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人気のある三角関数 >

tan^2(θ)=sec^2(θ)+1

解

tan^{2} (θ) = sec^{2} (θ) + 1

解

以下の解はない : θ \in R

解答ステップ

tan^{2} (θ) = sec^{2} (θ) + 1

両辺から

sec^{2} (θ) + 1

を引く

tan^{2} (θ) - sec^{2} (θ) - 1 = 0

サイン, コサインで表わす

- 1 - sec^{2} (θ) + tan^{2} (θ)

基本的な三角関数の公式を使用する:

sec (x) = \frac{1}{cos ( x )}

= - 1 - (\frac{1}{cos ( θ )})^{2} + tan^{2} (θ)

基本的な三角関数の公式を使用する:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

簡素化

- 1 - (\frac{1}{cos ( θ )})^{2} + (\frac{sin ( θ )}{cos ( θ )})^{2} : \frac{- cos ^{2} ( θ ) - 1 + sin ^{2} ( θ )}{cos ^{2} ( θ )}

- 1 - (\frac{1}{cos ( θ )})^{2} + (\frac{sin ( θ )}{cos ( θ )})^{2}

(\frac{1}{cos ( θ )})^{2} = \frac{1}{cos ^{2} ( θ )}

(\frac{1}{cos ( θ )})^{2}

指数の規則を適用する:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

規則を適用

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

指数の規則を適用する:

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

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

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

分数を組み合わせる

- \frac{1}{cos ^{2} ( θ )} + \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )} : \frac{- 1 + sin ^{2} ( θ )}{cos ^{2} ( θ )}

規則を適用

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

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

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

元を分数に変換する:

1 = \frac{1 cos ^{2} ( θ )}{cos ^{2} ( θ )}

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

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{- 1 \cdot cos ^{2} ( θ ) - 1 + sin ^{2} ( θ )}{cos ^{2} ( θ )}

乗算:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

\frac{- 1 - cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ )} = 0

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

- 1 - cos^{2} (θ) + sin^{2} (θ) = 0

三角関数の公式を使用して書き換える

- 1 - cos^{2} (θ) + sin^{2} (θ)

2倍角の公式を使用:

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

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

= - 1 - cos (2 θ)

- 1 - cos (2 θ) = 0

1

を右側に移動します

- 1 - cos (2 θ) = 0

両辺に

1

を足す

- 1 - cos (2 θ) + 1 = 0 + 1

簡素化

- cos (2 θ) = 1

- cos (2 θ) = 1

以下で両辺を割る

- 1

- cos (2 θ) = 1

以下で両辺を割る

- 1

\frac{- cos ( 2 θ )}{- 1} = \frac{1}{- 1}

簡素化

cos (2 θ) = - 1

cos (2 θ) = - 1

以下の一般解

cos (2 θ) = - 1

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}

2 θ = π + 2 πn

2 θ = π + 2 πn

解く

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

以下で両辺を割る

2

2 θ = π + 2 πn

以下で両辺を割る

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{2 πn}{2}

簡素化

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

equationは以下で未定義のため:

\frac{π}{2} + πn

以下の解はない : θ \in R

グラフ

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