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人気のある三角関数 >

証明する (sec(α))/(1-cos(α))=(sec(α)+1)/(sin^2(α))

解

証明する

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

解

真

解答ステップ

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

左側を操作する

\frac{sec ( α )}{1 - cos ( α )}

三角関数の公式を使用して書き換える

\frac{sec ( α )}{1 - cos ( α )}

以下を乗じる:

\frac{1 + cos ( α )}{1 + cos ( α )}

= \frac{sec ( α )}{\frac{( 1 - c o s ( α ) ) ( 1 + c o s ( α ) )}{1 + c o s ( α )}}

2乗の差の公式を適用する:

α^{2} - y^{2} = (α + y) (α - y)

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

= \frac{sec ( α )}{\frac{1 - c o s ^{2} ( α )}{1 + c o s ( α )}}

ピタゴラスの公式を使用する:

1 = cos^{2} (α) + sin^{2} (α)

1 - cos^{2} (α) = sin^{2} (α)

= \frac{sec ( α )}{\frac{s i n ^{2} ( α )}{1 + c o s ( α )}}

分数の規則を適用する:

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

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

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

拡張

\frac{( 1 + cos ( α ) ) sec ( α )}{sin ^{2} ( α )} : \frac{sec ( α )}{sin ^{2} ( α )} + \frac{sec ( α ) cos ( α )}{sin ^{2} ( α )}

\frac{( 1 + cos ( α ) ) sec ( α )}{sin ^{2} ( α )}

拡張

(1 + cos (α)) sec (α) : sec (α) + sec (α) cos (α)

(1 + cos (α)) sec (α)

= sec (α) (1 + cos (α))

分配法則を適用する:

a (b + c) = ab + a c

a = sec (α), b = 1, c = cos (α)

= sec (α) \cdot 1 + sec (α) cos (α)

= 1 \cdot sec (α) + sec (α) cos (α)

乗算:

1 \cdot sec (α) = sec (α)

= sec (α) + sec (α) cos (α)

= \frac{sec ( α ) + sec ( α ) cos ( α )}{sin ^{2} ( α )}

分数の規則を適用する:

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

\frac{sec ( α ) + sec ( α ) cos ( α )}{sin ^{2} ( α )} = \frac{sec ( α )}{sin ^{2} ( α )} + \frac{sec ( α ) cos ( α )}{sin ^{2} ( α )}

= \frac{sec ( α )}{sin ^{2} ( α )} + \frac{sec ( α ) cos ( α )}{sin ^{2} ( α )}

= \frac{cos ( α ) sec ( α )}{sin ^{2} ( α )} + \frac{sec ( α )}{sin ^{2} ( α )}

三角関数の公式を使用して書き換える

\frac{cos ( α ) sec ( α )}{sin ^{2} ( α )} + \frac{sec ( α )}{sin ^{2} ( α )}

sec (α) cos (α) = 1

sec (α) cos (α)

サイン, コサインで表わす

sec (α) cos (α)

基本的な三角関数の公式を使用する:

sec (α) = \frac{1}{cos ( α )}

= \frac{1}{cos ( α )} cos (α)

\frac{1}{cos ( α )} cos (α) = 1

\frac{1}{cos ( α )} cos (α)

分数を乗じる:

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

= \frac{1 cos ( α )}{cos ( α )}

共通因数を約分する:

cos (α)

= 1

= 1

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

規則を適用

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

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

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

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

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 cos(pi/4)=(sqrt(2))/2

p ro v e cos (\frac{π}{4}) = \frac{2}{2}

証明する cot(x)=(csc(x))/(sec(x))

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

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

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

証明する sec(2x)=(sec^2(x))/(1-tan^2(x))

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

証明する (sin(3x)-sin(5x))/(2cos(4x))=-sin(x)

p ro v e \frac{sin ( 3 x ) - sin ( 5 x )}{2 cos ( 4 x )} = - sin (x)