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人気のある三角関数 >

cot(b)=(tan(b)cot(b))/(csc(b))

解

cot (b) = \frac{tan ( b ) cot ( b )}{csc ( b )}

解

b = 0.90455 \dots + 2 πn, b = 2 π - 0.90455 \dots + 2 πn

+1

度

b = 51.82729 \dots^{\circ} + 36 0^{\circ} n, b = 308.17270 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

cot (b) = \frac{tan ( b ) cot ( b )}{csc ( b )}

両辺から

\frac{tan ( b ) cot ( b )}{csc ( b )}

を引く

cot (b) - \frac{tan ( b ) cot ( b )}{csc ( b )} = 0

簡素化

cot (b) - \frac{tan ( b ) cot ( b )}{csc ( b )} : \frac{cot ( b ) csc ( b ) - tan ( b ) cot ( b )}{csc ( b )}

cot (b) - \frac{tan ( b ) cot ( b )}{csc ( b )}

元を分数に変換する:

cot (b) = \frac{cot ( b ) csc ( b )}{csc ( b )}

= \frac{cot ( b ) csc ( b )}{csc ( b )} - \frac{tan ( b ) cot ( b )}{csc ( b )}

分母が等しいので, 分数を組み合わせる:

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

= \frac{cot ( b ) csc ( b ) - tan ( b ) cot ( b )}{csc ( b )}

\frac{cot ( b ) csc ( b ) - tan ( b ) cot ( b )}{csc ( b )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cot (b) csc (b) - tan (b) cot (b) = 0

サイン, コサインで表わす

\frac{cos ( b )}{sin ( b )} \cdot \frac{1}{sin ( b )} - \frac{sin ( b )}{cos ( b )} \cdot \frac{cos ( b )}{sin ( b )} = 0

簡素化

\frac{cos ( b )}{sin ( b )} \cdot \frac{1}{sin ( b )} - \frac{sin ( b )}{cos ( b )} \cdot \frac{cos ( b )}{sin ( b )} : \frac{cos ( b ) - sin ^{2} ( b )}{sin ^{2} ( b )}

\frac{cos ( b )}{sin ( b )} \cdot \frac{1}{sin ( b )} - \frac{sin ( b )}{cos ( b )} \cdot \frac{cos ( b )}{sin ( b )}

\frac{cos ( b )}{sin ( b )} \cdot \frac{1}{sin ( b )} = \frac{cos ( b )}{sin ^{2} ( b )}

\frac{cos ( b )}{sin ( b )} \cdot \frac{1}{sin ( b )}

分数を乗じる:

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

= \frac{cos ( b ) \cdot 1}{sin ( b ) sin ( b )}

乗算:

cos (b) \cdot 1 = cos (b)

= \frac{cos ( b )}{sin ( b ) sin ( b )}

sin (b) sin (b) = sin^{2} (b)

sin (b) sin (b)

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

sin (b) sin (b) = sin^{1 + 1} (b)

= sin^{1 + 1} (b)

数を足す:

1 + 1 = 2

= sin^{2} (b)

= \frac{cos ( b )}{sin ^{2} ( b )}

\frac{sin ( b )}{cos ( b )} \cdot \frac{cos ( b )}{sin ( b )} = 1

\frac{sin ( b )}{cos ( b )} \cdot \frac{cos ( b )}{sin ( b )}

分数を乗じる:

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

= \frac{sin ( b ) cos ( b )}{cos ( b ) sin ( b )}

共通因数を約分する:

sin (b)

= \frac{cos ( b )}{cos ( b )}

共通因数を約分する:

cos (b)

= 1

= \frac{cos ( b )}{sin ^{2} ( b )} - 1

元を分数に変換する:

1 = \frac{1 sin ^{2} ( b )}{sin ^{2} ( b )}

= \frac{cos ( b )}{sin ^{2} ( b )} - \frac{1 \cdot sin ^{2} ( b )}{sin ^{2} ( b )}

分母が等しいので, 分数を組み合わせる:

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

= \frac{cos ( b ) - 1 \cdot sin ^{2} ( b )}{sin ^{2} ( b )}

乗算:

1 \cdot sin^{2} (b) = sin^{2} (b)

= \frac{cos ( b ) - sin ^{2} ( b )}{sin ^{2} ( b )}

\frac{cos ( b ) - sin ^{2} ( b )}{sin ^{2} ( b )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos (b) - sin^{2} (b) = 0

両辺に

sin^{2} (b)

を足す

cos (b) = sin^{2} (b)

