ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

証明する 1+cos^4(x)-sin^4(x)=2cos^2(x)

解

証明する

1 + cos^{4} (x) - sin^{4} (x) = 2 cos^{2} (x)

解

真

解答ステップ

1 + cos^{4} (x) - sin^{4} (x) = 2 cos^{2} (x)

左側を操作する

1 + cos^{4} (x) - sin^{4} (x)

因数

1 - sin^{4} (x) : - (sin^{2} (x) + 1) (sin (x) + 1) (sin (x) - 1)

1 - sin^{4} (x)

共通項をくくり出す

- 1

= - (sin^{4} (x) - 1)

因数

sin^{4} (x) - 1 : (sin^{2} (x) + 1) (sin (x) + 1) (sin (x) - 1)

sin^{4} (x) - 1

sin^{4} (x) - 1

を書き換え

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

sin^{4} (x) - 1

1

を書き換え

1^{2}

= sin^{4} (x) - 1^{2}

指数の規則を適用する:

a^{b c} = (a^{b})^{c}

sin^{4} (x) = (sin^{2} (x))^{2}

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

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

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

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

= cos^{4} (x) - (- 1 + sin (x)) (1 + sin (x)) (1 + sin^{2} (x))

三角関数の公式を使用して書き換える

cos^{4} (x) - (- 1 + sin (x)) (1 + sin (x)) (1 + sin^{2} (x))

ピタゴラスの公式を使用する:

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

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

= cos^{4} (x) - (- 1 + sin (x)) (1 + sin (x)) (1 + 1 - cos^{2} (x))

簡素化

= cos^{4} (x) - (- 1 + sin (x)) (1 + sin (x)) (- cos^{2} (x) + 2)

拡張

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

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

2乗の差の公式を適用する:

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

a = sin (x), b = 1

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

規則を適用

1^{a} = 1

1^{2} = 1

= sin^{2} (x) - 1

= cos^{4} (x) - (sin^{2} (x) - 1) (2 - cos^{2} (x))

ピタゴラスの公式を使用する:

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

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

= cos^{4} (x) - (- cos^{2} (x)) (2 - cos^{2} (x))

簡素化

cos^{4} (x) - (- cos^{2} (x)) (2 - cos^{2} (x)) : 2 cos^{2} (x)

cos^{4} (x) - (- cos^{2} (x)) (2 - cos^{2} (x))

規則を適用

- (- a) = a

= cos^{4} (x) + cos^{2} (x) (2 - cos^{2} (x))

拡張

cos^{2} (x) (2 - cos^{2} (x)) : 2 cos^{2} (x) - cos^{4} (x)

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

分配法則を適用する:

a (b - c) = ab - a c

a = cos^{2} (x), b = 2, c = cos^{2} (x)

= cos^{2} (x) \cdot 2 - cos^{2} (x) cos^{2} (x)

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

cos^{2} (x) cos^{2} (x) = cos^{4} (x)

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

指数の規則を適用する:

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

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

= cos^{2 + 2} (x)

数を足す:

2 + 2 = 4

= cos^{4} (x)

= 2 cos^{2} (x) - cos^{4} (x)

= cos^{4} (x) + 2 cos^{2} (x) - cos^{4} (x)

類似した元を足す:

cos^{4} (x) - cos^{4} (x) = 0

= 2 cos^{2} (x)

= 2 cos^{2} (x)

= 2 cos^{2} (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 cot(x)-sec(x)csc(x)cos(2x)=tan(x)

p ro v e cot (x) - sec (x) csc (x) cos (2 x) = tan (x)

証明する 1/(tan(x))+1/(cot(x))=sec(x)csc(x)

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

証明する ((tan(x)cot(x)))/(cos(x))=sec(x)

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

証明する cot(t/2)=(sin(t))/(1-cos(t))

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

証明する 1-(cos^2(x))/(1+sin^2(x))=sin(x)

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