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人気のある三角関数 >

証明する (sec^2(x)-1)csc(x)=sin(x)sec^2(x)

解

証明する

(sec^{2} (x) - 1) csc (x) = sin (x) sec^{2} (x)

解

真

解答ステップ

(sec^{2} (x) - 1) csc (x) = sin (x) sec^{2} (x)

左側を操作する

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

サイン, コサインで表わす

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

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

分数を乗じる:

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

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

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

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

乗算:

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

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

括弧を削除する:

(- a) = - a

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

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

分数の規則を適用する:

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

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

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

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

三角関数の公式を使用して書き換える

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

共通因数を約分する:

sin (x)

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

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

三角関数の公式を使用して書き換える

基本的な三角関数の公式を使用する:

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

\frac{sin ( x )}{( \frac{1}{s e c ( x )} ) ^{2}}

簡素化

\frac{sin ( x )}{( \frac{1}{s e c ( x )} ) ^{2}}

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

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

指数の規則を適用する:

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

分数の規則を適用する:

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

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

規則を適用

\frac{a}{1} = a

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

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

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

= sin (x) sec^{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 (1+csc(x))/(sec(x))-cot(x)=cos(x)

p ro v e \frac{1 + csc ( x )}{sec ( x )} - cot (x) = cos (x)

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

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

証明する (csc(-x))/(sec(-x))=-cot(x)

p ro v e \frac{csc ( - x )}{sec ( - x )} = - cot (x)

証明する cos(x-y)+cos(x+y)=2cos(x)cos(y)

p ro v e cos (x - y) + cos (x + y) = 2 cos (x) cos (y)

証明する sin(pi/6+x)+sin(pi/6-x)=cos(x)

p ro v e sin (\frac{π}{6} + x) + sin (\frac{π}{6} - x) = cos (x)