zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

证明 (cot(4u)-1)/(cot(4u)+1)=(1-tan(4u))/(1+tan(4u))

解答

证明

\frac{cot ( 4 u ) - 1}{cot ( 4 u ) + 1} = \frac{1 - tan ( 4 u )}{1 + tan ( 4 u )}

解答

真

求解步骤

\frac{cot ( 4 u ) - 1}{cot ( 4 u ) + 1} = \frac{1 - tan ( 4 u )}{1 + tan ( 4 u )}

调整左侧

\frac{cot ( 4 u ) - 1}{cot ( 4 u ) + 1}

用 sin, cos 表示

\frac{- 1 + cot ( 4 u )}{1 + cot ( 4 u )}

使用基本三角恒等式:

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

= \frac{- 1 + \frac{c o s ( 4 u )}{s i n ( 4 u )}}{1 + \frac{c o s ( 4 u )}{s i n ( 4 u )}}

化简

\frac{- 1 + \frac{c o s ( 4 u )}{s i n ( 4 u )}}{1 + \frac{c o s ( 4 u )}{s i n ( 4 u )}} : \frac{- sin ( 4 u ) + cos ( 4 u )}{sin ( 4 u ) + cos ( 4 u )}

\frac{- 1 + \frac{c o s ( 4 u )}{s i n ( 4 u )}}{1 + \frac{c o s ( 4 u )}{s i n ( 4 u )}}

化简

1 + \frac{cos ( 4 u )}{sin ( 4 u )} : \frac{sin ( 4 u ) + cos ( 4 u )}{sin ( 4 u )}

1 + \frac{cos ( 4 u )}{sin ( 4 u )}

将项转换为分式:

1 = \frac{1 sin ( 4 u )}{sin ( 4 u )}

= \frac{1 \cdot sin ( 4 u )}{sin ( 4 u )} + \frac{cos ( 4 u )}{sin ( 4 u )}

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 ( 4 u ) + cos ( 4 u )}{sin ( 4 u )}

乘以:

1 \cdot sin (4 u) = sin (4 u)

= \frac{sin ( 4 u ) + cos ( 4 u )}{sin ( 4 u )}

= \frac{- 1 + \frac{c o s ( 4 u )}{s i n ( 4 u )}}{\frac{s i n ( 4 u ) + c o s ( 4 u )}{s i n ( 4 u )}}

化简

- 1 + \frac{cos ( 4 u )}{sin ( 4 u )} : \frac{- sin ( 4 u ) + cos ( 4 u )}{sin ( 4 u )}

- 1 + \frac{cos ( 4 u )}{sin ( 4 u )}

将项转换为分式:

1 = \frac{1 sin ( 4 u )}{sin ( 4 u )}

= - \frac{1 \cdot sin ( 4 u )}{sin ( 4 u )} + \frac{cos ( 4 u )}{sin ( 4 u )}

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 ( 4 u ) + cos ( 4 u )}{sin ( 4 u )}

乘以:

1 \cdot sin (4 u) = sin (4 u)

= \frac{- sin ( 4 u ) + cos ( 4 u )}{sin ( 4 u )}

= \frac{\frac{- s i n ( 4 u ) + c o s ( 4 u )}{s i n ( 4 u )}}{\frac{s i n ( 4 u ) + c o s ( 4 u )}{s i n ( 4 u )}}

分式相除:

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

= \frac{( - sin ( 4 u ) + cos ( 4 u ) ) sin ( 4 u )}{sin ( 4 u ) ( sin ( 4 u ) + cos ( 4 u ) )}

约分:

sin (4 u)

= \frac{- sin ( 4 u ) + cos ( 4 u )}{sin ( 4 u ) + cos ( 4 u )}

= \frac{- sin ( 4 u ) + cos ( 4 u )}{sin ( 4 u ) + cos ( 4 u )}

= \frac{cos ( 4 u ) - sin ( 4 u )}{cos ( 4 u ) + sin ( 4 u )}

调整右侧

\frac{1 - tan ( 4 u )}{1 + tan ( 4 u )}

用 sin, cos 表示

\frac{1 - tan ( 4 u )}{1 + tan ( 4 u )}

使用基本三角恒等式:

