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受欢迎的三角函数 >

证明 (1+cos(x))/(1-cos(x))-(1-cos(x))/(1+cos(x))=4cot(x)csc(x)

解答

证明

\frac{1 + cos ( x )}{1 - cos ( x )} - \frac{1 - cos ( x )}{1 + cos ( x )} = 4 cot (x) csc (x)

解答

真

求解步骤

\frac{1 + cos ( x )}{1 - cos ( x )} - \frac{1 - cos ( x )}{1 + cos ( x )} = 4 cot (x) csc (x)

调整左侧

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

化简

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

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

1 - cos (x), 1 + cos (x)

的最小公倍数:

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

1 - cos (x), 1 + cos (x)

最小公倍数 (LCM)

计算出由出现在

1 - cos (x)

或

1 + cos (x)

中的因子组成的表达式

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

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

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

对于

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

将分母和分子乘以

cos (x) + 1

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

对于

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

将分母和分子乘以

- cos (x) + 1

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

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

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

乘开

(1 + cos (x))^{2} - (1 - cos (x))^{2} : 4 cos (x)

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

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

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = cos (x)

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

数字相乘:

2 \cdot 1 = 2

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

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

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

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

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = cos (x)

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

数字相乘:

2 \cdot 1 = 2

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

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

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

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

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

打开括号

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

使用加减运算法则

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

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

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

化简

1 + 2 cos (x) + cos^{2} (x) - 1 + 2 cos (x) - cos^{2} (x) : 4 cos (x)

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

对同类项分组

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

同类项相加:

cos^{2} (x) - cos^{2} (x) = 0

= 2 cos (x) + 2 cos (x) + 1 - 1

同类项相加:

2 cos (x) + 2 cos (x) = 4 cos (x)

= 4 cos (x) + 1 - 1

1 - 1 = 0

= 4 cos (x)

= 4 cos (x)

= \frac{4 cos ( x )}{( - cos ( x ) + 1 ) ( cos ( x ) + 1 )}

= \frac{4 cos ( x )}{( 1 + cos ( x ) ) ( 1 - cos ( x ) )}

使用三角恒等式改写

\frac{4 cos ( x )}{( 1 + cos ( x ) ) ( 1 - cos ( x ) )}

乘开

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

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

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = cos (x)

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (x)

= \frac{4 cos ( x )}{1 - cos ^{2} ( x )}

使用毕达哥拉斯恒等式:

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

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

= \frac{4 cos ( x )}{sin ^{2} ( x )}

= \frac{4 cos ( x )}{sin ^{2} ( x )}

调整右侧

4 cot (x) csc (x)

用 sin, cos 表示

4 cot (x) csc (x)

使用基本三角恒等式:

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

= 4 \cdot \frac{cos ( x )}{sin ( x )} csc (x)

使用基本三角恒等式:

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

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

化简

4 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{sin ( x )} : \frac{4 cos ( x )}{sin ^{2} ( x )}

4 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{sin ( x )}

分式相乘:

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

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

数字相乘:

1 \cdot 4 = 4

= \frac{4 cos ( x )}{sin ( x ) sin ( 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{4 cos ( x )}{sin ^{2} ( x )}

= \frac{4 cos ( x )}{sin ^{2} ( x )}

= \frac{4 cos ( x )}{sin ^{2} ( x )}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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