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受欢迎的三角函数 >

sin(3v)=sin(18)

解答

sin (3 v) = sin (1 8^{\circ})

解答

v = \frac{0.31415 \dots}{3} + \frac{36 0 ^{\circ} n}{3}, v = 6 0^{\circ} - \frac{0.31415 \dots}{3} + \frac{36 0 ^{\circ} n}{3}

+1

弧度

v = \frac{0.31415 \dots}{3} + \frac{2 π}{3} n, v = \frac{π}{3} - \frac{0.31415 \dots}{3} + \frac{2 π}{3} n

求解步骤

sin (3 v) = sin (1 8^{\circ})

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

sin (1 8^{\circ})

使用三角恒等式改写:

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

sin (1 8^{\circ})

将

sin (1 8^{\circ})

写为

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

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

使用半角公式:

sin (\frac{θ}{2}) = \frac{1 - cos ( θ )}{2}

使用倍角公式

cos (2 θ) = 1 - 2 sin^{2} (θ)

用

\frac{θ}{2}

替代

θ

cos (θ) = 1 - 2 sin^{2} (\frac{θ}{2})

交换两边

2 sin^{2} (\frac{θ}{2}) = 1 - cos (θ)

两边除以

2

sin^{2} (\frac{θ}{2}) = \frac{( 1 - cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

:

range [0, 9 0^{\circ}] [9 0^{\circ}, 18 0^{\circ}] [18 0^{\circ}, 27 0^{\circ}] [27 0^{\circ}, 36 0^{\circ}] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

sin (\frac{θ}{2}) = \frac{( 1 - cos ( θ ) )}{2}

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

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

使用三角恒等式改写:

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

cos (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}

= \frac{5 + 1}{4}

= \frac{1 - \frac{5 + 1}{4}}{2}

化简

\frac{1 - \frac{5 + 1}{4}}{2} : \frac{2 3 - 5}{4}

\frac{1 - \frac{5 + 1}{4}}{2}

\frac{1 - \frac{5 + 1}{4}}{2} = \frac{3 - 5}{8}

\frac{1 - \frac{5 + 1}{4}}{2}

化简

1 - \frac{5 + 1}{4} : \frac{3 - 5}{4}

1 - \frac{5 + 1}{4}

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} - \frac{5 + 1}{4}

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{1 \cdot 4 - ( 5 + 1 )}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{4 - ( 1 + 5 )}{4}

乘开

4 - (5 + 1) : 3 - 5

4 - (5 + 1)

- (5 + 1) : - 5 - 1

- (5 + 1)

打开括号

= - (5) - (1)

使用加减运算法则

+ (- a) = - a

= - 5 - 1

= 4 - 5 - 1

数字相减:

4 - 1 = 3

= 3 - 5

= \frac{3 - 5}{4}

= \frac{\frac{3 - 5}{4}}{2}

使用分式法则:

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

= \frac{3 - 5}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{3 - 5}{8}

= \frac{3 - 5}{8}

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{3 - 5}{8}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= \frac{3 - 5}{2 2}

\frac{3 - 5}{2 2}

有理化:

\frac{2 3 - 5}{4}

\frac{3 - 5}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{3 - 5 2}{2 2 2}

222 = 4

222

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 3 - 5}{4}

= \frac{2 3 - 5}{4}

= \frac{2 3 - 5}{4}

sin (3 v) = \frac{2 3 - 5}{4}

使用反三角函数性质

sin (3 v) = \frac{2 3 - 5}{4}

sin (3 v) = \frac{2 3 - 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

3 v = arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n, 3 v = 18 0^{\circ} - arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n

3 v = arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n, 3 v = 18 0^{\circ} - arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n

解

3 v = arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n : v = \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

3 v = arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n

化简

arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n : arcsin (\frac{3 - 5}{2 2}) + 36 0^{\circ} n

arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n

\frac{2 3 - 5}{4} = \frac{3 - 5}{2 2}

\frac{2 3 - 5}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 3 - 5}{2 ^{2}}

消掉

\frac{2 3 - 5}{2 ^{2}} : \frac{3 - 5}{2 ^{\frac{3}{2}}}

\frac{2 3 - 5}{2 ^{2}}

使用根式运算法则:

n a = a^{\frac{1}{n}}

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

= \frac{2 ^{\frac{1}{2}} 3 - 5}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{3 - 5}{2 ^{2 - \frac{1}{2}}}

数字相减:

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

= \frac{3 - 5}{2 ^{\frac{3}{2}}}

= \frac{3 - 5}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

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

使用指数法则:

x^{a + b} = x^{a} x^{b}

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

整理后得

= 22

= \frac{3 - 5}{2 2}

= arcsin (\frac{3 - 5}{2 2}) + 36 0^{\circ} n

3 v = arcsin (\frac{3 - 5}{2 2}) + 36 0^{\circ} n

两边除以

3

3 v = arcsin (\frac{3 - 5}{2 2}) + 36 0^{\circ} n

两边除以

3

\frac{3 v}{3} = \frac{arcsin ( \frac{3 - 5}{2 2} )}{3} + \frac{36 0 ^{\circ} n}{3}

