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 >

tan^2(γ)+1=cos^2(γ)

解答

tan^{2} (γ) + 1 = cos^{2} (γ)

解答

γ = 2 πn, γ = π + 2 πn

+1

度数

γ = 0^{\circ} + 36 0^{\circ} n, γ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

tan^{2} (γ) + 1 = cos^{2} (γ)

两边减去

cos^{2} (γ)

tan^{2} (γ) + 1 - cos^{2} (γ) = 0

使用三角恒等式改写

1 - cos^{2} (γ) + tan^{2} (γ)

使用毕达哥拉斯恒等式:

tan^{2} (x) + 1 = sec^{2} (x)

= - cos^{2} (γ) + sec^{2} (γ)

使用基本三角恒等式:

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

= - (\frac{1}{sec ( γ )})^{2} + sec^{2} (γ)

(\frac{1}{sec ( γ )})^{2} = \frac{1}{sec ^{2} ( γ )}

(\frac{1}{sec ( γ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sec ^{2} ( γ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( γ )}

= - \frac{1}{sec ^{2} ( γ )} + sec^{2} (γ)

- \frac{1}{sec ^{2} ( γ )} + sec^{2} (γ) = 0

用替代法求解

- \frac{1}{sec ^{2} ( γ )} + sec^{2} (γ) = 0

令

: sec (γ) = u

- \frac{1}{u ^{2}} + u^{2} = 0

- \frac{1}{u ^{2}} + u^{2} = 0 : u = 1, u = - 1, u = i, u = - i

- \frac{1}{u ^{2}} + u^{2} = 0

在两边乘以

u^{2}

- \frac{1}{u ^{2}} + u^{2} = 0

在两边乘以

u^{2}

- \frac{1}{u ^{2}} u^{2} + u^{2} u^{2} = 0 \cdot u^{2}

化简

- \frac{1}{u ^{2}} u^{2} + u^{2} u^{2} = 0 \cdot u^{2}

化简

- \frac{1}{u ^{2}} u^{2} : - 1

- \frac{1}{u ^{2}} u^{2}

分式相乘:

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

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

约分:

u^{2}

= - 1

化简

u^{2} u^{2} : u^{4}

u^{2} u^{2}

使用指数法则:

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

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

数字相加:

2 + 2 = 4

= u^{4}

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

- 1 + u^{4} = 0

- 1 + u^{4} = 0

- 1 + u^{4} = 0

解

- 1 + u^{4} = 0 : u = 1, u = - 1, u = i, u = - i

- 1 + u^{4} = 0

将

1

到右边

- 1 + u^{4} = 0

两边加上

1

- 1 + u^{4} + 1 = 0 + 1

化简

u^{4} = 1

u^{4} = 1

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} = 1

解

v^{2} = 1 : v = 1, v = - 1

v^{2} = 1

对于

(g (x))^{2} = f (a)

解为

g (x) = f (a), - f (a)

v = 1, v = - 1

v = 1, v = - 1

代回

v = u^{2}

，求解

u

解

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

u^{2} = 1

使用法则

1 = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

解

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

u^{2} = - 1

使用法则

1 = 1

u^{2} = - 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = - 1, u = - - 1

化简

- 1 : i

- 1

使用虚数运算法则:

- 1 = i

= i

化简

- - 1 : - i

- - 1

使用虚数运算法则:

- 1 = i

= - i

u = i, u = - i

解为

u = 1, u = - 1, u = i, u = - i

u = 1, u = - 1, u = i, u = - i

验证解

找到无定义的点（奇点）:

u = 0

取

- \frac{1}{u ^{2}} + u^{2}

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 1, u = - 1, u = i, u = - i

u = sec (γ)

代回

sec (γ) = 1, sec (γ) = - 1, sec (γ) = i, sec (γ) = - i

sec (γ) = 1, sec (γ) = - 1, sec (γ) = i, sec (γ) = - i

sec (γ) = 1 : γ = 2 πn

sec (γ) = 1

sec (γ) = 1

的通解

sec (x)

周期表（周期为

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

γ = 0 + 2 πn

γ = 0 + 2 πn

解

γ = 0 + 2 πn : γ = 2 πn

γ = 0 + 2 πn

0 + 2 πn = 2 πn

γ = 2 πn

γ = 2 πn

sec (γ) = - 1 : γ = π + 2 πn

sec (γ) = - 1

sec (γ) = - 1

的通解

sec (x)

周期表（周期为

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

γ = π + 2 πn

γ = π + 2 πn

sec (γ) = i :

无解

sec (γ) = i

无解

sec (γ) = - i :

无解

sec (γ) = - i

无解

合并所有解

γ = 2 πn, γ = π + 2 πn

作图

流行的例子

sin^2(x)+8sin(x)-9=0

sin^{2} (x) + 8 sin (x) - 9 = 0

-2sin^2(θ)+10sin(θ)-3=3sin(θ),0<= θ<360

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ), 0^{\circ} \leq θ < 36 0^{\circ}

cot^2(θ)+4csc(θ)=-5

cot^{2} (θ) + 4 csc (θ) = - 5

10sin(a)+7sin(a)+1sin(a)+8sin(a)=48

10 sin (a) + 7 sin (a) + 1 sin (a) + 8 sin (a) = 48

2cos^2(θ)+7sin(θ)=5

2 cos^{2} (θ) + 7 sin (θ) = 5