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(2θ)=sin(θ),0<= θ<2pi

解答

cos (2 θ) = sin (θ), 0 \leq θ < 2 π

解答

θ = \frac{3 π}{2}, θ = \frac{π}{6}, θ = \frac{5 π}{6}

+1

度数

θ = 27 0^{\circ}, θ = 3 0^{\circ}, θ = 15 0^{\circ}

求解步骤

cos (2 θ) = sin (θ), 0 \leq θ < 2 π

两边减去

sin (θ)

cos (2 θ) - sin (θ) = 0

使用三角恒等式改写

cos (2 θ) - sin (θ)

使用倍角公式:

cos (2 x) = 1 - 2 sin^{2} (x)

= 1 - 2 sin^{2} (θ) - sin (θ)

1 - sin (θ) - 2 sin^{2} (θ) = 0

用替代法求解

1 - sin (θ) - 2 sin^{2} (θ) = 0

令

: sin (θ) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 2, b = - 1, c = 1

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

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

(- 1)^{2} - 4 (- 2) \cdot 1 = 3

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 3}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 3}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 3}{2 ( - 2 )} : - 1

\frac{- ( - 1 ) + 3}{2 ( - 2 )}

去除括号:

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

= \frac{1 + 3}{- 2 \cdot 2}

数字相加:

1 + 3 = 4

= \frac{4}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{4}{- 4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 3}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 1 ) - 3}{2 ( - 2 )}

去除括号:

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

= \frac{1 - 3}{- 2 \cdot 2}

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

u = - 1, u = \frac{1}{2}

u = sin (θ)

代回

sin (θ) = - 1, sin (θ) = \frac{1}{2}

sin (θ) = - 1, sin (θ) = \frac{1}{2}

sin (θ) = - 1, 0 \leq θ < 2 π : θ = \frac{3 π}{2}

sin (θ) = - 1, 0 \leq θ < 2 π

sin (θ) = - 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

在

0 \leq θ < 2 π

范围内的解

θ = \frac{3 π}{2}

sin (θ) = \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{π}{6}, θ = \frac{5 π}{6}

sin (θ) = \frac{1}{2}, 0 \leq θ < 2 π

sin (θ) = \frac{1}{2}

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

在

0 \leq θ < 2 π

范围内的解

θ = \frac{π}{6}, θ = \frac{5 π}{6}

合并所有解

θ = \frac{3 π}{2}, θ = \frac{π}{6}, θ = \frac{5 π}{6}

作图

流行的例子

sin(x)(1+cos(x))=sin(x)+cos(x)

90^2=60^2+70^2-2(60)(70)(cos(x))

sin^2(x)+3sin(x)+7=5

cos(2x)=-0.6,-180<= x<= 180

1/(cos^2(θ/2))=12cos(θ)