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受欢迎的三角函数 >

sin(216)=sin(θ)

解答

sin (21 6^{\circ}) = sin (θ)

解答

θ = - 0.62831 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} + 0.62831 \dots + 36 0^{\circ} n

+1

弧度

θ = - 0.62831 \dots + 2 πn, θ = π + 0.62831 \dots + 2 πn

求解步骤

sin (21 6^{\circ}) = sin (θ)

sin (21 6^{\circ}) = - \frac{2 5 - 5}{4}

sin (21 6^{\circ})

使用三角恒等式改写:

- sin (14 4^{\circ})

sin (21 6^{\circ})

利用以下特性:

sin (x) = - sin (36 0^{\circ} - x)

sin (x)

利用以下特性:

sin (θ) = - sin (- θ)

sin (x) = - sin (- x)

= - sin (- x)

使用周期

sin

:

sin (36 0^{\circ} + θ) = sin (θ)

- sin (- x) = - sin (36 0^{\circ} - x)

= - sin (36 0^{\circ} - x)

= - sin (36 0^{\circ} - 21 6^{\circ})

化简:

36 0^{\circ} - 21 6^{\circ} = 14 4^{\circ}

36 0^{\circ} - 21 6^{\circ}

将项转换为分式:

36 0^{\circ} = 36 0^{\circ}

= 36 0^{\circ} - 21 6^{\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{36 0 ^{\circ} 5 - 108 0 ^{\circ}}{5}

36 0^{\circ} 5 - 108 0^{\circ} = 72 0^{\circ}

36 0^{\circ} 5 - 108 0^{\circ}

数字相乘:

2 \cdot 5 = 10

= 180 0^{\circ} - 108 0^{\circ}

同类项相加:

180 0^{\circ} - 108 0^{\circ} = 72 0^{\circ}

= 72 0^{\circ}

= 14 4^{\circ}

= - sin (14 4^{\circ})

= - sin (14 4^{\circ})

使用三角恒等式改写:

sin (14 4^{\circ}) = \frac{2 5 - 5}{4}

sin (14 4^{\circ})

使用三角恒等式改写:

sin (3 6^{\circ})

sin (14 4^{\circ})

使用基本三角恒等式:

sin (x) = sin (18 0^{\circ} - x)

= sin (18 0^{\circ} - 14 4^{\circ})

化简:

18 0^{\circ} - 14 4^{\circ} = 3 6^{\circ}

18 0^{\circ} - 14 4^{\circ}

将项转换为分式:

18 0^{\circ} = 18 0^{\circ}

= 18 0^{\circ} - 14 4^{\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{18 0 ^{\circ} 5 - 72 0 ^{\circ}}{5}

同类项相加:

90 0^{\circ} - 72 0^{\circ} = 18 0^{\circ}

= 3 6^{\circ}

= sin (3 6^{\circ})

= sin (3 6^{\circ})

使用三角恒等式改写:

\frac{2 5 - 5}{4}

sin (3 6^{\circ})

显示:

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

使用以下积化和差公式:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = sin (5 4^{\circ}) - sin (1 8^{\circ})

显示:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

利用以下特性:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

两边除以

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

代入

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

\frac{1}{2} = sin (5 4^{\circ}) - sin (1 8^{\circ})

sin (5 4^{\circ}) = cos (9 0^{\circ} - 5 4^{\circ})

\frac{1}{2} = cos (9 0^{\circ} - 5 4^{\circ}) - sin (1 8^{\circ})

\frac{1}{2} = cos (3 6^{\circ}) - sin (1 8^{\circ})

显示:

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

使用因式分解法则:

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

a = cos (3 6^{\circ}) + sin (1 8^{\circ})

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = ((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) ((cos (3 6^{\circ}) + sin (1 8^{\circ})) - (cos (3 6^{\circ}) - sin (1 8^{\circ})))

整理后得

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 2 (2 cos (3 6^{\circ}) sin (1 8^{\circ}))

显示:

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (7 2^{\circ}) = 2 sin (3 6^{\circ}) cos (3 6^{\circ})

sin (7 2^{\circ}) sin (3 6^{\circ}) = 4 sin (3 6^{\circ}) sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

sin (3 6^{\circ})

sin (7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

利用以下特性:

sin (x) = cos (9 0^{\circ} - x)

sin (7 2^{\circ}) = cos (9 0^{\circ} - 7 2^{\circ})

cos (9 0^{\circ} - 7 2^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

cos (1 8^{\circ}) = 4 sin (1 8^{\circ}) cos (3 6^{\circ}) cos (1 8^{\circ})

