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 >

(1+csc(A))(1-csc(A))=-cot(A)

解答

(1 + csc (A)) (1 - csc (A)) = - cot (A)

解答

A = \frac{π}{2} + πn, A = \frac{π}{4} + πn

+1

度数

A = 9 0^{\circ} + 18 0^{\circ} n, A = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

(1 + csc (A)) (1 - csc (A)) = - cot (A)

两边减去

- cot (A)

(1 + csc (A)) (1 - csc (A)) + cot (A) = 0

使用三角恒等式改写

cot (A) + (1 + csc (A)) (1 - csc (A))

(1 + csc (A)) (1 - csc (A)) = - cot^{2} (A)

(1 + csc (A)) (1 - csc (A))

乘开

(1 + csc (A)) (1 - csc (A)) : 1 - csc^{2} (A)

(1 + csc (A)) (1 - csc (A))

使用平方差公式:

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

a = 1, b = csc (A)

= 1^{2} - csc^{2} (A)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - csc^{2} (A)

= 1 - csc^{2} (A)

使用毕达哥拉斯恒等式:

csc^{2} (A) = 1 + cot^{2} (A)

csc^{2} (A) - 1 = cot^{2} (A)

= - cot^{2} (A)

= cot (A) - cot^{2} (A)

cot (A) - cot^{2} (A) = 0

用替代法求解

cot (A) - cot^{2} (A) = 0

令

: cot (A) = u

u - u^{2} = 0

u - u^{2} = 0 : u = 0, u = 1

u - u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

- u^{2} + u = 0

使用求根公式求解

- u^{2} + u = 0

二次方程求根公式:

若

a = - 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 0 = 1

1^{2} - 4 (- 1) \cdot 0

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 0

= 1 + 0

数字相加:

1 + 0 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- 1 + 1}{2 ( - 1 )}, u_{2} = \frac{- 1 - 1}{2 ( - 1 )}

u = \frac{- 1 + 1}{2 ( - 1 )} : 0

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

去除括号:

(- a) = - a

= \frac{- 1 + 1}{- 2 \cdot 1}

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{- 2}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{2}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

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

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

去除括号:

(- a) = - a

= \frac{- 1 - 1}{- 2 \cdot 1}

数字相减:

- 1 - 1 = - 2

= \frac{- 2}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 0, u = 1

u = cot (A)

代回

cot (A) = 0, cot (A) = 1

cot (A) = 0, cot (A) = 1

cot (A) = 0 : A = \frac{π}{2} + πn

cot (A) = 0

cot (A) = 0

的通解

cot (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

A = \frac{π}{2} + πn

A = \frac{π}{2} + πn

cot (A) = 1 : A = \frac{π}{4} + πn

cot (A) = 1

cot (A) = 1

的通解

cot (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

A = \frac{π}{4} + πn

A = \frac{π}{4} + πn

合并所有解

A = \frac{π}{2} + πn, A = \frac{π}{4} + πn

作图

流行的例子

6cos^2(3x)-cos(3x)-2=0

6 cos^{2} (3 x) - cos (3 x) - 2 = 0

cos(2x)-3cos(x)+1=0

cos (2 x) - 3 cos (x) + 1 = 0

2cos^2(x)-cos(x)-1=0,0<= x<= 360

2 cos^{2} (x) - cos (x) - 1 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

solvefor θ,x=tan(θ)

so l v e f or θ, x = tan (θ)

sin(5X)*cos(5X)=1

sin (5 X) \cdot cos (5 X) = 1