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 >

solvefor θ,cos(θ)+cos(90-θ)=0

解答

求解

θ, cos (θ) + cos (9 0^{\circ} - θ) = 0

解答

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

+1

弧度

θ = \frac{3 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

求解步骤

cos (θ) + cos (9 0^{\circ} - θ) = 0

使用三角恒等式改写

cos (θ) + cos (9 0^{\circ} - θ)

使用和差化积恒等式:

cos (s) + cos (t) = 2 cos (\frac{s + t}{2}) cos (\frac{s - t}{2})

= 2 cos (\frac{θ + 9 0 ^{\circ} - θ}{2}) cos (\frac{θ - ( 9 0 ^{\circ} - θ )}{2})

化简

2 cos (\frac{θ + 9 0 ^{\circ} - θ}{2}) cos (\frac{θ - ( 9 0 ^{\circ} - θ )}{2}) : 2 cos (\frac{4 θ - 18 0 ^{\circ}}{4})

2 cos (\frac{θ + 9 0 ^{\circ} - θ}{2}) cos (\frac{θ - ( 9 0 ^{\circ} - θ )}{2})

\frac{θ + 9 0 ^{\circ} - θ}{2} = 4 5^{\circ}

\frac{θ + 9 0 ^{\circ} - θ}{2}

θ + 9 0^{\circ} - θ = 9 0^{\circ}

θ + 9 0^{\circ} - θ

对同类项分组

= θ - θ + 9 0^{\circ}

同类项相加:

θ - θ = 0

= 9 0^{\circ}

= \frac{9 0 ^{\circ}}{2}

使用分式法则:

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

= \frac{18 0 ^{\circ}}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= 4 5^{\circ}

= 2 cos (4 5^{\circ}) cos (\frac{θ - ( - θ + 9 0 ^{\circ} )}{2})

\frac{θ - ( 9 0 ^{\circ} - θ )}{2} = \frac{4 θ - 18 0 ^{\circ}}{4}

\frac{θ - ( 9 0 ^{\circ} - θ )}{2}

乘开

θ - (9 0^{\circ} - θ) : 2 θ - 9 0^{\circ}

θ - (9 0^{\circ} - θ)

- (9 0^{\circ} - θ) : - 9 0^{\circ} + θ

- (9 0^{\circ} - θ)

打开括号

= - (9 0^{\circ}) - (- θ)

使用加减运算法则

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

= - 9 0^{\circ} + θ

= θ - 9 0^{\circ} + θ

化简

θ - 9 0^{\circ} + θ : 2 θ - 9 0^{\circ}

θ - 9 0^{\circ} + θ

对同类项分组

= θ + θ - 9 0^{\circ}

同类项相加:

θ + θ = 2 θ

= 2 θ - 9 0^{\circ}

= 2 θ - 9 0^{\circ}

= \frac{2 θ - 9 0 ^{\circ}}{2}

化简

2 θ - 9 0^{\circ} : \frac{4 θ - 18 0 ^{\circ}}{2}

2 θ - 9 0^{\circ}

将项转换为分式:

2 θ = \frac{2 θ 2}{2}

= \frac{2 θ \cdot 2}{2} - 9 0^{\circ}

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{2 θ \cdot 2 - 18 0 ^{\circ}}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 θ - 18 0 ^{\circ}}{2}

= \frac{\frac{4 θ - 18 0 ^{\circ}}{2}}{2}

使用分式法则:

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

= \frac{4 θ - 18 0 ^{\circ}}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 θ - 18 0 ^{\circ}}{4}

= 2 cos (4 5^{\circ}) cos (\frac{4 θ - 18 0 ^{\circ}}{4})

化简

cos (4 5^{\circ}) : \frac{2}{2}

cos (4 5^{\circ})

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= 2 \cdot \frac{2}{2} cos (\frac{4 θ - 18 0 ^{\circ}}{4})

分式相乘:

