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人気のある三角関数 >

(sin(x)+cos(x))/(cos(x)+1)=tan(x)

解

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} = tan (x)

解

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

+1

度

x = 38.17270 \dots^{\circ} + 36 0^{\circ} n, x = 141.82729 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} = tan (x)

両辺から

tan (x)

を引く

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} - tan (x) = 0

簡素化

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} - tan (x) : \frac{sin ( x ) + cos ( x ) - tan ( x ) ( cos ( x ) + 1 )}{cos ( x ) + 1}

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} - tan (x)

元を分数に変換する:

tan (x) = \frac{tan ( x ) ( cos ( x ) + 1 )}{cos ( x ) + 1}

= \frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} - \frac{tan ( x ) ( cos ( x ) + 1 )}{cos ( x ) + 1}

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{sin ( x ) + cos ( x ) - tan ( x ) ( cos ( x ) + 1 )}{cos ( x ) + 1}

\frac{sin ( x ) + cos ( x ) - tan ( x ) ( cos ( x ) + 1 )}{cos ( x ) + 1} = 0

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

sin (x) + cos (x) - tan (x) (cos (x) + 1) = 0

サイン, コサインで表わす

sin (x) + cos (x) - \frac{sin ( x )}{cos ( x )} (cos (x) + 1) = 0

簡素化

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

sin (x) + cos (x) - \frac{sin ( x )}{cos ( x )} (cos (x) + 1)

乗じる

\frac{sin ( x )}{cos ( x )} (cos (x) + 1) : \frac{sin ( x ) ( cos ( x ) + 1 )}{cos ( x )}

\frac{sin ( x )}{cos ( x )} (cos (x) + 1)

分数を乗じる:

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

= \frac{sin ( x ) ( cos ( x ) + 1 )}{cos ( x )}

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

元を分数に変換する:

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

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

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

sin (x) cos (x) + cos (x) cos (x) - sin (x) (cos (x) + 1)

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)

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

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

拡張

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

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

拡張

- sin (x) (cos (x) + 1) : - sin (x) cos (x) - sin (x)

- sin (x) (cos (x) + 1)

分配法則を適用する:

a (b + c) = ab + a c

a = - sin (x), b = cos (x), c = 1

= - sin (x) cos (x) + (- sin (x)) \cdot 1

マイナス・プラスの規則を適用する

+ (- a) = - a

= - sin (x) cos (x) - 1 \cdot sin (x)

乗算:

1 \cdot sin (x) = sin (x)

= - sin (x) cos (x) - sin (x)

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

類似した元を足す:

sin (x) cos (x) - sin (x) cos (x) = 0

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

両辺に

sin (x)

を足す

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

両辺を2乗する

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

両辺から

sin^{2} (x)

を引く

cos^{4} (x) - sin^{2} (x) = 0

因数

cos^{4} (x) - sin^{2} (x) : (cos^{2} (x) + sin (x)) (cos^{2} (x) - sin (x))

cos^{4} (x) - sin^{2} (x)

指数の規則を適用する:

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

cos^{4} (x) = (cos^{2} (x))^{2}

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

2乗の差の公式を適用する:

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

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

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

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

各部分を別個に解く

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

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

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

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

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

解なし

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

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

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{5 - 1}{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

- (- a) = a

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

(- 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 1 = 4

4 \cdot 1 \cdot 1

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 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, - (- 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{5 - 1}{2}

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

括弧を削除する:

(- a) = - a, - (- 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 = - (5 - 1)

= \frac{5 - 1}{2}

二次equationの解:

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

代用を戻す

u = sin (x)

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

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

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

解なし

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

- 1 \leq sin (x) \leq 1

解なし

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

すべての解を組み合わせる

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

すべての解を組み合わせる

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

元のequationに当てはめて解を検算する

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} = tan (x)

に当てはめて解を確認する

equationに一致しないものを削除する。

解答を確認する

arcsin (- \frac{- 1 + 5}{2}) + 2 πn :

偽

arcsin (- \frac{- 1 + 5}{2}) + 2 πn

挿入

n = 1

arcsin (- \frac{- 1 + 5}{2}) + 2 π 1

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} = tan (x)

の挿入向け

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

\frac{sin ( arcsin ( - \frac{- 1 + 5}{2} ) + 2 π 1 ) + cos ( arcsin ( - \frac{- 1 + 5}{2} ) + 2 π 1 )}{cos ( arcsin ( - \frac{- 1 + 5}{2} ) + 2 π 1 ) + 1} = tan (arcsin (- \frac{- 1 + 5}{2}) + 2 π 1)

改良

0.09412 \dots = - 0.78615 \dots

\Rightarrow 偽

解答を確認する

π + arcsin (\frac{- 1 + 5}{2}) + 2 πn :

偽

π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

挿入

n = 1

π + arcsin (\frac{- 1 + 5}{2}) + 2 π 1

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} = tan (x)

の挿入向け

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

\frac{sin ( π + arcsin ( \frac{- 1 + 5}{2} ) + 2 π 1 ) + cos ( π + arcsin ( \frac{- 1 + 5}{2} ) + 2 π 1 )}{cos ( π + arcsin ( \frac{- 1 + 5}{2} ) + 2 π 1 ) + 1} = tan (π + arcsin (\frac{- 1 + 5}{2}) + 2 π 1)

改良

- 6.56625 \dots = 0.78615 \dots

\Rightarrow 偽

解答を確認する

arcsin (\frac{5 - 1}{2}) + 2 πn :

真

arcsin (\frac{5 - 1}{2}) + 2 πn

挿入

n = 1

arcsin (\frac{5 - 1}{2}) + 2 π 1

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} = tan (x)

の挿入向け

x = arcsin (\frac{5 - 1}{2}) + 2 π 1

\frac{sin ( arcsin ( \frac{5 - 1}{2} ) + 2 π 1 ) + cos ( arcsin ( \frac{5 - 1}{2} ) + 2 π 1 )}{cos ( arcsin ( \frac{5 - 1}{2} ) + 2 π 1 ) + 1} = tan (arcsin (\frac{5 - 1}{2}) + 2 π 1)

改良

0.78615 \dots = 0.78615 \dots

\Rightarrow 真

解答を確認する

π - arcsin (\frac{5 - 1}{2}) + 2 πn :

真

π - arcsin (\frac{5 - 1}{2}) + 2 πn

挿入

n = 1

π - arcsin (\frac{5 - 1}{2}) + 2 π 1

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} = tan (x)

の挿入向け

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

\frac{sin ( π - arcsin ( \frac{5 - 1}{2} ) + 2 π 1 ) + cos ( π - arcsin ( \frac{5 - 1}{2} ) + 2 π 1 )}{cos ( π - arcsin ( \frac{5 - 1}{2} ) + 2 π 1 ) + 1} = tan (π - arcsin (\frac{5 - 1}{2}) + 2 π 1)

改良

- 0.78615 \dots = - 0.78615 \dots

\Rightarrow 真

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

10進法形式で解を証明する

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

グラフ

人気の例

cos(x)-cos(2x)=1

cos (x) - cos (2 x) = 1

tan(θ)= 1/(sqrt(2))

tan (θ) = \frac{1}{2}

-cos(x)-sin(x)=1

- cos (x) - sin (x) = 1

sin(2x)=sin(0.5x)

sin (2 x) = sin (0.5 x)

2sin(2x+15)= 1/2

2 sin (2 x + 15) = \frac{1}{2}