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人気のある三角関数 >

証明する (sec^2(x)-1)cot^2(x)=1

解

証明する

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

解

真

解答ステップ

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

左側を操作する

(sec^{2} (x) - 1) cot^{2} (x)

サイン, コサインで表わす

(- 1 + sec^{2} (x)) cot^{2} (x)

基本的な三角関数の公式を使用する:

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

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

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

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

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

指数の規則を適用する:

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

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

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

指数の規則を適用する:

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

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

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

分数を乗じる:

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

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

結合

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

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

元を分数に変換する:

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

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

乗算:

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

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

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

乗じる

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

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

分数を乗じる:

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

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

共通因数を約分する:

cos^{2} (x)

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

規則を適用

\frac{a}{a} = 1

= 1

= 1

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 cos(x)=-1/2

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

証明する (1+cos(θ))/(tan(θ)+sin(θ))=cot(θ)

p ro v e \frac{1 + cos ( θ )}{tan ( θ ) + sin ( θ )} = cot (θ)

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

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

証明する (tan(x)sec(-x))cos(x)=sin(x)-1

p ro v e (tan (x) sec (- x)) cos (x) = sin (x) - 1

証明する (1-cos(x))*(1-cos(x))=(1-cos(x))^2

p ro v e (1 - cos (x)) \cdot (1 - cos (x)) = (1 - cos (x))^{2}