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

= \frac{2 \cdot 2 cos ( \frac{4 θ - 18 0 ^{\circ}}{4} )}{2}

约分:

2

= 2 cos (\frac{4 θ - 18 0 ^{\circ}}{4})

= 2 cos (\frac{4 θ - 18 0 ^{\circ}}{4})

2 cos (\frac{4 θ - 18 0 ^{\circ}}{4}) = 0

两边除以

2

2 cos (\frac{4 θ - 18 0 ^{\circ}}{4}) = 0

两边除以

2

\frac{2 cos ( \frac{4 θ - 18 0 ^{\circ}}{4} )}{2} = \frac{0}{2}

化简

cos (\frac{4 θ - 18 0 ^{\circ}}{4}) = 0

cos (\frac{4 θ - 18 0 ^{\circ}}{4}) = 0

cos (\frac{4 θ - 18 0 ^{\circ}}{4}) = 0

的通解

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

\frac{4 θ - 18 0 ^{\circ}}{4} = 9 0^{\circ} + 36 0^{\circ} n, \frac{4 θ - 18 0 ^{\circ}}{4} = 27 0^{\circ} + 36 0^{\circ} n

\frac{4 θ - 18 0 ^{\circ}}{4} = 9 0^{\circ} + 36 0^{\circ} n, \frac{4 θ - 18 0 ^{\circ}}{4} = 27 0^{\circ} + 36 0^{\circ} n

解

\frac{4 θ - 18 0 ^{\circ}}{4} = 9 0^{\circ} + 36 0^{\circ} n : θ = 13 5^{\circ} + 36 0^{\circ} n

\frac{4 θ - 18 0 ^{\circ}}{4} = 9 0^{\circ} + 36 0^{\circ} n

在两边乘以

4

\frac{4 θ - 18 0 ^{\circ}}{4} = 9 0^{\circ} + 36 0^{\circ} n

在两边乘以

4

\frac{4 ( 4 θ - 18 0 ^{\circ} )}{4} = 4 \cdot 9 0^{\circ} + 4 \cdot 36 0^{\circ} n

化简

\frac{4 ( 4 θ - 18 0 ^{\circ} )}{4} = 4 \cdot 9 0^{\circ} + 4 \cdot 36 0^{\circ} n

化简

\frac{4 ( 4 θ - 18 0 ^{\circ} )}{4} : 4 θ - 18 0^{\circ}

\frac{4 ( 4 θ - 18 0 ^{\circ} )}{4}

数字相除:

\frac{4}{4} = 1

= 4 θ - 18 0^{\circ}

化简

4 \cdot 9 0^{\circ} + 4 \cdot 36 0^{\circ} n : 36 0^{\circ} + 144 0^{\circ} n

4 \cdot 9 0^{\circ} + 4 \cdot 36 0^{\circ} n

4 \cdot 9 0^{\circ} = 36 0^{\circ}

4 \cdot 9 0^{\circ}

分式相乘:

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

= 36 0^{\circ}

数字相除:

\frac{4}{2} = 2

= 36 0^{\circ}

4 \cdot 36 0^{\circ} n = 144 0^{\circ} n

4 \cdot 36 0^{\circ} n

数字相乘:

4 \cdot 2 = 8

= 144 0^{\circ} n

= 36 0^{\circ} + 144 0^{\circ} n

4 θ - 18 0^{\circ} = 36 0^{\circ} + 144 0^{\circ} n

4 θ - 18 0^{\circ} = 36 0^{\circ} + 144 0^{\circ} n

4 θ - 18 0^{\circ} = 36 0^{\circ} + 144 0^{\circ} n

将

18 0^{\circ}

到右边

4 θ - 18 0^{\circ} = 36 0^{\circ} + 144 0^{\circ} n

两边加上

18 0^{\circ}

4 θ - 18 0^{\circ} + 18 0^{\circ} = 36 0^{\circ} + 144 0^{\circ} n + 18 0^{\circ}

