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人気のある三角関数 >

証明する cos(2β)=(1-tan^2(β))/(1+tan^2(β))

解

証明する

cos (2 β) = \frac{1 - tan ^{2} ( β )}{1 + tan ^{2} ( β )}

解

真

解答ステップ

cos (2 β) = \frac{1 - tan ^{2} ( β )}{1 + tan ^{2} ( β )}

右側を操作する

\frac{1 - tan ^{2} ( β )}{1 + tan ^{2} ( β )}

サイン, コサインで表わす

\frac{1 - tan ^{2} ( β )}{1 + tan ^{2} ( β )}

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 - ( \frac{s i n ( β )}{c o s ( β )} ) ^{2}}{1 + ( \frac{s i n ( β )}{c o s ( β )} ) ^{2}}

簡素化

\frac{1 - ( \frac{s i n ( β )}{c o s ( β )} ) ^{2}}{1 + ( \frac{s i n ( β )}{c o s ( β )} ) ^{2}} : \frac{cos ^{2} ( β ) - sin ^{2} ( β )}{cos ^{2} ( β ) + sin ^{2} ( β )}

\frac{1 - ( \frac{s i n ( β )}{c o s ( β )} ) ^{2}}{1 + ( \frac{s i n ( β )}{c o s ( β )} ) ^{2}}

指数の規則を適用する:

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

= \frac{1 - ( \frac{s i n ( β )}{c o s ( β )} ) ^{2}}{1 + \frac{s i n ^{2} ( β )}{c o s ^{2} ( β )}}

指数の規則を適用する:

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

= \frac{1 - \frac{s i n ^{2} ( β )}{c o s ^{2} ( β )}}{1 + \frac{s i n ^{2} ( β )}{c o s ^{2} ( β )}}

結合

1 + \frac{sin ^{2} ( β )}{cos ^{2} ( β )} : \frac{cos ^{2} ( β ) + sin ^{2} ( β )}{cos ^{2} ( β )}

1 + \frac{sin ^{2} ( β )}{cos ^{2} ( β )}

元を分数に変換する:

1 = \frac{1 cos ^{2} ( β )}{cos ^{2} ( β )}

= \frac{1 \cdot cos ^{2} ( β )}{cos ^{2} ( β )} + \frac{sin ^{2} ( β )}{cos ^{2} ( β )}

分母が等しいので, 分数を組み合わせる:

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

= \frac{1 \cdot cos ^{2} ( β ) + sin ^{2} ( β )}{cos ^{2} ( β )}

乗算:

1 \cdot cos^{2} (β) = cos^{2} (β)

= \frac{cos ^{2} ( β ) + sin ^{2} ( β )}{cos ^{2} ( β )}

= \frac{1 - \frac{s i n ^{2} ( β )}{c o s ^{2} ( β )}}{\frac{c o s ^{2} ( β ) + s i n ^{2} ( β )}{c o s ^{2} ( β )}}

結合

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

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

元を分数に変換する:

1 = \frac{1 cos ^{2} ( β )}{cos ^{2} ( β )}

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

分母が等しいので, 分数を組み合わせる:

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

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

乗算:

1 \cdot cos^{2} (β) = cos^{2} (β)

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

= \frac{\frac{c o s ^{2} ( β ) - s i n ^{2} ( β )}{c o s ^{2} ( β )}}{\frac{c o s ^{2} ( β ) + s i n ^{2} ( β )}{c o s ^{2} ( β )}}

分数を割る:

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

= \frac{( cos ^{2} ( β ) - sin ^{2} ( β ) ) cos ^{2} ( β )}{cos ^{2} ( β ) ( cos ^{2} ( β ) + sin ^{2} ( β ) )}

共通因数を約分する:

cos^{2} (β)

= \frac{cos ^{2} ( β ) - sin ^{2} ( β )}{cos ^{2} ( β ) + sin ^{2} ( β )}

= \frac{cos ^{2} ( β ) - sin ^{2} ( β )}{cos ^{2} ( β ) + sin ^{2} ( β )}

= \frac{cos ^{2} ( β ) - sin ^{2} ( β )}{cos ^{2} ( β ) + sin ^{2} ( β )}

三角関数の公式を使用して書き換える

\frac{cos ^{2} ( β ) - sin ^{2} ( β )}{cos ^{2} ( β ) + sin ^{2} ( β )}

ピタゴラスの公式を使用する:

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

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

規則を適用

\frac{a}{1} = a

= cos^{2} (β) - sin^{2} (β)

2倍角の公式を使用:

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

= cos (2 β)

= cos (2 β)

両辺を同じ形式にできることを証明した

\Rightarrow 真

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