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人気のある三角関数 >

証明する ((sin(x)))/((1-cos(x)))-((1))/((sin(x)))=cot(x)

解

証明する

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

解

真

解答ステップ

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

左側を操作する

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

三角関数の公式を使用して書き換える

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

以下を乗じる:

\frac{1 + cos ( x )}{1 + cos ( x )}

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

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

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

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

簡素化

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

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

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

\frac{sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}}

分数の規則を適用する:

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

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

共通因数を約分する:

sin (x)

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

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

規則を適用

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

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

1 + cos (x) - 1 = cos (x)

1 + cos (x) - 1

条件のようなグループ

= cos (x) + 1 - 1

1 - 1 = 0

= cos (x)

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

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

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

基本的な三角関数の公式を使用する:

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

= cot (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(0)sec(0)=tan(0)

p ro v e sin (0) sec (0) = tan (0)

証明する cos(2a)=(cot(a)-tan(a))/(csc(a)sec(a))

p ro v e cos (2 a) = \frac{cot ( a ) - tan ( a )}{csc ( a ) sec ( a )}

証明する cos(a)cos(b)= 1/2 (cos(a-b)+cos(a+b))

p ro v e cos (a) cos (b) = \frac{1}{2} (cos (a - b) + cos (a + b))

証明する sin(a-b)=sin(a)cos(b)-cos(a)sin(b)

p ro v e sin (a - b) = sin (a) cos (b) - cos (a) sin (b)

証明する 2/(cos(x))=2sec(x)

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