (x))^{2} cos (x) + 2 i (2 - cos (x) + i sin (x))^{2} i sin (x) : i (8 cos (x) + 2 cos (- x) - 8 cos^{2} (x) - 8 sin^{2} (x)) - 6 sin (x)

= i (8 cos (x) + 2 cos (- x) - 8 cos^{2} (x) - 8 sin^{2} (x)) - 6 sin (x)

拡張

i (8 cos (x) + 2 cos (- x) - 8 cos^{2} (x) - 8 sin^{2} (x)) : 8 cos (x) i + 2 cos (- x) i - 8 cos^{2} (x) i - 8 sin^{2} (x) i

= 8 cos (x) i + 2 cos (- x) i - 8 cos^{2} (x) i - 8 sin^{2} (x) i - 6 sin (x)

標準的な複素数形式で書き直す:

- 6 sin (x) + (8 cos (x) + 2 cos (- x) - 8 cos^{2} (x) - 8 sin^{2} (x)) i

= - 6 sin (x) + (8 cos (x) + 2 cos (- x) - 8 cos^{2} (x) - 8 sin^{2} (x)) i