証明する (1+cos(2θ))(1-cos(2θ))=sin^2(2θ)

解

証明する

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

真

解答ステップ

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

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

(1 + cos (x)) (1 - cos (x))

拡張

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

(1 + cos (x)) (1 - cos (x))

2乗の差の公式を適用する:

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

a = 1, b = cos (x)

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

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

ピタゴラスの公式を使用する:

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

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

= sin^{2} (x)

\Rightarrow 真