両辺を2乗する

cos^{2} (b) = (sin^{2} (b))^{2}

両辺から

(sin^{2} (b))^{2}

を引く

cos^{2} (b) - sin^{4} (b) = 0

因数

cos^{2} (b) - sin^{4} (b) : (cos (b) + sin^{2} (b)) (cos (b) - sin^{2} (b))

cos^{2} (b) - sin^{4} (b)

指数の規則を適用する:

a^{b c} = (a^{b})^{c}

sin^{4} (b) = (sin^{2} (b))^{2}

= cos^{2} (b) - (sin^{2} (b))^{2}

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

cos^{2} (b) - (sin^{2} (b))^{2} = (cos (b) + sin^{2} (b)) (cos (b) - sin^{2} (b))

= (cos (b) + sin^{2} (b)) (cos (b) - sin^{2} (b))

(cos (b) + sin^{2} (b)) (cos (b) - sin^{2} (b)) = 0

各部分を別個に解く

cos (b) + sin^{2} (b) = 0 or cos (b) - sin^{2} (b) = 0

cos (b) + sin^{2} (b) = 0 : b = arccos (- \frac{- 1 + 5}{2}) + 2 πn, b = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (b) + sin^{2} (b) = 0

三角関数の公式を使用して書き換える

cos (b) + sin^{2} (b)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= cos (b) + 1 - cos^{2} (b)

1 + cos (b) - cos^{2} (b) = 0

置換で解く

1 + cos (b) - cos^{2} (b) = 0

仮定:

cos (b) = u

1 + u - u^{2} = 0

1 + u - u^{2} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

1 + u - u^{2} = 0

標準的な形式で書く

a x^{2} + b x + c = 0

- u^{2} + u + 1 = 0

解くとthe二次式

- u^{2} + u + 1 = 0

二次Equationの公式:

次の場合:

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

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 1 = 5

1^{2} - 4 (- 1) \cdot 1

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

規則を適用

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数を足す:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 1 )}

解を分離する

u_{1} = \frac{- 1 + 5}{2 ( - 1 )}, u_{2} = \frac{- 1 - 5}{2 ( - 1 )}

u = \frac{- 1 + 5}{2 ( - 1 )} : - \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 ( - 1 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

分数の規則を適用する:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 ( - 1 )} : \frac{1 + 5}{2}

\frac{- 1 - 5}{2 ( - 1 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 1 - 5}{- 2}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

二次equationの解:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

代用を戻す

u = cos (b)

cos (b) = - \frac{- 1 + 5}{2}, cos (b) = \frac{1 + 5}{2}

cos (b) = - \frac{- 1 + 5}{2}, cos (b) = \frac{1 + 5}{2}

cos (b) = - \frac{- 1 + 5}{2} : b = arccos (- \frac{- 1 + 5}{2}) + 2 πn, b = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (b) = - \frac{- 1 + 5}{2}

三角関数の逆数プロパティを適用する

cos (b) = - \frac{- 1 + 5}{2}

以下の一般解

cos (b) = - \frac{- 1 + 5}{2}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

b = arccos (- \frac{- 1 + 5}{2}) + 2 πn, b = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

b = arccos (- \frac{- 1 + 5}{2}) + 2 πn, b = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (b) = \frac{1 + 5}{2} :

解なし

cos (b) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

b = arccos (- \frac{- 1 + 5}{2}) + 2 πn, b = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (b) - sin^{2} (b) = 0 : b = arccos (\frac{- 1 + 5}{2}) + 2 πn, b = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (b) - sin^{2} (b) = 0

三角関数の公式を使用して書き換える

cos (b) - sin^{2} (b)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= cos (b) - (1 - cos^{2} (b))

- (1 - cos^{2} (b)) : - 1 + cos^{2} (b)

- (1 - cos^{2} (b))

括弧を分配する

= - (1) - (- cos^{2} (b))

マイナス・プラスの規則を適用する

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (b)

= cos (b) - 1 + cos^{2} (b)

- 1 + cos (b) + cos^{2} (b) = 0

置換で解く

- 1 + cos (b) + cos^{2} (b) = 0

仮定:

cos (b) = u

- 1 + u + u^{2} = 0

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

- 1 + u + u^{2} = 0

標準的な形式で書く

a x^{2} + b x + c = 0

u^{2} + u - 1 = 0

解くとthe二次式

u^{2} + u - 1 = 0

二次Equationの公式:

次の場合:

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

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

1^{2} - 4 \cdot 1 \cdot (- 1)

規則を適用

1^{a} = 1

1^{2} = 1

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

規則を適用

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数を足す:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

解を分離する

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

二次equationの解:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

代用を戻す

u = cos (b)

cos (b) = \frac{- 1 + 5}{2}, cos (b) = \frac{- 1 - 5}{2}

cos (b) = \frac{- 1 + 5}{2}, cos (b) = \frac{- 1 - 5}{2}

cos (b) = \frac{- 1 + 5}{2} : b = arccos (\frac{- 1 + 5}{2}) + 2 πn, b = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (b) = \frac{- 1 + 5}{2}

