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人気のある三角関数 >

証明する (tan(X)-sin(X))/(sin^3(X))=(sec(X))/(1+cos(X))

解

証明する

\frac{tan ( X ) - sin ( X )}{sin ^{3} ( X )} = \frac{sec ( X )}{1 + cos ( X )}

解

真

解答ステップ

\frac{tan ( X ) - sin ( X )}{sin ^{3} ( X )} = \frac{sec ( X )}{1 + cos ( X )}

左側を操作する

\frac{tan ( X ) - sin ( X )}{sin ^{3} ( X )}

サイン, コサインで表わす

\frac{- sin ( X ) + tan ( X )}{sin ^{3} ( X )}

基本的な三角関数の公式を使用する:

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

= \frac{- sin ( X ) + \frac{s i n ( X )}{c o s ( X )}}{sin ^{3} ( X )}

簡素化

\frac{- sin ( X ) + \frac{s i n ( X )}{c o s ( X )}}{sin ^{3} ( X )} : \frac{- cos ( X ) + 1}{sin ^{2} ( X ) cos ( X )}

\frac{- sin ( X ) + \frac{s i n ( X )}{c o s ( X )}}{sin ^{3} ( X )}

結合

- sin (X) + \frac{sin ( X )}{cos ( X )} : \frac{- sin ( X ) cos ( X ) + sin ( X )}{cos ( X )}

- sin (X) + \frac{sin ( X )}{cos ( X )}

元を分数に変換する:

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

= - \frac{sin ( X ) cos ( X )}{cos ( X )} + \frac{sin ( X )}{cos ( 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{- sin ( X ) cos ( X ) + sin ( X )}{cos ( X )}

= \frac{\frac{- s i n ( X ) c o s ( X ) + s i n ( X )}{c o s ( X )}}{sin ^{3} ( X )}

分数の規則を適用する:

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

= \frac{- sin ( X ) cos ( X ) + sin ( X )}{cos ( X ) sin ^{3} ( X )}

共通項をくくり出す

sin (X)

= \frac{sin ( X ) ( - cos ( X ) + 1 )}{cos ( X ) sin ^{3} ( X )}

共通因数を約分する:

sin (X)

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

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

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

簡素化

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

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

キャンセル

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

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

cos (X) - 1 = - (- cos (X) + 1)

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

改良

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

共通因数を約分する:

1 - cos (X)

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

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

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

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

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:

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

\frac{sec ( X )}{( cos ( X ) + 1 )}

簡素化

\frac{sec ( X )}{( cos ( X ) + 1 )} : \frac{sec ( X )}{cos ( X ) + 1}

\frac{sec ( X )}{( cos ( X ) + 1 )}

括弧を削除する:

(a) = a

= \frac{sec ( X )}{cos ( X ) + 1}

\frac{sec ( X )}{cos ( X ) + 1}

\frac{sec ( X )}{cos ( X ) + 1}

= \frac{sec ( X )}{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 sin(x)+cos(x)=(cot(x)+1)/(csc(x))

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

証明する sin^2(t)=2sin(t)cos(t)

p ro v e sin^{2} (t) = 2 sin (t) cos (t)

証明する 1-cos(θ)=(sin^2(θ))/(1+cos(θ))

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

証明する (sin(2x))/(sin(x))= 2/(sec(x))

p ro v e \frac{sin ( 2 x )}{sin ( x )} = \frac{2}{sec ( x )}

証明する 2/(1+cos(2x))=sec^2(x)

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