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人気のある三角関数 >

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

解

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

解

x = 2 πn, x = π + 2 πn

+1

度

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

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

を引く

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

簡素化

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

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

以下の最小公倍数:

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

cos (x), 1 - sin (x)

最小公倍数 (LCM)

cos (x)

または以下のいずれかに現れる因数で構成された式を計算する:

1 - sin (x)

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

LCMに基づいて分数を調整する

該当する分母を乗じてLCMに変えるために

必要な量で各分子を乗じる

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

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

の場合

:

分母と分子に以下を乗じる:

- sin (x) + 1

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

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

の場合

:

分母と分子に以下を乗じる:

cos (x)

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

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

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

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

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

三角関数の公式を使用して書き換える

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

簡素化

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

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

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

完全平方式を適用する:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = sin (x)

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

簡素化

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

数を乗じる:

2 \cdot 1 = 2

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

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

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

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

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

1 - 1 = 0

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

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

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

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

置換で解く

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

仮定:

sin (x) = u

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

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

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

標準的な形式で書く

a x^{2} + b x + c = 0

2 u^{2} - 2 u = 0

解くとthe二次式

2 u^{2} - 2 u = 0

二次Equationの公式:

次の場合:

a = 2, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 2)^{2} - 4 \cdot 2 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 2 \cdot 0

指数の規則を適用する:

n

が偶数であれば

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

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

= 2^{2} - 4 \cdot 2 \cdot 0

規則を適用

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a,

, 以下を想定

a \geq 0

= 2

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

解を分離する

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

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

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

規則を適用

- (- a) = a

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

数を足す:

2 + 2 = 4

= \frac{4}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 2}{2 \cdot 2} : 0

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

規則を適用

- (- a) = a

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

数を引く:

2 - 2 = 0

= \frac{0}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{0}{4}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= 0

二次equationの解:

u = 1, u = 0

代用を戻す

u = sin (x)

sin (x) = 1, sin (x) = 0

sin (x) = 1, sin (x) = 0

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

以下の一般解

sin (x) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

以下の一般解

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

すべての解を組み合わせる

x = \frac{π}{2} + 2 πn, x = 2 πn, x = π + 2 πn

equationは以下で未定義のため:

\frac{π}{2} + 2 πn

x = 2 πn, x = π + 2 πn

グラフ

人気の例

5 sec (x) + 7 = - 3

cos(2θ)+2cos(θ)=-1

cos (2 θ) + 2 cos (θ) = - 1

((tan(x))/1)^{-1}=(20)/(9.7)

(\frac{tan ( x )}{1})^{- 1} = \frac{20}{9.7}

1-tan(x)=sec(x)

1 - tan (x) = sec (x)

sec((3θ)/2)=-2,0<= θ<2pi

sec (\frac{3 θ}{2}) = - 2, 0 \leq θ < 2 π