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人気のある三角関数 >

(sqrt(3))/2 =sin(arcsin(0)+c_{1})

解

\frac{3}{2} = sin (arcsin (0) + c_{1})

解

c_{1} = \frac{π}{3} + 2 πn, c_{1} = \frac{2 π}{3} + 2 πn

解答ステップ

\frac{3}{2} = sin (arcsin (0) + c_{1})

辺を交換する

sin (arcsin (0) + c_{1}) = \frac{3}{2}

arcsin (0) = 0

arcsin (0)

次の自明恒等式を使用する:

arcsin (0) = 0

arcsin (0)

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= 0

= 0

sin (0 + c_{1}) = \frac{3}{2}

以下の一般解

sin (0 + c_{1}) = \frac{3}{2}

sin (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

0 + c_{1} = \frac{π}{3} + 2 πn, 0 + c_{1} = \frac{2 π}{3} + 2 πn

0 + c_{1} = \frac{π}{3} + 2 πn, 0 + c_{1} = \frac{2 π}{3} + 2 πn

解く

0 + c_{1} = \frac{π}{3} + 2 πn : c_{1} = \frac{π}{3} + 2 πn

0 + c_{1} = \frac{π}{3} + 2 πn

0 + c_{1} = c_{1}

c_{1} = \frac{π}{3} + 2 πn

解く

0 + c_{1} = \frac{2 π}{3} + 2 πn : c_{1} = \frac{2 π}{3} + 2 πn

0 + c_{1} = \frac{2 π}{3} + 2 πn

0 + c_{1} = c_{1}

c_{1} = \frac{2 π}{3} + 2 πn

c_{1} = \frac{π}{3} + 2 πn, c_{1} = \frac{2 π}{3} + 2 πn

グラフ

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