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人気のある三角関数 >

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

解

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

解

a = 0.90455 \dots + πn, a = 2.23703 \dots + πn

+1

度

a = 51.82729 \dots^{\circ} + 18 0^{\circ} n, a = 128.17270 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

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

両辺から

tan^{2} (a)

を引く

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

三角関数の公式を使用して書き換える

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

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} (a) - (\frac{1}{cot ( a )})^{2}

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

(\frac{1}{cot ( a )})^{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} ( a )}

規則を適用

1^{a} = 1

1^{2} = 1

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

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

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

置換で解く

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

仮定:

cot (a) = u

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

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

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

v_{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

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

解を分離する

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

v = \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}

v = \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:

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

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

代用を戻す

u = cot (a)

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

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

a = arccot \frac{- 1 + 5}{2} + πn

a = arccot \frac{- 1 + 5}{2} + πn

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

a = arccot - \frac{- 1 + 5}{2} + πn

a = arccot - \frac{- 1 + 5}{2} + πn

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

a = arccot \frac{- 1 - 5}{2} + πn

a = arccot \frac{- 1 - 5}{2} + πn

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

a = arccot - \frac{- 1 - 5}{2} + πn

a = arccot - \frac{- 1 - 5}{2} + πn

すべての解を組み合わせる

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

equationは以下で未定義のため:

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

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

10進法形式で解を証明する

a = 0.90455 \dots + πn, a = 2.23703 \dots + πn

グラフ

人気の例

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