化简

\frac{3 v}{3} = \frac{arcsin ( \frac{3 - 5}{2 2} )}{3} + \frac{36 0 ^{\circ} n}{3}

化简

\frac{3 v}{3} : v

\frac{3 v}{3}

数字相除:

\frac{3}{3} = 1

= v

化简

\frac{arcsin ( \frac{3 - 5}{2 2} )}{3} + \frac{36 0 ^{\circ} n}{3} : \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

\frac{arcsin ( \frac{3 - 5}{2 2} )}{3} + \frac{36 0 ^{\circ} n}{3}

arcsin (\frac{3 - 5}{2 2}) = arcsin (\frac{2 3 - 5}{4})

arcsin (\frac{3 - 5}{2 2})

= arcsin (\frac{2 3 - 5}{4})

= \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

v = \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

v = \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

v = \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

解

3 v = 18 0^{\circ} - arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n : v = 6 0^{\circ} - \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

3 v = 18 0^{\circ} - arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n

化简

18 0^{\circ} - arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n : 18 0^{\circ} - arcsin (\frac{3 - 5}{2 2}) + 36 0^{\circ} n

18 0^{\circ} - arcsin (\frac{2 3 - 5}{4}) + 36 0^{\circ} n

\frac{2 3 - 5}{4} = \frac{3 - 5}{2 2}

\frac{2 3 - 5}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 3 - 5}{2 ^{2}}

消掉

\frac{2 3 - 5}{2 ^{2}} : \frac{3 - 5}{2 ^{\frac{3}{2}}}

\frac{2 3 - 5}{2 ^{2}}

使用根式运算法则:

n a = a^{\frac{1}{n}}

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

= \frac{2 ^{\frac{1}{2}} 3 - 5}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{3 - 5}{2 ^{2 - \frac{1}{2}}}

数字相减:

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

= \frac{3 - 5}{2 ^{\frac{3}{2}}}

= \frac{3 - 5}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

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

使用指数法则:

x^{a + b} = x^{a} x^{b}

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

整理后得

= 22

= \frac{3 - 5}{2 2}

= 18 0^{\circ} - arcsin (\frac{3 - 5}{2 2}) + 36 0^{\circ} n

3 v = 18 0^{\circ} - arcsin (\frac{3 - 5}{2 2}) + 36 0^{\circ} n

两边除以

3

3 v = 18 0^{\circ} - arcsin (\frac{3 - 5}{2 2}) + 36 0^{\circ} n

两边除以

3

\frac{3 v}{3} = 6 0^{\circ} - \frac{arcsin ( \frac{3 - 5}{2 2} )}{3} + \frac{36 0 ^{\circ} n}{3}

化简

\frac{3 v}{3} = 6 0^{\circ} - \frac{arcsin ( \frac{3 - 5}{2 2} )}{3} + \frac{36 0 ^{\circ} n}{3}

化简

\frac{3 v}{3} : v

\frac{3 v}{3}

数字相除:

\frac{3}{3} = 1

= v

化简

6 0^{\circ} - \frac{arcsin ( \frac{3 - 5}{2 2} )}{3} + \frac{36 0 ^{\circ} n}{3} : 6 0^{\circ} - \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

6 0^{\circ} - \frac{arcsin ( \frac{3 - 5}{2 2} )}{3} + \frac{36 0 ^{\circ} n}{3}

arcsin (\frac{3 - 5}{2 2}) = arcsin (\frac{2 3 - 5}{4})

arcsin (\frac{3 - 5}{2 2})

= arcsin (\frac{2 3 - 5}{4})

= 6 0^{\circ} - \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

v = 6 0^{\circ} - \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

v = 6 0^{\circ} - \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

v = 6 0^{\circ} - \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

v = \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}, v = 6 0^{\circ} - \frac{arcsin ( \frac{2 3 - 5}{4} )}{3} + \frac{36 0 ^{\circ} n}{3}

以小数形式表示解

v = \frac{0.31415 \dots}{3} + \frac{36 0 ^{\circ} n}{3}, v = 6 0^{\circ} - \frac{0.31415 \dots}{3} + \frac{36 0 ^{\circ} n}{3}

作图

流行的例子

2sqrt(2)cos(θ)-2=0

22 cos (θ) - 2 = 0

tan (x) = \frac{22}{326}

3 cos (6 x) = 2

2cos(t)=1+sin(t)

2 cos (t) = 1 + sin (t)

sin(x)-sqrt(3)cos(x)=-1

sin (x) - 3 cos (x) = - 1