cot(675)

解

cot (67 5^{\circ})

- 1

解答ステップ

cot (67 5^{\circ})

cot (67 5^{\circ}) = cot (13 5^{\circ})

cot (67 5^{\circ})

67 5^{\circ}

を書き換え

18 0^{\circ} \cdot 3 + 13 5^{\circ}

= cot (18 0^{\circ} 3 + 13 5^{\circ})

以下の周期性を適用する:

cot

cot (x + 18 0^{\circ} \cdot k) = cot (x)

cot (18 0^{\circ} \cdot 3 + 13 5^{\circ}) = cot (13 5^{\circ})

= cot (13 5^{\circ})

= cot (13 5^{\circ})

次の自明恒等式を使用する:

cot (13 5^{\circ}) = - 1

cot (13 5^{\circ})

cot (x)

18 0^{\circ} n

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

= - 1

= - 1