展開する sqrt((-2sin(t))^2+(2cos(t))^2+(4cos(t)-2sin(t))^2)

解

展開する

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

2 2 sin^{2} (t) + 5 cos^{2} (t) - 2 sin (2 t)

解答ステップ

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

(- 2 sin (t))^{2} = 2^{2} sin^{2} (t)

(2 cos (t))^{2} = 2^{2} cos^{2} (t)

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

2^{2} = 4

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

拡張

4 sin^{2} (t) + 4 cos^{2} (t) + (4 cos (t) - 2 sin (t))^{2} : 8 sin^{2} (t) + 20 cos^{2} (t) - 16 cos (t) sin (t)

= 8 sin^{2} (t) + 20 cos^{2} (t) - 16 cos (t) sin (t)

因数

8 sin^{2} (t) + 20 cos^{2} (t) - 16 cos (t) sin (t) : 4 (2 sin^{2} (t) + 5 cos^{2} (t) - 4 cos (t) sin (t))

= 4 (2 sin^{2} (t) + 5 cos^{2} (t) - 4 cos (t) sin (t))

累乗根の規則を適用する:

n ab = n a n b,

, 以下を想定

a \geq 0, b \geq 0

= 4 2 sin^{2} (t) + 5 cos^{2} (t) - 4 cos (t) sin (t)

4 = 2

= 2 2 sin^{2} (t) + 5 cos^{2} (t) - 4 cos (t) sin (t)

2 sin^{2} (t) + 5 cos^{2} (t) - 4 cos (t) sin (t) = 2 sin^{2} (t) + 5 cos^{2} (t) - 2 sin (2 t)

= 2 2 sin^{2} (t) + 5 cos^{2} (t) - 2 sin (2 t)