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人気のある三角関数 >

tan^2(θ)sec^2(θ)=3

解

tan^{2} (θ) sec^{2} (θ) = 3

解

θ = 0.85133 \dots + 2 πn, θ = 2 π - 0.85133 \dots + 2 πn, θ = 2.29026 \dots + 2 πn, θ = - 2.29026 \dots + 2 πn

+1

度

θ = 48.77764 \dots^{\circ} + 36 0^{\circ} n, θ = 311.22235 \dots^{\circ} + 36 0^{\circ} n, θ = 131.22235 \dots^{\circ} + 36 0^{\circ} n, θ = - 131.22235 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

tan^{2} (θ) sec^{2} (θ) = 3

両辺から

3

を引く

tan^{2} (θ) sec^{2} (θ) - 3 = 0

三角関数の公式を使用して書き換える

- 3 + sec^{2} (θ) tan^{2} (θ)

ピタゴラスの公式を使用する:

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

tan^{2} (x) = sec^{2} (x) - 1

= - 3 + sec^{2} (θ) (sec^{2} (θ) - 1)

- 3 + (- 1 + sec^{2} (θ)) sec^{2} (θ) = 0

置換で解く

- 3 + (- 1 + sec^{2} (θ)) sec^{2} (θ) = 0

仮定:

sec (θ) = u

- 3 + (- 1 + u^{2}) u^{2} = 0

- 3 + (- 1 + u^{2}) u^{2} = 0 : u = \frac{1 + 13}{2}, u = - \frac{1 + 13}{2}, u = \frac{1 - 13}{2}, u = - \frac{1 - 13}{2}

- 3 + (- 1 + u^{2}) u^{2} = 0

拡張

- 3 + (- 1 + u^{2}) u^{2} : - 3 - u^{2} + u^{4}

- 3 + (- 1 + u^{2}) u^{2}

= - 3 + u^{2} (- 1 + u^{2})

拡張

u^{2} (- 1 + u^{2}) : - u^{2} + u^{4}

u^{2} (- 1 + u^{2})

分配法則を適用する:

a (b + c) = ab + a c

a = u^{2}, b = - 1, c = u^{2}

= u^{2} (- 1) + u^{2} u^{2}

マイナス・プラスの規則を適用する

+ (- a) = - a

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

簡素化

- 1 \cdot u^{2} + u^{2} u^{2} : - u^{2} + u^{4}

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

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

1 \cdot u^{2}

乗算:

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

= u^{2}

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}

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

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

= - 3 - u^{2} + u^{4}

- 3 - u^{2} + u^{4} = 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} - 3 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

v^{2} - v - 3 = 0

解く

v^{2} - v - 3 = 0 : v = \frac{1 + 13}{2}, v = \frac{1 - 13}{2}

v^{2} - v - 3 = 0

解くとthe二次式

v^{2} - v - 3 = 0

二次Equationの公式:

次の場合:

a = 1, b = - 1, c = - 3

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 3) = 13

(- 1)^{2} - 4 \cdot 1 \cdot (- 3)

規則を適用

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 3

(- 1)^{2} = 1

(- 1)^{2}

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

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

= 1^{2}

規則を適用

1^{a} = 1

= 1

4 \cdot 1 \cdot 3 = 12

4 \cdot 1 \cdot 3

数を乗じる:

4 \cdot 1 \cdot 3 = 12

= 12

= 1 + 12

数を足す:

1 + 12 = 13

= 13

v_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 \cdot 1}

解を分離する

v_{1} = \frac{- ( - 1 ) + 13}{2 \cdot 1}, v_{2} = \frac{- ( - 1 ) - 13}{2 \cdot 1}

v = \frac{- ( - 1 ) + 13}{2 \cdot 1} : \frac{1 + 13}{2}

\frac{- ( - 1 ) + 13}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{1 + 13}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{1 + 13}{2}

v = \frac{- ( - 1 ) - 13}{2 \cdot 1} : \frac{1 - 13}{2}

\frac{- ( - 1 ) - 13}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{1 - 13}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{1 - 13}{2}

二次equationの解:

v = \frac{1 + 13}{2}, v = \frac{1 - 13}{2}

v = \frac{1 + 13}{2}, v = \frac{1 - 13}{2}

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = \frac{1 + 13}{2} : u = \frac{1 + 13}{2}, u = - \frac{1 + 13}{2}

u^{2} = \frac{1 + 13}{2}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1 + 13}{2}, u = - \frac{1 + 13}{2}

解く

u^{2} = \frac{1 - 13}{2} : u = \frac{1 - 13}{2}, u = - \frac{1 - 13}{2}

u^{2} = \frac{1 - 13}{2}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1 - 13}{2}, u = - \frac{1 - 13}{2}

解答は

u = \frac{1 + 13}{2}, u = - \frac{1 + 13}{2}, u = \frac{1 - 13}{2}, u = - \frac{1 - 13}{2}

