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受欢迎的三角函数 >

sin((3pi)/(10))-sin(pi/(10))

解答

sin (\frac{3 π}{10}) - sin (\frac{π}{10})

解答

\frac{2 ( 1 + 5 ) 3 - 5}{8}

+1

十进制

0.5

求解步骤

sin (\frac{3 π}{10}) - sin (\frac{π}{10})

使用三角恒等式改写:

2 sin (\frac{π}{10}) cos (\frac{π}{5})

sin (\frac{3 π}{10}) - sin (\frac{π}{10})

使用和差化积恒等式:

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

= 2 sin (\frac{\frac{3 π}{10} - \frac{π}{10}}{2}) cos (\frac{\frac{3 π}{10} + \frac{π}{10}}{2})

化简:

\frac{\frac{3 π}{10} - \frac{π}{10}}{2} = \frac{π}{10}

\frac{\frac{3 π}{10} - \frac{π}{10}}{2}

合并分式

\frac{3 π}{10} - \frac{π}{10} : \frac{π}{5}

使用法则

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

= \frac{3 π - π}{10}

同类项相加:

3 π - π = 2 π

= \frac{2 π}{10}

约分:

2

= \frac{π}{5}

= \frac{\frac{π}{5}}{2}

使用分式法则:

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

= \frac{π}{5 \cdot 2}

数字相乘:

5 \cdot 2 = 10

= \frac{π}{10}

化简:

\frac{\frac{3 π}{10} + \frac{π}{10}}{2} = \frac{π}{5}

\frac{\frac{3 π}{10} + \frac{π}{10}}{2}

合并分式

\frac{3 π}{10} + \frac{π}{10} : \frac{2 π}{5}

使用法则

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

= \frac{3 π + π}{10}

同类项相加:

3 π + π = 4 π

= \frac{4 π}{10}

约分:

2

= \frac{2 π}{5}

= \frac{\frac{2 π}{5}}{2}

使用分式法则:

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

= \frac{2 π}{5 \cdot 2}

数字相乘:

5 \cdot 2 = 10

= \frac{2 π}{10}

约分:

2

= \frac{π}{5}

= 2 sin (\frac{π}{10}) cos (\frac{π}{5})

= 2 sin (\frac{π}{10}) cos (\frac{π}{5})

使用三角恒等式改写:

sin (\frac{π}{10}) = \frac{2 3 - 5}{4}

sin (\frac{π}{10})

使用三角恒等式改写:

\frac{1 - cos ( \frac{π}{5} )}{2}

sin (\frac{π}{10})

将

sin (\frac{π}{10})

写为

sin (\frac{\frac{π}{5}}{2})

= sin (\frac{\frac{π}{5}}{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, \frac{π}{2}] [\frac{π}{2}, π] [π, \frac{3 π}{2}] [\frac{3 π}{2}, 2 π] 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 ( \frac{π}{5} )}{2}

= \frac{1 - cos ( \frac{π}{5} )}{2}

使用三角恒等式改写:

cos (\frac{π}{5}) = \frac{5 + 1}{4}

cos (\frac{π}{5})

显示:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \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 (\frac{π}{5}) sin (\frac{π}{10}) = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

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

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

\frac{1}{2} = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin (\frac{3 π}{10}) = cos (\frac{π}{2} - \frac{3 π}{10})

\frac{1}{2} = cos (\frac{π}{2} - \frac{3 π}{10}) - sin (\frac{π}{10})

\frac{1}{2} = cos (\frac{π}{5}) - sin (\frac{π}{10})

显示:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

使用因式分解法则:

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

a = cos (\frac{π}{5}) + sin (\frac{π}{10})

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = ((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) ((cos (\frac{π}{5}) + sin (\frac{π}{10})) - (cos (\frac{π}{5}) - sin (\frac{π}{10})))

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 2 (2 cos (\frac{π}{5}) sin (\frac{π}{10}))

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

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

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 1

代入

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (\frac{1}{2})^{2} = 1

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} = 1

两边加上

\frac{1}{4}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} = \frac{5}{4}

在两侧开平方

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \pm \frac{5}{4}

cos (\frac{π}{5})

不能为负

sin (\frac{π}{10})

不能为负

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

以下方程式相加

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{2}

((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) = (\frac{5}{2} + \frac{1}{2})

整理后得

cos (\frac{π}{5}) = \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:

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:

= 2 2^{2}

使用根式运算法则:

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}

使用三角恒等式改写:

cos (\frac{π}{5}) = \frac{5 + 1}{4}

cos (\frac{π}{5})

显示:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \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 (\frac{π}{5}) sin (\frac{π}{10}) = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

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

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

\frac{1}{2} = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin (\frac{3 π}{10}) = cos (\frac{π}{2} - \frac{3 π}{10})

\frac{1}{2} = cos (\frac{π}{2} - \frac{3 π}{10}) - sin (\frac{π}{10})

\frac{1}{2} = cos (\frac{π}{5}) - sin (\frac{π}{10})

显示:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

使用因式分解法则:

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

a = cos (\frac{π}{5}) + sin (\frac{π}{10})

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = ((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) ((cos (\frac{π}{5}) + sin (\frac{π}{10})) - (cos (\frac{π}{5}) - sin (\frac{π}{10})))

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 2 (2 cos (\frac{π}{5}) sin (\frac{π}{10}))

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

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

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 1

代入

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (\frac{1}{2})^{2} = 1

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} = 1

两边加上

\frac{1}{4}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} = \frac{5}{4}

在两侧开平方

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \pm \frac{5}{4}

cos (\frac{π}{5})

不能为负

sin (\frac{π}{10})

不能为负

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

以下方程式相加

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{2}

((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) = (\frac{5}{2} + \frac{1}{2})

整理后得

cos (\frac{π}{5}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

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

化简

2 \cdot \frac{2 3 - 5}{4} \cdot \frac{5 + 1}{4} : \frac{2 ( 1 + 5 ) 3 - 5}{8}

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

分式相乘:

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

= \frac{2 3 - 5 ( 5 + 1 ) \cdot 2}{4 \cdot 4}

数字相乘:

4 \cdot 4 = 16

= \frac{2 2 ( 1 + 5 ) 3 - 5}{16}

约分:

2

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

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

流行的例子