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 >

cos(5θ)=sin(3θ)

解答

cos (5 θ) = sin (3 θ)

解答

θ = \frac{π + 4 πn}{16}, θ = - \frac{π + 4 πn}{4}

+1

度数

θ = 11.2 5^{\circ} + 4 5^{\circ} n, θ = - 4 5^{\circ} - 18 0^{\circ} n

求解步骤

cos (5 θ) = sin (3 θ)

使用三角恒等式改写

cos (5 θ) = sin (3 θ)

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

cos (5 θ) = sin (\frac{π}{2} - 5 θ)

cos (5 θ) = sin (\frac{π}{2} - 5 θ)

使用反三角函数性质

cos (5 θ) = sin (\frac{π}{2} - 5 θ)

sin (x) = sin (y) \Rightarrow x = y + 2 π n, x = π - y + 2 π n

3 θ = \frac{π}{2} - 5 θ + 2 πn, 3 θ = π - (\frac{π}{2} - 5 θ) + 2 πn

3 θ = \frac{π}{2} - 5 θ + 2 πn, 3 θ = π - (\frac{π}{2} - 5 θ) + 2 πn

3 θ = \frac{π}{2} - 5 θ + 2 πn : θ = \frac{π + 4 πn}{16}

3 θ = \frac{π}{2} - 5 θ + 2 πn

将

5 θ

para o lado esquerdo

3 θ = \frac{π}{2} - 5 θ + 2 πn

两边加上

5 θ

3 θ + 5 θ = \frac{π}{2} - 5 θ + 2 πn + 5 θ

化简

8 θ = \frac{π}{2} + 2 πn

8 θ = \frac{π}{2} + 2 πn

两边除以

8

8 θ = \frac{π}{2} + 2 πn

两边除以

8

\frac{8 θ}{8} = \frac{\frac{π}{2}}{8} + \frac{2 πn}{8}

化简

\frac{8 θ}{8} = \frac{\frac{π}{2}}{8} + \frac{2 πn}{8}

化简

\frac{8 θ}{8} : θ

\frac{8 θ}{8}

数字相除:

\frac{8}{8} = 1

= θ

化简

\frac{\frac{π}{2}}{8} + \frac{2 πn}{8} : \frac{π + 4 πn}{16}

\frac{\frac{π}{2}}{8} + \frac{2 πn}{8}

使用法则

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

= \frac{\frac{π}{2} + 2 πn}{8}

化简

\frac{π}{2} + 2 πn : \frac{π + 4 πn}{2}

\frac{π}{2} + 2 πn

将项转换为分式:

2 πn = \frac{2 πn 2}{2}

= \frac{π}{2} + \frac{2 πn \cdot 2}{2}

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{π + 2 πn \cdot 2}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{π + 4 πn}{2}

= \frac{\frac{π + 4 πn}{2}}{8}

使用分式法则:

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

= \frac{π + 4 πn}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{π + 4 πn}{16}

θ = \frac{π + 4 πn}{16}

θ = \frac{π + 4 πn}{16}

θ = \frac{π + 4 πn}{16}

3 θ = π - (\frac{π}{2} - 5 θ) + 2 πn : θ = - \frac{π + 4 πn}{4}

3 θ = π - (\frac{π}{2} - 5 θ) + 2 πn

展开

π - (\frac{π}{2} - 5 θ) + 2 πn : π - \frac{π}{2} + 5 θ + 2 πn

π - (\frac{π}{2} - 5 θ) + 2 πn

- (\frac{π}{2} - 5 θ) : - \frac{π}{2} + 5 θ

- (\frac{π}{2} - 5 θ)

打开括号

= - (\frac{π}{2}) - (- 5 θ)

使用加减运算法则

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

= - \frac{π}{2} + 5 θ

= π - \frac{π}{2} + 5 θ + 2 πn

3 θ = π - \frac{π}{2} + 5 θ + 2 πn

将

5 θ

para o lado esquerdo

3 θ = π - \frac{π}{2} + 5 θ + 2 πn

两边减去

5 θ

3 θ - 5 θ = π - \frac{π}{2} + 5 θ + 2 πn - 5 θ

化简

- 2 θ = π - \frac{π}{2} + 2 πn

- 2 θ = π - \frac{π}{2} + 2 πn

两边除以

- 2

- 2 θ = π - \frac{π}{2} + 2 πn

两边除以

- 2

\frac{- 2 θ}{- 2} = \frac{π}{- 2} - \frac{\frac{π}{2}}{- 2} + \frac{2 πn}{- 2}

化简

\frac{- 2 θ}{- 2} = \frac{π}{- 2} - \frac{\frac{π}{2}}{- 2} + \frac{2 πn}{- 2}

化简

\frac{- 2 θ}{- 2} : θ

\frac{- 2 θ}{- 2}

使用分式法则:

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

= \frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{π}{- 2} - \frac{\frac{π}{2}}{- 2} + \frac{2 πn}{- 2} : - \frac{π + 4 πn}{4}

\frac{π}{- 2} - \frac{\frac{π}{2}}{- 2} + \frac{2 πn}{- 2}

使用法则

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

= \frac{π - \frac{π}{2} + 2 πn}{- 2}

使用分式法则:

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

= - \frac{π - \frac{π}{2} + 2 πn}{2}

化简

π - \frac{π}{2} + 2 πn : \frac{π + 4 πn}{2}

π - \frac{π}{2} + 2 πn

将项转换为分式:

π = \frac{π 2}{2}, 2 πn = \frac{2 πn 2}{2}

= \frac{π 2}{2} - \frac{π}{2} + \frac{2 πn \cdot 2}{2}

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{π 2 - π + 2 πn \cdot 2}{2}

π 2 - π + 2 πn \cdot 2 = π + 4 πn

π 2 - π + 2 πn \cdot 2

同类项相加:

2 π - π = π

= π + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= π + 4 πn

= \frac{π + 4 πn}{2}

= - \frac{\frac{π + 4 πn}{2}}{2}

化简

\frac{\frac{π + 4 πn}{2}}{2} : \frac{π + 4 πn}{4}

\frac{\frac{π + 4 πn}{2}}{2}

使用分式法则:

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

= \frac{π + 4 πn}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π + 4 πn}{4}

= - \frac{π + 4 πn}{4}

θ = - \frac{π + 4 πn}{4}

θ = - \frac{π + 4 πn}{4}

θ = - \frac{π + 4 πn}{4}

θ = \frac{π + 4 πn}{16}, θ = - \frac{π + 4 πn}{4}

θ = \frac{π + 4 πn}{16}, θ = - \frac{π + 4 πn}{4}

作图

流行的例子

(2sin(2x)+2cos(2x))^2=4

(2 sin (2 x) + 2 cos (2 x))^{2} = 4

sin (x) = \frac{7}{3}

0 = sin (4 x)

solvefor x,sin(x)=sin(2x)

so l v e f or x, sin (x) = sin (2 x)

-7tan(A)-14=tan(A)-6

- 7 tan (A) - 14 = tan (A) - 6