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受欢迎的三角函数 >

证明 cos(2x)=(cot^2(x)-1)/(cot^2(x)+1)

解答

证明

cos (2 x) = \frac{cot ^{2} ( x ) - 1}{cot ^{2} ( x ) + 1}

解答

真

求解步骤

cos (2 x) = \frac{cot ^{2} ( x ) - 1}{cot ^{2} ( x ) + 1}

调整右侧

\frac{cot ^{2} ( x ) - 1}{cot ^{2} ( x ) + 1}

用 sin, cos 表示

\frac{- 1 + cot ^{2} ( x )}{1 + cot ^{2} ( x )}

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

1 = \frac{1 sin ^{2} ( x )}{sin ^{2} ( x )}

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

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

乘以:

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

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

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

化简

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

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

将项转换为分式:

1 = \frac{1 sin ^{2} ( x )}{sin ^{2} ( x )}

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

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

乘以:

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

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

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

分式相除:

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

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

约分:

sin^{2} (x)

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

= \frac{cos ( 2 x )}{1}

使用法则

\frac{a}{1} = a

= cos (2 x)

= cos (2 x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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