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人気のある三角関数 >

証明する 1/(1+cos(x))=csc^2(x)-csc(x)cot(x)

解

証明する

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

解

真

解答ステップ

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

右側を操作する

csc^{2} (x) - csc (x) cot (x)

サイン, コサインで表わす

csc^{2} (x) - cot (x) csc (x)

基本的な三角関数の公式を使用する:

csc (x) = \frac{1}{sin ( x )}

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

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

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

(\frac{1}{sin ( x )})^{2}

指数の規則を適用する:

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

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

分数を乗じる:

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

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

乗算:

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

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

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

sin (x) sin (x)

指数の規則を適用する:

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

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

= sin^{1 + 1} (x)

数を足す:

1 + 1 = 2

= sin^{2} (x)

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

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

規則を適用

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

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

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

因数

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

1 - cos^{2} (x)

共通項をくくり出す

- 1

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

因数

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

cos^{2} (x) - 1

1

を書き換え

1^{2}

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

2乗の差の公式を適用する:

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

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

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

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

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

キャンセル

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

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

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

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

改良

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

共通因数を約分する:

1 - cos (x)

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

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

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

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

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

両辺を同じ形式にできることを証明した

\Rightarrow 真

人気の例

証明する 1/(1-sin(x))-1/(1+sin(x))=(2tan(x))/(cos(x))

p ro v e \frac{1}{1 - sin ( x )} - \frac{1}{1 + sin ( x )} = \frac{2 tan ( x )}{cos ( x )}

証明する cos(t)cot(t)=(1-sin^2(t))/(sin(t))

p ro v e cos (t) cot (t) = \frac{1 - sin ^{2} ( t )}{sin ( t )}

証明する cos^2(2u)-sin^2(2u)=cos(4u)

p ro v e cos^{2} (2 u) - sin^{2} (2 u) = cos (4 u)

証明する tan(u)cot(u)-cos^2(u)=sin^2(u)

p ro v e tan (u) cot (u) - cos^{2} (u) = sin^{2} (u)

証明する cos(2θ)=1-2sin^2(θ)

p ro v e cos (2 θ) = 1 - 2 sin^{2} (θ)