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受欢迎的三角函数 >

证明 1/(1-cos(x))-1/(1+cos(x))=2cot(x)csc(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

化简

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

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

将分母和分子乘以

- cos (x) + 1

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

对于

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

将分母和分子乘以

cos (x) + 1

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

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

乘开

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

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

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

- (- cos (x) + 1)

打开括号

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

使用加减运算法则

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

= cos (x) - 1

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

化简

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

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

对同类项分组

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

同类项相加:

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

= 2 cos (x) - 1 + 1

- 1 + 1 = 0

= 2 cos (x)

= 2 cos (x)

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

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

使用三角恒等式改写

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

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

调整右侧

2 cot (x) csc (x)

用 sin, cos 表示

2 cot (x) csc (x)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

2 \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 2}{sin ( x ) sin ( x )}

数字相乘:

1 \cdot 2 = 2

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

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

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

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 tan^2(t)= 1/(cos^2(t))-1

p ro v e tan^{2} (t) = \frac{1}{cos ^{2} ( t )} - 1

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p ro v e \frac{cot ( θ ) - tan ( θ )}{cot ( θ ) + tan ( θ )} = cos (2 θ)

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p ro v e cos (x) (sec (x) - 1) = 1 - cos (x)

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p ro v e csc^{2} (θ) (1 - cos^{2} (θ)) = - cos (\frac{π}{2})