三角関数の逆数プロパティを適用する

cos (b) = \frac{- 1 + 5}{2}

以下の一般解

cos (b) = \frac{- 1 + 5}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

b = arccos (\frac{- 1 + 5}{2}) + 2 πn, b = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

b = arccos (\frac{- 1 + 5}{2}) + 2 πn, b = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (b) = \frac{- 1 - 5}{2} :

解なし

cos (b) = \frac{- 1 - 5}{2}

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

b = arccos (\frac{- 1 + 5}{2}) + 2 πn, b = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

すべての解を組み合わせる

b = arccos (- \frac{- 1 + 5}{2}) + 2 πn, b = - arccos (- \frac{- 1 + 5}{2}) + 2 πn, b = arccos (\frac{- 1 + 5}{2}) + 2 πn, b = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

元のequationに当てはめて解を検算する

cot (b) = \frac{tan ( b ) cot ( b )}{csc ( b )}

に当てはめて解を確認する

equationに一致しないものを削除する。

解答を確認する

arccos (- \frac{- 1 + 5}{2}) + 2 πn :

偽

arccos (- \frac{- 1 + 5}{2}) + 2 πn

挿入

n = 1

arccos (- \frac{- 1 + 5}{2}) + 2 π 1

cot (b) = \frac{tan ( b ) cot ( b )}{csc ( b )}

の挿入向け

b = arccos (- \frac{- 1 + 5}{2}) + 2 π 1

cot (arccos (- \frac{- 1 + 5}{2}) + 2 π 1) = \frac{tan ( arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 ) cot ( arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )}{csc ( arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )}

改良

- 0.78615 \dots = 0.78615 \dots

\Rightarrow 偽

解答を確認する

- arccos (- \frac{- 1 + 5}{2}) + 2 πn :

偽

- arccos (- \frac{- 1 + 5}{2}) + 2 πn

挿入

n = 1

- arccos (- \frac{- 1 + 5}{2}) + 2 π 1

cot (b) = \frac{tan ( b ) cot ( b )}{csc ( b )}

の挿入向け

b = - arccos (- \frac{- 1 + 5}{2}) + 2 π 1

cot (- arccos (- \frac{- 1 + 5}{2}) + 2 π 1) = \frac{tan ( - arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 ) cot ( - arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )}{csc ( - arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )}

改良

0.78615 \dots = - 0.78615 \dots

\Rightarrow 偽

解答を確認する

arccos (\frac{- 1 + 5}{2}) + 2 πn :

真

arccos (\frac{- 1 + 5}{2}) + 2 πn

挿入

n = 1

arccos (\frac{- 1 + 5}{2}) + 2 π 1

cot (b) = \frac{tan ( b ) cot ( b )}{csc ( b )}

の挿入向け

b = arccos (\frac{- 1 + 5}{2}) + 2 π 1

cot (arccos (\frac{- 1 + 5}{2}) + 2 π 1) = \frac{tan ( arccos ( \frac{- 1 + 5}{2} ) + 2 π 1 ) cot ( arccos ( \frac{- 1 + 5}{2} ) + 2 π 1 )}{csc ( arccos ( \frac{- 1 + 5}{2} ) + 2 π 1 )}

改良

0.78615 \dots = 0.78615 \dots

\Rightarrow 真

解答を確認する

2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn :

真

2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

挿入

n = 1

2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1

cot (b) = \frac{tan ( b ) cot ( b )}{csc ( b )}

の挿入向け

b = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1

cot (2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1) = \frac{tan ( 2 π - arccos ( \frac{- 1 + 5}{2} ) + 2 π 1 ) cot ( 2 π - arccos ( \frac{- 1 + 5}{2} ) + 2 π 1 )}{csc ( 2 π - arccos ( \frac{- 1 + 5}{2} ) + 2 π 1 )}

改良

- 0.78615 \dots = - 0.78615 \dots

\Rightarrow 真

b = arccos (\frac{- 1 + 5}{2}) + 2 πn, b = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

10進法形式で解を証明する

b = 0.90455 \dots + 2 πn, b = 2 π - 0.90455 \dots + 2 πn

グラフ

人気の例

cos(x)+2sin(x)=-1

cos (x) + 2 sin (x) = - 1

csc (x) = - 7

cos(4t)+sin(4t)=0

cos (4 t) + sin (4 t) = 0

cos (x - 3 0^{\circ}) = 0.22

sin(4x-22)=cos(6x-13)

sin (4 x - 22) = cos (6 x - 13)