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受欢迎的三角函数 >

证明 (cot^2(x)-1)/(2cot(x))=cot(2x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

用 sin, cos 表示

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

化简

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

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

乘

2 \cdot \frac{cos ( x )}{sin ( x )} : \frac{2 cos ( x )}{sin ( x )}

2 \cdot \frac{cos ( x )}{sin ( x )}

分式相乘:

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

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

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

使用分式法则:

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

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

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{c o s ^{2} ( x ) - s i n ^{2} ( x )}{s i n ^{2} ( x )} sin ( x )}{2 cos ( x )}

乘

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

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

分式相乘:

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

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

约分:

sin (x)

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

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

使用分式法则:

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

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

使用倍角公式:

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

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

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

使用基本三角恒等式:

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

= cot (2 x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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