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 1/(1+cos(x))+1/(1-cos(x))=2csc^2(x)

解答

证明

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

解答

真

求解步骤

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

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

- 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) = 0

= 1 + 1

数字相加:

1 + 1 = 2

= 2

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

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

使用三角恒等式改写

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

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

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

化简

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

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

使用分式法则:

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

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

使用法则

\frac{a}{1} = a

= 2 csc^{2} (x)

2 csc^{2} (x)

2 csc^{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 (csc(-x))/(sec(-x))=-cot(x)

p ro v e \frac{csc ( - x )}{sec ( - x )} = - cot (x)

证明 cos(x-y)+cos(x+y)=2cos(x)cos(y)

p ro v e cos (x - y) + cos (x + y) = 2 cos (x) cos (y)

证明 sin(pi/6+x)+sin(pi/6-x)=cos(x)

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

证明 (cot(x)-csc(x))(cos(x)+1)=-sin(x)

p ro v e (cot (x) - csc (x)) (cos (x) + 1) = - sin (x)

证明 csc^4(θ)-cot^4(θ)=2csc^2(θ)-1

p ro v e csc^{4} (θ) - cot^{4} (θ) = 2 csc^{2} (θ) - 1