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受欢迎的三角函数 >

证明 (1-tan^2(A))/(1+tan^2(A))=cos(2A)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

化简

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

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

使用指数法则:

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

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

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

因为分母相等，所以合并分式:

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

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

乘以:

1 \cdot cos^{2} (A) = cos^{2} (A)

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

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

化简

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

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

将项转换为分式:

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

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

因为分母相等，所以合并分式:

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

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

乘以:

1 \cdot cos^{2} (A) = cos^{2} (A)

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

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

分式相除:

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

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

约分:

cos^{2} (A)

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

使用法则

\frac{a}{1} = a

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

使用倍角公式:

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

= cos (2 A)

= cos (2 A)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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