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人気のある三角関数 >

証明する ((cos(x)+1))/(sin^3(x))=((csc(x)))/(1-cos(x))

解

証明する

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

解

真

解答ステップ

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

右側を操作する

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

サイン, コサインで表わす

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

基本的な三角関数の公式を使用する:

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

= \frac{\frac{1}{s i n ( x )}}{1 - cos ( x )}

簡素化

\frac{\frac{1}{s i n ( x )}}{1 - cos ( x )} : \frac{1}{sin ( x ) ( 1 - cos ( x ) )}

\frac{\frac{1}{s i n ( x )}}{1 - cos ( x )}

分数の規則を適用する:

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

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

サイン, コサインで表わす

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

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

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

分数を組み合わせる

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

規則を適用

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

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

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

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

因数

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

- cos^{2} (x) + 1

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

因数

sin^{2} (x) (1 - cos (x)) sin (x) : - sin^{3} (x) (cos (x) - 1)

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

因数

1 - cos (x) : - (cos (x) - 1)

1 - cos (x)

共通項をくくり出す

- 1

= - (cos (x) - 1)

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

改良

= - sin^{3} (x) (cos (x) - 1)

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

共通因数を約分する:

cos (x) - 1

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

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

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

= \frac{cos ( x ) + 1}{sin ^{3} ( 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+cot^2(x))/(sec^2(x))=cot^2(x)

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

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

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

証明する sin(a)cos(a)=cos^2(a)tan(a)

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

証明する sin(-pi/4)=-sin(pi/4)

p ro v e sin (- \frac{π}{4}) = - sin (\frac{π}{4})

証明する sin(z+w)=sin(z)cos(w)+cos(z)sin(w)

p ro v e sin (z + w) = sin (z) cos (w) + cos (z) sin (w)