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人気のある三角関数 >

証明する csc^2(x)*tan^2(x)-1=tan^2(x)

解

証明する

csc^{2} (x) \cdot tan^{2} (x) - 1 = tan^{2} (x)

解

真

解答ステップ

csc^{2} (x) tan^{2} (x) - 1 = tan^{2} (x)

左側を操作する

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

サイン, コサインで表わす

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

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

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

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

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

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

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

指数の規則を適用する:

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

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

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

分数を乗じる:

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

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

共通因数を約分する:

sin^{2} (x)

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

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

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 )}

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

三角関数の公式を使用して書き換える

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

基本的な三角関数の公式を使用する:

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

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

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

基本的な三角関数の公式を使用する:

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

tan (x) tan (x)

簡素化

tan (x) tan (x) : tan^{2} (x)

tan (x) tan (x)

指数の規則を適用する:

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

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

= tan^{1 + 1} (x)

数を足す:

1 + 1 = 2

= tan^{2} (x)

tan^{2} (x)

tan^{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 sec^2(x)-cot^2(pi/2-x)=1

p ro v e sec^{2} (x) - cot^{2} (\frac{π}{2} - x) = 1

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p ro v e sin (2 x) = - 2 sin (- x) cos (- x)

証明する sin(θ)csc(θ)-sin^2(θ)=cos^2(θ)

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

証明する cos^2(x)csc(x)sec(x)=cot(x)

p ro v e cos^{2} (x) csc (x) sec (x) = cot (x)

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

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