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受欢迎的三角函数 >

证明 csc^2(x/2)=(2sec(x))/(sec(x)-1)

解答

证明

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

解答

真

求解步骤

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

令

: u = \frac{x}{2}

csc^{2} (u) = \frac{2 sec ( 2 u )}{sec ( 2 u ) - 1}

证明

csc^{2} (u) = \frac{2 sec ( 2 u )}{sec ( 2 u ) - 1} :

真

csc^{2} (u) = \frac{2 sec ( 2 u )}{sec ( 2 u ) - 1}

调整右侧

\frac{2 sec ( 2 u )}{sec ( 2 u ) - 1}

用 sin, cos 表示

\frac{2 sec ( 2 u )}{- 1 + sec ( 2 u )}

使用基本三角恒等式:

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

= \frac{2 \cdot \frac{1}{c o s ( 2 u )}}{- 1 + \frac{1}{c o s ( 2 u )}}

化简

\frac{2 \cdot \frac{1}{c o s ( 2 u )}}{- 1 + \frac{1}{c o s ( 2 u )}} : \frac{2}{- cos ( 2 u ) + 1}

\frac{2 \cdot \frac{1}{c o s ( 2 u )}}{- 1 + \frac{1}{c o s ( 2 u )}}

化简

- 1 + \frac{1}{cos ( 2 u )} : \frac{- cos ( 2 u ) + 1}{cos ( 2 u )}

- 1 + \frac{1}{cos ( 2 u )}

将项转换为分式:

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

= - \frac{1 \cdot cos ( 2 u )}{cos ( 2 u )} + \frac{1}{cos ( 2 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 ( 2 u ) + 1}{cos ( 2 u )}

乘以:

1 \cdot cos (2 u) = cos (2 u)

= \frac{- cos ( 2 u ) + 1}{cos ( 2 u )}

= \frac{2 \cdot \frac{1}{c o s ( 2 u )}}{\frac{- c o s ( 2 u ) + 1}{c o s ( 2 u )}}

乘

2 \cdot \frac{1}{cos ( 2 u )} : \frac{2}{cos ( 2 u )}

2 \cdot \frac{1}{cos ( 2 u )}

分式相乘:

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

= \frac{1 \cdot 2}{cos ( 2 u )}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{cos ( 2 u )}

= \frac{\frac{2}{c o s ( 2 u )}}{\frac{- c o s ( 2 u ) + 1}{c o s ( 2 u )}}

分式相除:

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

= \frac{2 cos ( 2 u )}{cos ( 2 u ) ( - cos ( 2 u ) + 1 )}

约分:

cos (2 u)

= \frac{2}{- cos ( 2 u ) + 1}

= \frac{2}{1 - cos ( 2 u )}

= \frac{2}{1 - cos ( 2 u )}

使用三角恒等式改写

\frac{2}{1 - cos ( 2 u )}

使用倍角公式:

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

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

化简

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

\frac{2}{1 - ( 1 - 2 sin ^{2} ( u ) )}

乘开

1 - (1 - 2 sin^{2} (u)) : 2 sin^{2} (u)

1 - (1 - 2 sin^{2} (u))

- (1 - 2 sin^{2} (u)) : - 1 + 2 sin^{2} (u)

- (1 - 2 sin^{2} (u))

打开括号

= - (1) - (- 2 sin^{2} (u))

使用加减运算法则

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (u)

= 1 - 1 + 2 sin^{2} (u)

1 - 1 = 0

= 2 sin^{2} (u)

= \frac{2}{2 sin ^{2} ( u )}

数字相除:

\frac{2}{2} = 1

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

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

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

使用三角恒等式改写

使用基本三角恒等式:

sin (x) = \frac{1}{csc ( x )}

\frac{1}{( \frac{1}{c s c ( u )} ) ^{2}}

化简

\frac{1}{( \frac{1}{c s c ( u )} ) ^{2}}

(\frac{1}{csc ( u )})^{2} = \frac{1}{csc ^{2} ( u )}

(\frac{1}{csc ( u )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{csc ^{2} ( u )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( u )}

= \frac{1}{\frac{1}{c s c ^{2} ( u )}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ^{2} ( u )}{1}

使用法则

\frac{a}{1} = a

= csc^{2} (u)

csc^{2} (u)

csc^{2} (u)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

因此

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

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 (sec(x))/(-csc(x))+(-sin(x))/(cos(x))=-2tan(x)

p ro v e \frac{sec ( x )}{- csc ( x )} + \frac{- sin ( x )}{cos ( x )} = - 2 tan (x)

证明 1=(sec(u)+tan(u))/(sec(u)+tan(u))

p ro v e 1 = \frac{sec ( u ) + tan ( u )}{sec ( u ) + tan ( u )}

证明 (sin^2(x))/1*1/(sin^2(x))=1

p ro v e \frac{sin ^{2} ( x )}{1} \cdot \frac{1}{sin ^{2} ( x )} = 1

证明 1/2 cot(x)-1/2 tan(x)=cot(2x)

p ro v e \frac{1}{2} cot (x) - \frac{1}{2} tan (x) = cot (2 x)

证明 cos(x+pi/6)=(sqrt(3))/2 cos(x)-1/2 sin(x)

p ro v e cos (x + \frac{π}{6}) = \frac{3}{2} cos (x) - \frac{1}{2} sin (x)