化简

4 θ = 54 0^{\circ} + 144 0^{\circ} n

4 θ = 54 0^{\circ} + 144 0^{\circ} n

两边除以

4

4 θ = 54 0^{\circ} + 144 0^{\circ} n

两边除以

4

\frac{4 θ}{4} = 13 5^{\circ} + \frac{144 0 ^{\circ} n}{4}

化简

θ = 13 5^{\circ} + 36 0^{\circ} n

θ = 13 5^{\circ} + 36 0^{\circ} n

解

\frac{4 θ - 18 0 ^{\circ}}{4} = 27 0^{\circ} + 36 0^{\circ} n : θ = 31 5^{\circ} + 36 0^{\circ} n

\frac{4 θ - 18 0 ^{\circ}}{4} = 27 0^{\circ} + 36 0^{\circ} n

在两边乘以

4

\frac{4 θ - 18 0 ^{\circ}}{4} = 27 0^{\circ} + 36 0^{\circ} n

在两边乘以

4

\frac{4 ( 4 θ - 18 0 ^{\circ} )}{4} = 4 \cdot 27 0^{\circ} + 4 \cdot 36 0^{\circ} n

化简

\frac{4 ( 4 θ - 18 0 ^{\circ} )}{4} = 4 \cdot 27 0^{\circ} + 4 \cdot 36 0^{\circ} n

化简

\frac{4 ( 4 θ - 18 0 ^{\circ} )}{4} : 4 θ - 18 0^{\circ}

\frac{4 ( 4 θ - 18 0 ^{\circ} )}{4}

数字相除:

\frac{4}{4} = 1

= 4 θ - 18 0^{\circ}

化简

4 \cdot 27 0^{\circ} + 4 \cdot 36 0^{\circ} n : 108 0^{\circ} + 144 0^{\circ} n

4 \cdot 27 0^{\circ} + 4 \cdot 36 0^{\circ} n

4 \cdot 27 0^{\circ} = 108 0^{\circ}

4 \cdot 27 0^{\circ}

分式相乘:

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

= 108 0^{\circ}

数字相乘:

3 \cdot 4 = 12

= 108 0^{\circ}

数字相除:

\frac{12}{2} = 6

= 108 0^{\circ}

4 \cdot 36 0^{\circ} n = 144 0^{\circ} n

4 \cdot 36 0^{\circ} n

数字相乘:

4 \cdot 2 = 8

= 144 0^{\circ} n

= 108 0^{\circ} + 144 0^{\circ} n

4 θ - 18 0^{\circ} = 108 0^{\circ} + 144 0^{\circ} n

4 θ - 18 0^{\circ} = 108 0^{\circ} + 144 0^{\circ} n

4 θ - 18 0^{\circ} = 108 0^{\circ} + 144 0^{\circ} n

将

18 0^{\circ}

到右边

4 θ - 18 0^{\circ} = 108 0^{\circ} + 144 0^{\circ} n

两边加上

18 0^{\circ}

4 θ - 18 0^{\circ} + 18 0^{\circ} = 108 0^{\circ} + 144 0^{\circ} n + 18 0^{\circ}

化简

4 θ = 126 0^{\circ} + 144 0^{\circ} n

4 θ = 126 0^{\circ} + 144 0^{\circ} n

两边除以

4

4 θ = 126 0^{\circ} + 144 0^{\circ} n

两边除以

4

\frac{4 θ}{4} = 31 5^{\circ} + \frac{144 0 ^{\circ} n}{4}

化简

θ = 31 5^{\circ} + 36 0^{\circ} n

θ = 31 5^{\circ} + 36 0^{\circ} n

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

作图

流行的例子

sin(A)-0.6*cos(A)=(2.77)/(9.8)

sin (A) - 0.6 \cdot cos (A) = \frac{2.77}{9.8}

sin (x) = - 0.397

2cos(x)*tan(x)-1=0

2 cos (x) \cdot tan (x) - 1 = 0

sin (x) = - 0.465

cos (x) = \frac{10}{17}