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受欢迎的三角函数 >

证明 (sin(x)tan(x))/(cos(x)+1)=sec(x)-1

解答

证明

\frac{sin ( x ) tan ( x )}{cos ( x ) + 1} = sec (x) - 1

解答

真

求解步骤

\frac{sin ( x ) tan ( x )}{cos ( x ) + 1} = sec (x) - 1

调整左侧

\frac{sin ( x ) tan ( x )}{cos ( x ) + 1}

用 sin, cos 表示

\frac{sin ( x ) tan ( x )}{1 + cos ( x )}

使用基本三角恒等式:

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

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

化简

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

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

乘

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

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

分式相乘:

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

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

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

sin (x) sin (x)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

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

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

使用分式法则:

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

分解

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

1 - cos^{2} (x)

因式分解出通项

- 1

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

分解

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

cos^{2} (x) - 1

将

1

改写为

1^{2}

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (x) + 1) (cos (x) - 1)

= - (cos (x) + 1) (cos (x) - 1)

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

约分:

cos (x) + 1

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

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

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

乘开

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

- \frac{- 1 + cos ( x )}{cos ( x )}

使用分式法则:

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

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

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

去除括号:

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

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

使用法则

\frac{a}{a} = 1

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

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

化简

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

- 1 + \frac{1}{cos ( x )}

将项转换为分式:

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

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

乘以:

1 \cdot cos (x) = cos (x)

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

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

使用三角恒等式改写

使用基本三角恒等式:

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

\frac{1 - \frac{1}{s e c ( x )}}{\frac{1}{s e c ( x )}}

化简

\frac{1 - \frac{1}{s e c ( x )}}{\frac{1}{s e c ( x )}}

使用分式法则:

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

= \frac{( 1 - \frac{1}{s e c ( x )} ) sec ( x )}{1}

化简

1 - \frac{1}{sec ( x )} : \frac{sec ( x ) - 1}{sec ( x )}

1 - \frac{1}{sec ( x )}

将项转换为分式:

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

= \frac{1 \cdot sec ( x )}{sec ( x )} - \frac{1}{sec ( 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 sec ( x ) - 1}{sec ( x )}

乘以:

1 \cdot sec (x) = sec (x)

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

= \frac{\frac{s e c ( x ) - 1}{s e c ( x )} sec ( x )}{1}

使用分式法则:

\frac{a}{1} = a

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

分式相乘:

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

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

约分:

sec (x)

= sec (x) - 1

sec (x) - 1

sec (x) - 1

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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