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人気のある三角関数 >

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

解

証明する

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

解

真

解答ステップ

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

左側を操作する

\frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{1 - sin ( A )}

三角関数の公式を使用して書き換える

\frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{1 - sin ( A )}

以下を乗じる:

\frac{1 + sin ( A )}{1 + sin ( A )}

= \frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{\frac{( 1 - s i n ( A ) ) ( 1 + s i n ( A ) )}{1 + s i n ( A )}}

2乗の差の公式を適用する:

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

(1 - sin (A)) (1 + sin (A)) = 1 - sin^{2} (A)

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

ピタゴラスの公式を使用する:

1 = cos^{2} (A) + sin^{2} (A)

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

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

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

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

分数の規則を適用する:

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

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

共通因数を約分する:

cos (A)

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

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

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

簡素化

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

\frac{1 + sin ( A )}{cos ( A )} + \frac{cos ( A )}{1 + sin ( A )}

以下の最小公倍数:

cos (A), 1 + sin (A) : cos (A) (sin (A) + 1)

cos (A), 1 + sin (A)

最小公倍数 (LCM)

cos (A)

または以下のいずれかに現れる因数で構成された式を計算する:

1 + sin (A)

= cos (A) (sin (A) + 1)

LCMに基づいて分数を調整する

該当する分母を乗じてLCMに変えるために

必要な量で各分子を乗じる

cos (A) (sin (A) + 1)

\frac{1 + sin ( A )}{cos ( A )}

の場合

:

分母と分子に以下を乗じる:

sin (A) + 1

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

\frac{cos ( A )}{1 + sin ( A )}

の場合

:

分母と分子に以下を乗じる:

cos (A)

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

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

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 + sin ( A ) ) ^{2} + cos ^{2} ( A )}{cos ( A ) ( sin ( A ) + 1 )}

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

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

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

拡張

(1 + sin (A))^{2} + 1 - sin^{2} (A) : 2 sin (A) + 2

(1 + sin (A))^{2} + 1 - sin^{2} (A)

(1 + sin (A))^{2} : 1 + 2 sin (A) + sin^{2} (A)

完全平方式を適用する:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = sin (A)

= 1^{2} + 2 \cdot 1 \cdot sin (A) + sin^{2} (A)

簡素化

1^{2} + 2 \cdot 1 \cdot sin (A) + sin^{2} (A) : 1 + 2 sin (A) + sin^{2} (A)

1^{2} + 2 \cdot 1 \cdot sin (A) + sin^{2} (A)

規則を適用

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot sin (A) + sin^{2} (A)

数を乗じる:

2 \cdot 1 = 2

= 1 + 2 sin (A) + sin^{2} (A)

= 1 + 2 sin (A) + sin^{2} (A)

= 1 + 2 sin (A) + sin^{2} (A) + 1 - sin^{2} (A)

簡素化

1 + 2 sin (A) + sin^{2} (A) + 1 - sin^{2} (A) : 2 sin (A) + 2

1 + 2 sin (A) + sin^{2} (A) + 1 - sin^{2} (A)

条件のようなグループ

= 2 sin (A) + sin^{2} (A) - sin^{2} (A) + 1 + 1

類似した元を足す:

sin^{2} (A) - sin^{2} (A) = 0

= 2 sin (A) + 1 + 1

数を足す:

1 + 1 = 2

= 2 sin (A) + 2

= 2 sin (A) + 2

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

因数

2 sin (A) + 2 : 2 (sin (A) + 1)

2 sin (A) + 2

書き換え

= 2 sin (A) + 2 \cdot 1

共通項をくくり出す

2

= 2 (sin (A) + 1)

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

共通因数を約分する:

sin (A) + 1

= \frac{2}{cos ( A )}

= \frac{2}{cos ( A )}

= \frac{2}{cos ( A )}

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:

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

\frac{2}{\frac{1}{s e c ( A )}}

簡素化

\frac{2}{\frac{1}{s e c ( A )}}

分数の規則を適用する:

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

= \frac{2 sec ( A )}{1}

規則を適用

\frac{a}{1} = a

= 2 sec (A)

2 sec (A)

2 sec (A)

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 cot(x/2)=(sin(x))/(1-cos(x))

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

証明する-tan(y)+sec(y)=(cos(y))/(1+sin(y))

p ro v e - tan (y) + sec (y) = \frac{cos ( y )}{1 + sin ( y )}

証明する sin(2z)=2sin(z)cos(z)

p ro v e sin (2 z) = 2 sin (z) cos (z)

証明する cot(x/2)=((1+cos(x)))/(sin(x))

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

証明する 1/(tan(b))+tan(b)=sec(b)csc(b)

p ro v e \frac{1}{tan ( b )} + tan (b) = sec (b) csc (b)