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人気のある三角関数 >

証明する sin^4(x)-cos^4(x)=2sin^2(x)-1

解

証明する

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

解

真

解答ステップ

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

左側を操作する

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

因数

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

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

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

を書き換え

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

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

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} - cos^{4} (x)

指数の規則を適用する:

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

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

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

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

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

因数

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

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

2乗の差の公式を適用する:

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

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

= (sin (x) + cos (x)) (sin (x) - cos (x))

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

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

三角関数の公式を使用して書き換える

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

拡張

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

(sin (x) + cos (x)) (sin (x) - cos (x))

2乗の差の公式を適用する:

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

a = sin (x), b = cos (x)

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

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

2倍角の公式を使用:

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

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

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

簡素化

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

2倍角の公式を使用:

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

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

ピタゴラスの公式を使用する:

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

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

簡素化

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

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

乗算:

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

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

括弧を分配する

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

マイナス・プラスの規則を適用する

- (- a) = a, - (a) = - a

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

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

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

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

両辺を同じ形式にできることを証明した

\Rightarrow 真

人気の例

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