代用を戻す

u = sec (θ)

sec (θ) = \frac{1 + 13}{2}, sec (θ) = - \frac{1 + 13}{2}, sec (θ) = \frac{1 - 13}{2}, sec (θ) = - \frac{1 - 13}{2}

sec (θ) = \frac{1 + 13}{2}, sec (θ) = - \frac{1 + 13}{2}, sec (θ) = \frac{1 - 13}{2}, sec (θ) = - \frac{1 - 13}{2}

sec (θ) = \frac{1 + 13}{2} : θ = arcsec \frac{1 + 13}{2} + 2 πn, θ = 2 π - arcsec \frac{1 + 13}{2} + 2 πn

sec (θ) = \frac{1 + 13}{2}

三角関数の逆数プロパティを適用する

sec (θ) = \frac{1 + 13}{2}

以下の一般解

sec (θ) = \frac{1 + 13}{2}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

θ = arcsec \frac{1 + 13}{2} + 2 πn, θ = 2 π - arcsec \frac{1 + 13}{2} + 2 πn

θ = arcsec \frac{1 + 13}{2} + 2 πn, θ = 2 π - arcsec \frac{1 + 13}{2} + 2 πn

sec (θ) = - \frac{1 + 13}{2} : θ = arcsec - \frac{1 + 13}{2} + 2 πn, θ = - arcsec - \frac{1 + 13}{2} + 2 πn

sec (θ) = - \frac{1 + 13}{2}

三角関数の逆数プロパティを適用する

sec (θ) = - \frac{1 + 13}{2}

以下の一般解

sec (θ) = - \frac{1 + 13}{2}

sec (x) = - a \Rightarrow x = arcsec (- a) + 2 πn, x = - arcsec (- a) + 2 πn

θ = arcsec - \frac{1 + 13}{2} + 2 πn, θ = - arcsec - \frac{1 + 13}{2} + 2 πn

θ = arcsec - \frac{1 + 13}{2} + 2 πn, θ = - arcsec - \frac{1 + 13}{2} + 2 πn

sec (θ) = \frac{1 - 13}{2} : θ = arcsec \frac{1 - 13}{2} + 2 πn, θ = - arcsec \frac{1 - 13}{2} + 2 πn

sec (θ) = \frac{1 - 13}{2}

三角関数の逆数プロパティを適用する

sec (θ) = \frac{1 - 13}{2}

以下の一般解

sec (θ) = \frac{1 - 13}{2}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = - arcsec (a) + 2 πn

θ = arcsec \frac{1 - 13}{2} + 2 πn, θ = - arcsec \frac{1 - 13}{2} + 2 πn

θ = arcsec \frac{1 - 13}{2} + 2 πn, θ = - arcsec \frac{1 - 13}{2} + 2 πn

sec (θ) = - \frac{1 - 13}{2} : θ = arcsec - \frac{1 - 13}{2} + 2 πn, θ = - arcsec - \frac{1 - 13}{2} + 2 πn

sec (θ) = - \frac{1 - 13}{2}

三角関数の逆数プロパティを適用する

sec (θ) = - \frac{1 - 13}{2}

以下の一般解

sec (θ) = - \frac{1 - 13}{2}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = - arcsec (a) + 2 πn

θ = arcsec - \frac{1 - 13}{2} + 2 πn, θ = - arcsec - \frac{1 - 13}{2} + 2 πn

θ = arcsec - \frac{1 - 13}{2} + 2 πn, θ = - arcsec - \frac{1 - 13}{2} + 2 πn

すべての解を組み合わせる

θ = arcsec \frac{1 + 13}{2} + 2 πn, θ = 2 π - arcsec \frac{1 + 13}{2} + 2 πn, θ = arcsec - \frac{1 + 13}{2} + 2 πn, θ = - arcsec - \frac{1 + 13}{2} + 2 πn, θ = arcsec \frac{1 - 13}{2} + 2 πn, θ = - arcsec \frac{1 - 13}{2} + 2 πn, θ = arcsec - \frac{1 - 13}{2} + 2 πn, θ = - arcsec - \frac{1 - 13}{2} + 2 πn

equationは以下で未定義のため:

arcsec \frac{1 - 13}{2} + 2 πn, - arcsec \frac{1 - 13}{2} + 2 πn, arcsec - \frac{1 - 13}{2} + 2 πn, - arcsec - \frac{1 - 13}{2} + 2 πn

θ = arcsec \frac{1 + 13}{2} + 2 πn, θ = 2 π - arcsec \frac{1 + 13}{2} + 2 πn, θ = arcsec - \frac{1 + 13}{2} + 2 πn, θ = - arcsec - \frac{1 + 13}{2} + 2 πn

10進法形式で解を証明する

θ = 0.85133 \dots + 2 πn, θ = 2 π - 0.85133 \dots + 2 πn, θ = 2.29026 \dots + 2 πn, θ = - 2.29026 \dots + 2 πn

グラフ

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