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人気のある三角関数 >

証明する cos^4(A)-sin^4(A)=cos(2A)

解

証明する

cos^{4} (A) - sin^{4} (A) = cos (2 A)

解

真

解答ステップ

cos^{4} (A) - sin^{4} (A) = cos (2 A)

左側を操作する

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

因数

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

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

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

を書き換え

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

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

指数の規則を適用する:

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

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

= cos^{4} (A) - (sin^{2} (A))^{2}

指数の規則を適用する:

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

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

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

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

2乗の差の公式を適用する:

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

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

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

因数

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

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

2乗の差の公式を適用する:

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

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

= (cos (A) + sin (A)) (cos (A) - sin (A))

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

= (cos (A) + sin (A)) (cos (A) - sin (A)) \cdot 1

簡素化

= (cos (A) + sin (A)) (cos (A) - sin (A))

拡張

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

(cos (A) + sin (A)) (cos (A) - sin (A))

2乗の差の公式を適用する:

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

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

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

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

2倍角の公式を使用:

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

= cos (2 A)

= cos (2 A)

両辺を同じ形式にできることを証明した

\Rightarrow 真

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