两边除以

cos (1 8^{\circ})

1 = 4 sin (1 8^{\circ}) cos (3 6^{\circ})

两边除以

2

\frac{1}{2} = 2 sin (1 8^{\circ}) cos (3 6^{\circ})

代入

2 cos (3 6^{\circ}) sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (cos (3 6^{\circ}) - sin (1 8^{\circ}))^{2} = 1

代入

cos (3 6^{\circ}) - sin (1 8^{\circ}) = \frac{1}{2}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - (\frac{1}{2})^{2} = 1

整理后得

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} = 1

两边加上

\frac{1}{4}

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

整理后得

(cos (3 6^{\circ}) + sin (1 8^{\circ}))^{2} = \frac{5}{4}

在两侧开平方

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \pm \frac{5}{4}

cos (3 6^{\circ})

不能为负

sin (1 8^{\circ})

不能为负

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{4}

以下方程式相加

cos (3 6^{\circ}) + sin (1 8^{\circ}) = \frac{5}{2}

((cos (3 6^{\circ}) + sin (1 8^{\circ})) + (cos (3 6^{\circ}) - sin (1 8^{\circ}))) = (\frac{5}{2} + \frac{1}{2})

整理后得

cos (3 6^{\circ}) = \frac{5 + 1}{4}

两边进行平方

(cos (3 6^{\circ}))^{2} = (\frac{5 + 1}{4})^{2}

利用以下特性:

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

sin^{2} (3 6^{\circ}) = 1 - cos^{2} (3 6^{\circ})

代入

cos (3 6^{\circ}) = \frac{5 + 1}{4}

sin^{2} (3 6^{\circ}) = 1 - (\frac{5 + 1}{4})^{2}

整理后得

sin^{2} (3 6^{\circ}) = \frac{5 - 5}{8}

在两侧开平方

sin (3 6^{\circ}) = \pm \frac{5 - 5}{8}

sin (3 6^{\circ})

不能为负

sin (3 6^{\circ}) = \frac{5 - 5}{8}

整理后得

sin (3 6^{\circ}) = \frac{\frac{5 - 5}{2}}{2}

= \frac{\frac{5 - 5}{2}}{2}

\frac{\frac{5 - 5}{2}}{2} = \frac{2 5 - 5}{4}

\frac{\frac{5 - 5}{2}}{2}

\frac{5 - 5}{2} = \frac{5 - 5}{2}

\frac{5 - 5}{2}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{5 - 5}{2}

= \frac{\frac{5 - 5}{2}}{2}

使用分式法则:

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

= \frac{5 - 5}{2 \cdot 2}

\frac{5 - 5}{2 2}

有理化:

\frac{2 5 - 5}{4}

\frac{5 - 5}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{5 - 5 2}{2 \cdot 2 2}

2 \cdot 22 = 4

2 \cdot 22

使用指数法则:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

= 2^{1 + \frac{1}{2} + \frac{1}{2}}

同类项相加:

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

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

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

2 \cdot \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 5 - 5}{4}

= \frac{2 5 - 5}{4}

= \frac{2 5 - 5}{4}

= \frac{2 5 - 5}{4}

= - \frac{2 5 - 5}{4}

- \frac{2 5 - 5}{4} = sin (θ)

交换两边

sin (θ) = - \frac{2 5 - 5}{4}

使用反三角函数性质

sin (θ) = - \frac{2 5 - 5}{4}

sin (θ) = - \frac{2 5 - 5}{4}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 36 0^{\circ} n, x = 18 0^{\circ} + arcsin (a) + 36 0^{\circ} n

θ = arcsin (- \frac{2 5 - 5}{4}) + 36 0^{\circ} n, θ = 18 0^{\circ} + arcsin (\frac{2 5 - 5}{4}) + 36 0^{\circ} n

θ = arcsin (- \frac{2 5 - 5}{4}) + 36 0^{\circ} n, θ = 18 0^{\circ} + arcsin (\frac{2 5 - 5}{4}) + 36 0^{\circ} n

以小数形式表示解

θ = - 0.62831 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} + 0.62831 \dots + 36 0^{\circ} n

作图

流行的例子

cos (θ) = - 0.2

sin^2(x)-5cos(x)=3

sin^{2} (x) - 5 cos (x) = 3

cos (x) = - \frac{9}{41}

cos(θ)=-0.42,0<= θ<360

cos (θ) = - 0.42, 0^{\circ} \leq θ < 36 0^{\circ}

sin(2x)= 1/2 ,0<x<2pi

sin (2 x) = \frac{1}{2}, 0 < x < 2 π