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人気のある三角関数 >

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

解

証明する

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

解

真

解答ステップ

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

右側を操作する

(sec (x) - tan (x)) sec (x)

サイン, コサインで表わす

(sec (x) - tan (x)) sec (x)

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

分数を組み合わせる

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

規則を適用

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

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

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

括弧を削除する:

(a) = a

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

分数を乗じる:

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

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

(1 - sin (x)) \cdot 1 = 1 - sin (x)

(1 - sin (x)) \cdot 1

乗算:

(1 - sin (x)) \cdot 1 = (1 - sin (x))

= (1 - sin (x))

括弧を削除する:

(a) = a

= 1 - sin (x)

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

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

数を足す:

1 + 1 = 2

= cos^{2} (x)

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

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

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

三角関数の公式を使用して書き換える

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

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

簡素化

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

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

因数

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

1 - sin^{2} (x)

共通項をくくり出す

- 1

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

因数

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

sin^{2} (x) - 1

1

を書き換え

1^{2}

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

2乗の差の公式を適用する:

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

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

= (sin (x) + 1) (sin (x) - 1)

= - (sin (x) + 1) (sin (x) - 1)

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

キャンセル

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

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

sin (x) - 1 = - (- sin (x) + 1)

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

改良

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

共通因数を約分する:

1 - sin (x)

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

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

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

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

= \frac{1}{1 + sin ( 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 tan(a)+cot(a)= 2/(sin(2a))

p ro v e tan (a) + cot (a) = \frac{2}{sin ( 2 a )}

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

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

証明する 2csc^2(y)= 1/(1-cos(y))+1/(1+cos(y))

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

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

証明する sin(θ-pi/2)=-cos(θ)

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