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人気のある三角関数 >

(sqrt(2)(cos(pi/2)+isin(pi/2)))/(1/2 (cos(pi/4)+isin(pi/4)))

解

\frac{2 ( cos ( \frac{π}{2} ) + i sin ( \frac{π}{2} ) )}{\frac{1}{2} ( cos ( \frac{π}{4} ) + i sin ( \frac{π}{4} ) )}

解

2 + 2 i

解答ステップ

\frac{2 ( cos ( \frac{π}{2} ) + i sin ( \frac{π}{2} ) )}{\frac{1}{2} ( cos ( \frac{π}{4} ) + i sin ( \frac{π}{4} ) )}

次の自明恒等式を使用する:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

次の自明恒等式を使用する:

sin (\frac{π}{2}) = 1

sin (\frac{π}{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}

= 1

次の自明恒等式を使用する:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

次の自明恒等式を使用する:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

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}

= \frac{2}{2}

= \frac{2 ( 0 + i 1 )}{\frac{1}{2} ( \frac{2}{2} + i \frac{2}{2} )}

簡素化

\frac{2 ( 0 + i 1 )}{\frac{1}{2} ( \frac{2}{2} + i \frac{2}{2} )} : 2 + 2 i

\frac{2 ( 0 + i 1 )}{\frac{1}{2} ( \frac{2}{2} + i \frac{2}{2} )}

2 (0 + i 1) = 2 i

2 (0 + i 1)

乗算:

i 1 = i

= 2 (0 + i)

0 + i = i

= 2 i

= \frac{2 i}{\frac{1}{2} ( i \frac{2}{2} + \frac{2}{2} )}

乗じる

\frac{1}{2} (\frac{2}{2} + i \frac{2}{2}) : \frac{1 + i}{2 2}

\frac{1}{2} (\frac{2}{2} + i \frac{2}{2})

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( \frac{2}{2} + i \frac{2}{2} )}{2}

1 \cdot (\frac{2}{2} + i \frac{2}{2}) = \frac{2}{2} + i \frac{2}{2}

1 \cdot (\frac{2}{2} + i \frac{2}{2})

乗算:

1 \cdot (\frac{2}{2} + i \frac{2}{2}) = (\frac{2}{2} + i \frac{2}{2})

= (\frac{2}{2} + i \frac{2}{2})

括弧を削除する:

(a) = a

= \frac{2}{2} + i \frac{2}{2}

= \frac{\frac{2}{2} + i \frac{2}{2}}{2}

\frac{2}{2} = \frac{1}{2}

\frac{2}{2}

キャンセル

\frac{2}{2} : \frac{1}{2}

\frac{2}{2}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}}}{2}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{1}} = \frac{1}{2 ^{1 - \frac{1}{2}}}

= \frac{1}{2 ^{1 - \frac{1}{2}}}

数を引く:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{1}{2 ^{\frac{1}{2}}}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

2^{\frac{1}{2}} = 2

= \frac{1}{2}

= \frac{1}{2}

i \frac{2}{2} = \frac{i}{2}

i \frac{2}{2}

\frac{2}{2} = \frac{1}{2}

\frac{2}{2}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}}}{2}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{1}} = \frac{1}{2 ^{1 - \frac{1}{2}}}

= \frac{1}{2 ^{1 - \frac{1}{2}}}

数を引く:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{1}{2 ^{\frac{1}{2}}}

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

2^{\frac{1}{2}} = 2

= \frac{1}{2}

= i \frac{1}{2}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2}

乗算:

1 i = i

= \frac{i}{2}

= \frac{\frac{1}{2} + \frac{i}{2}}{2}

分数を組み合わせる

\frac{1}{2} + \frac{i}{2} : \frac{1 + i}{2}

規則を適用

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 + i}{2}

= \frac{\frac{1 + i}{2}}{2}

分数の規則を適用する:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{1 + i}{2 \cdot 2}

= \frac{2 i}{\frac{1 + i}{2 2}}

分数の規則を適用する:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{2 i 2 \cdot 2}{1 + i}

2 i 2 \cdot 2 = 4 i

2 i 2 \cdot 2

累乗根の規則を適用する:

a a = a

22 = 2

= 2 \cdot 2 i

数を乗じる:

2 \cdot 2 = 4

= 4 i

= \frac{4 i}{1 + i}

有理化する

\frac{4 i}{1 + i} : 2 (1 + i)

\frac{4 i}{1 + i}

共役で乗じる

\frac{1 - i}{1 - i}

= \frac{4 i ( 1 - i )}{( 1 + i ) ( 1 - i )}

簡素化

4 i (1 - i) : 4 + 4 i

4 i (1 - i)

複素数の算術規則を適用する:

(ai) (b + c i) = - a c + abi

a = 4, b = 1, c = - 1

= - 4 (- 1) + 4 \cdot 1 i

- 4 (- 1) = 4

- 4 (- 1)

規則を適用

- (- a) = a

= 4 \cdot 1

数を乗じる:

4 \cdot 1 = 4

= 4

4 \cdot 1 = 4

4 \cdot 1

数を乗じる:

4 \cdot 1 = 4

= 4

= 4 + 4 i

(1 + i) (1 - i) = 2

(1 + i) (1 - i)

複素数の算術規則を適用する:

(a + bi) (a - bi) = a^{2} + b^{2}

a = 1, b = 1

= 1^{2} + 1^{2}

1^{2} + 1^{2} = 2

1^{2} + 1^{2}

規則を適用

1^{a} = 1

1^{2} = 1

= 1 + 1

数を足す:

1 + 1 = 2

= 2

= 2

= \frac{4 + 4 i}{2}

因数

4 + 4 i : 4 (1 + i)

4 + 4 i

書き換え

= 4 \cdot 1 + 4 i

共通項をくくり出す

4

= 4 (1 + i)

= \frac{4 ( 1 + i )}{2}

数を割る:

\frac{4}{2} = 2

= 2 (1 + i)

= 2 (1 + i)

標準的な複素数形式で

2 (1 + i)

を書き換える:

2 + 2 i

2 (1 + i)

分配法則を適用する:

a (b + c) = ab + a c

a = 2, b = 1, c = i

= 2 \cdot 1 + 2 i

数を乗じる:

2 \cdot 1 = 2

= 2 + 2 i

= 2 + 2 i

= 2 + 2 i

人気の例

cos^{2} (\frac{π}{20})

- cos (15 0^{\circ})

cos (5. 4^{\circ})

2 \cdot sin (1 0^{\circ})

(sec(32)+csc(58))/(3csc(58)+sec(32))

\frac{sec ( 3 2 ^{\circ} ) + csc ( 5 8 ^{\circ} )}{3 csc ( 5 8 ^{\circ} ) + sec ( 3 2 ^{\circ} )}