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人気のある三角関数 >

証明する cot^2(x)sec^2(x)=cot^2(x)+1

解

証明する

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

解

真

解答ステップ

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

左側を操作する

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

サイン, コサインで表わす

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

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

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

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

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

分数を乗じる:

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

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

共通因数を約分する:

cos^{2} (x)

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

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

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

右側を操作する

cot^{2} (x) + 1

サイン, コサインで表わす

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

簡素化

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

1 + (\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}}

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

元を分数に変換する:

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

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

乗算:

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

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

証明する cos(x)=(cot(x))/(tan(x))

p ro v e cos (x) = \frac{cot ( x )}{tan ( x )}

証明する cos^2(2a)-sin^2(2a)=cos(4a)

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

証明する tan(2x)-2cos(x)=0

p ro v e tan (2 x) - 2 cos (x) = 0

証明する cos^2(a)cot^2(a)=cot^2(a)-cos^2(a)

p ro v e cos^{2} (a) cot^{2} (a) = cot^{2} (a) - cos^{2} (a)