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人気のある三角関数 >

sin^2(x)=1-tan^2(x),0<= x<= pi/2

解

sin^{2} (x) = 1 - tan^{2} (x), 0 \leq x \leq \frac{π}{2}

解

x = 0.66623 \dots

+1

度

x = 38.17270 \dots^{\circ}

解答ステップ

sin^{2} (x) = 1 - tan^{2} (x), 0 \leq x \leq \frac{π}{2}

両辺から

1 - tan^{2} (x)

を引く

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

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

因数

- cos^{2} (x) + tan^{2} (x) : (tan (x) + cos (x)) (tan (x) - cos (x))

- cos^{2} (x) + tan^{2} (x)

2乗の差の公式を適用する:

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

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

= (tan (x) + cos (x)) (tan (x) - cos (x))

(tan (x) + cos (x)) (tan (x) - cos (x)) = 0

各部分を別個に解く

tan (x) + cos (x) = 0 or tan (x) - cos (x) = 0

tan (x) + cos (x) = 0, 0 \leq x \leq \frac{π}{2} :

解なし

tan (x) + cos (x) = 0, 0 \leq x \leq \frac{π}{2}

サイン, コサインで表わす

cos (x) + tan (x)

基本的な三角関数の公式を使用する:

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

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

簡素化

cos (x) + \frac{sin ( x )}{cos ( x )} : \frac{cos ^{2} ( x ) + sin ( x )}{cos ( x )}

cos (x) + \frac{sin ( x )}{cos ( x )}

元を分数に変換する:

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

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

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 ( x ) cos ( x ) + sin ( x )}{cos ( x )}

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

cos (x) cos (x) + sin (x)

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

cos (x) cos (x)

指数の規則を適用する:

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

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

= cos^{1 + 1} (x)

数を足す:

1 + 1 = 2

= cos^{2} (x)

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

= \frac{cos ^{2} ( x ) + sin ( x )}{cos ( x )}

= \frac{cos ^{2} ( x ) + sin ( x )}{cos ( x )}

\frac{cos ^{2} ( x ) + sin ( x )}{cos ( x )} = 0

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

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

置換で解く

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

仮定:

sin (x) = 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 = sin (x)

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

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

sin (x) = - \frac{- 1 + 5}{2}, 0 \leq x \leq \frac{π}{2} :

解なし

sin (x) = - \frac{- 1 + 5}{2}, 0 \leq x \leq \frac{π}{2}

三角関数の逆数プロパティを適用する

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

以下の一般解

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

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

範囲の解答

0 \leq x \leq \frac{π}{2}

解なし

sin (x) = \frac{1 + 5}{2}, 0 \leq x \leq \frac{π}{2} :

解なし

sin (x) = \frac{1 + 5}{2}, 0 \leq x \leq \frac{π}{2}

- 1 \leq sin (x) \leq 1

解なし

すべての解を組み合わせる

解なし

tan (x) - cos (x) = 0, 0 \leq x \leq \frac{π}{2} : x = arcsin (\frac{5 - 1}{2})

tan (x) - cos (x) = 0, 0 \leq x \leq \frac{π}{2}

サイン, コサインで表わす

- cos (x) + tan (x)

基本的な三角関数の公式を使用する:

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

= - cos (x) + \frac{sin ( x )}{cos ( x )}

簡素化

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

- cos (x) + \frac{sin ( x )}{cos ( x )}

元を分数に変換する:

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

= - \frac{cos ( x ) cos ( x )}{cos ( x )} + \frac{sin ( x )}{cos ( x )}

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 ( x ) cos ( x ) + sin ( x )}{cos ( x )}

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

- cos (x) cos (x) + sin (x)

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

cos (x) cos (x)

指数の規則を適用する:

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

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

= cos^{1 + 1} (x)

数を足す:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

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

括弧を分配する

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

マイナス・プラスの規則を適用する

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

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

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

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

置換で解く

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

仮定:

sin (x) = 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 = sin (x)

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

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

sin (x) = \frac{- 1 + 5}{2}, 0 \leq x \leq \frac{π}{2} : x = arcsin (\frac{5 - 1}{2})

sin (x) = \frac{- 1 + 5}{2}, 0 \leq x \leq \frac{π}{2}

三角関数の逆数プロパティを適用する

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

以下の一般解

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

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

範囲の解答

0 \leq x \leq \frac{π}{2}

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

sin (x) = \frac{- 1 - 5}{2}, 0 \leq x \leq \frac{π}{2} :

解なし

sin (x) = \frac{- 1 - 5}{2}, 0 \leq x \leq \frac{π}{2}

- 1 \leq sin (x) \leq 1

解なし

すべての解を組み合わせる

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 0.66623 \dots

グラフ

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21.16 = 19.6 sin (x) - 29.4 cos (x)

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