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

= \frac{1 - \frac{s i n ( 4 u )}{c o s ( 4 u )}}{1 + \frac{s i n ( 4 u )}{c o s ( 4 u )}}

化简

\frac{1 - \frac{s i n ( 4 u )}{c o s ( 4 u )}}{1 + \frac{s i n ( 4 u )}{c o s ( 4 u )}} : \frac{cos ( 4 u ) - sin ( 4 u )}{cos ( 4 u ) + sin ( 4 u )}

\frac{1 - \frac{s i n ( 4 u )}{c o s ( 4 u )}}{1 + \frac{s i n ( 4 u )}{c o s ( 4 u )}}

化简

1 + \frac{sin ( 4 u )}{cos ( 4 u )} : \frac{cos ( 4 u ) + sin ( 4 u )}{cos ( 4 u )}

1 + \frac{sin ( 4 u )}{cos ( 4 u )}

将项转换为分式:

1 = \frac{1 cos ( 4 u )}{cos ( 4 u )}

= \frac{1 \cdot cos ( 4 u )}{cos ( 4 u )} + \frac{sin ( 4 u )}{cos ( 4 u )}

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 ( 4 u ) + sin ( 4 u )}{cos ( 4 u )}

乘以:

1 \cdot cos (4 u) = cos (4 u)

= \frac{cos ( 4 u ) + sin ( 4 u )}{cos ( 4 u )}

= \frac{1 - \frac{s i n ( 4 u )}{c o s ( 4 u )}}{\frac{c o s ( 4 u ) + s i n ( 4 u )}{c o s ( 4 u )}}

化简

1 - \frac{sin ( 4 u )}{cos ( 4 u )} : \frac{cos ( 4 u ) - sin ( 4 u )}{cos ( 4 u )}

1 - \frac{sin ( 4 u )}{cos ( 4 u )}

将项转换为分式:

1 = \frac{1 cos ( 4 u )}{cos ( 4 u )}

= \frac{1 \cdot cos ( 4 u )}{cos ( 4 u )} - \frac{sin ( 4 u )}{cos ( 4 u )}

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 ( 4 u ) - sin ( 4 u )}{cos ( 4 u )}

乘以:

1 \cdot cos (4 u) = cos (4 u)

= \frac{cos ( 4 u ) - sin ( 4 u )}{cos ( 4 u )}

= \frac{\frac{c o s ( 4 u ) - s i n ( 4 u )}{c o s ( 4 u )}}{\frac{c o s ( 4 u ) + s i n ( 4 u )}{c o s ( 4 u )}}

分式相除:

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

= \frac{( cos ( 4 u ) - sin ( 4 u ) ) cos ( 4 u )}{cos ( 4 u ) ( cos ( 4 u ) + sin ( 4 u ) )}

约分:

cos (4 u)

= \frac{cos ( 4 u ) - sin ( 4 u )}{cos ( 4 u ) + sin ( 4 u )}

= \frac{cos ( 4 u ) - sin ( 4 u )}{cos ( 4 u ) + sin ( 4 u )}

= \frac{cos ( 4 u ) - sin ( 4 u )}{cos ( 4 u ) + sin ( 4 u )}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 sin(-x)=sin(x-pi)

p ro v e sin (- x) = sin (x - π)

证明 sin(4a)=4sin(a)cos(a)cos(2a)

p ro v e sin (4 a) = 4 sin (a) cos (a) cos (2 a)

证明 3cos^2(x)-2=1-3sin^2(x)

p ro v e 3 cos^{2} (x) - 2 = 1 - 3 sin^{2} (x)

证明 tan(a/2)=(sin(a))/(1+cos(a))

p ro v e tan (\frac{a}{2}) = \frac{sin ( a )}{1 + cos ( a )}

证明 cos(3t)=cos^3(t)-3sin^2(t)cos(t)

p ro v e cos (3 t) = cos^{3} (t) - 3 sin^{2} (t) cos (t)