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人気のある三角関数 >

1+tan^2(x)=cot^2(x)

解

1 + tan^{2} (x) = cot^{2} (x)

解

x = 0.66623 \dots + πn, x = 2.47535 \dots + πn

+1

度

x = 38.17270 \dots^{\circ} + 18 0^{\circ} n, x = 141.82729 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

1 + tan^{2} (x) = cot^{2} (x)

両辺から

cot^{2} (x)

を引く

1 + tan^{2} (x) - cot^{2} (x) = 0

三角関数の公式を使用して書き換える

1 - cot^{2} (x) + tan^{2} (x)

基本的な三角関数の公式を使用する:

tan (x) = \frac{1}{cot ( x )}

= 1 - cot^{2} (x) + (\frac{1}{cot ( x )})^{2}

(\frac{1}{cot ( x )})^{2} = \frac{1}{cot ^{2} ( x )}

(\frac{1}{cot ( x )})^{2}

指数の規則を適用する:

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

= \frac{1 ^{2}}{cot ^{2} ( x )}

規則を適用

1^{a} = 1

1^{2} = 1

= \frac{1}{cot ^{2} ( x )}

= 1 - cot^{2} (x) + \frac{1}{cot ^{2} ( x )}

1 - cot^{2} (x) + \frac{1}{cot ^{2} ( x )} = 0

置換で解く

1 - cot^{2} (x) + \frac{1}{cot ^{2} ( x )} = 0

仮定:

cot (x) = u

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

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

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

以下で両辺を乗じる:

u^{2}

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

以下で両辺を乗じる:

u^{2}

1 \cdot u^{2} - u^{2} u^{2} + \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

簡素化

1 \cdot u^{2} - u^{2} u^{2} + \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

簡素化

1 \cdot u^{2} : u^{2}

1 \cdot u^{2}

乗算:

1 \cdot u^{2} = u^{2}

= u^{2}

簡素化

- u^{2} u^{2} : - u^{4}

- u^{2} u^{2}

指数の規則を適用する:

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

u^{2} u^{2} = u^{2 + 2}

= - u^{2 + 2}

数を足す:

2 + 2 = 4

= - u^{4}

簡素化

\frac{1}{u ^{2}} u^{2} : 1

\frac{1}{u ^{2}} u^{2}

分数を乗じる:

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

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

共通因数を約分する:

u^{2}

= 1

簡素化

0 \cdot u^{2} : 0

0 \cdot u^{2}

規則を適用

0 \cdot a = 0

= 0

u^{2} - u^{4} + 1 = 0

u^{2} - u^{4} + 1 = 0

u^{2} - u^{4} + 1 = 0

解く

u^{2} - u^{4} + 1 = 0 : u = i \frac{- 1 + 5}{2}, u = - i \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

u^{2} - u^{4} + 1 = 0

標準的な形式で書く

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{4} + u^{2} + 1 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

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

解く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

v_{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

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

解を分離する

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

v = \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}

v = \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:

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

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

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

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

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

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

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

簡素化

- \frac{- 1 + 5}{2} : i \frac{- 1 + 5}{2}

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

累乗根の規則を適用する:

- a = - 1 a

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

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

虚数の規則を適用する:

- 1 = i

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

簡素化

- - \frac{- 1 + 5}{2} : - i \frac{- 1 + 5}{2}

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

簡素化

- \frac{- 1 + 5}{2} : i \frac{5 - 1}{2}

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

累乗根の規則を適用する:

- a = - 1 a

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

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

虚数の規則を適用する:

- 1 = i

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

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

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

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

解く

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

u^{2} = \frac{1 + 5}{2}

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

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

解答は

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

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

解を検算する

未定義の (特異) 点を求める:

u = 0

1 - u^{2} + \frac{1}{u ^{2}}

の分母をゼロに比較する

解く

u^{2} = 0 : u = 0

u^{2} = 0

規則を適用

x^{n} = 0 \Rightarrow x = 0

u = 0

以下の点は定義されていない

u = 0

未定義のポイントを解に組み合わせる:

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

代用を戻す

u = cot (x)

cot (x) = i \frac{- 1 + 5}{2}, cot (x) = - i \frac{- 1 + 5}{2}, cot (x) = \frac{1 + 5}{2}, cot (x) = - \frac{1 + 5}{2}

cot (x) = i \frac{- 1 + 5}{2}, cot (x) = - i \frac{- 1 + 5}{2}, cot (x) = \frac{1 + 5}{2}, cot (x) = - \frac{1 + 5}{2}

cot (x) = i \frac{- 1 + 5}{2} :

解なし

cot (x) = i \frac{- 1 + 5}{2}

解なし

cot (x) = - i \frac{- 1 + 5}{2} :

解なし

cot (x) = - i \frac{- 1 + 5}{2}

解なし

cot (x) = \frac{1 + 5}{2} : x = arccot \frac{1 + 5}{2} + πn

cot (x) = \frac{1 + 5}{2}

三角関数の逆数プロパティを適用する

cot (x) = \frac{1 + 5}{2}

以下の一般解

cot (x) = \frac{1 + 5}{2}

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot \frac{1 + 5}{2} + πn

x = arccot \frac{1 + 5}{2} + πn

cot (x) = - \frac{1 + 5}{2} : x = arccot - \frac{1 + 5}{2} + πn

cot (x) = - \frac{1 + 5}{2}

三角関数の逆数プロパティを適用する

cot (x) = - \frac{1 + 5}{2}

以下の一般解

cot (x) = - \frac{1 + 5}{2}

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot - \frac{1 + 5}{2} + πn

x = arccot - \frac{1 + 5}{2} + πn

すべての解を組み合わせる

x = arccot \frac{1 + 5}{2} + πn, x = arccot - \frac{1 + 5}{2} + πn

10進法形式で解を証明する

x = 0.66623 \dots + πn, x = 2.47535 \dots + πn

グラフ

人気の例

2-2sin^2(x)=-3cos(x)+2

2 - 2 sin^{2} (x) = - 3 cos (x) + 2

4 tan (θ) + 7 = 0

arctan(x)+arctan(3x)+arctan(7x)=180

arctan (x) + arctan (3 x) + arctan (7 x) = 18 0^{\circ}

2 cos (a) = 1

16 sin^{2} (θ) = 12