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受欢迎的三角函数 >

((254picos^2(4pi0.3)))/(1000)

解答

\frac{( 2 \cdot 5 \cdot 4 π \cdot cos ^{2} ( 4 π \cdot 0.3 ) )}{1000}

解答

\frac{π ( 3 + 5 )}{200}

+1

十进制

0.08224 \dots

求解步骤

\frac{( 2 \cdot 5 \cdot 4 π cos ^{2} ( 4 π 0.3 ) )}{1000}

= \frac{2 \cdot 5 \cdot 4 π cos ^{2} ( 4 π \frac{3}{10} )}{1000}

化简:

4 π \frac{3}{10} = \frac{6 π}{5}

4 π \frac{3}{10}

分式相乘:

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

= \frac{3 \cdot 4 π}{10}

数字相乘:

3 \cdot 4 = 12

= \frac{12 π}{10}

约分:

2

= \frac{6 π}{5}

\frac{2 \cdot 5 \cdot 4 π cos ^{2} ( \frac{6 π}{5} )}{1000} = \frac{π cos ^{2} ( \frac{6 π}{5} )}{25}

\frac{2 \cdot 5 \cdot 4 π cos ^{2} ( \frac{6 π}{5} )}{1000}

数字相乘:

2 \cdot 5 \cdot 4 = 40

= \frac{40 π cos ^{2} ( \frac{6 π}{5} )}{1000}

约分:

40

= \frac{π cos ^{2} ( \frac{6 π}{5} )}{25}

= \frac{π cos ^{2} ( \frac{6 π}{5} )}{25}

使用三角恒等式改写:

cos (\frac{6 π}{5}) = - \frac{5 + 1}{4}

cos (\frac{6 π}{5})

使用三角恒等式改写:

cos (\frac{4 π}{5})

cos (\frac{6 π}{5})

利用以下特性:

cos (x) = cos (2 π - x)

cos (x)

利用以下特性:

cos (θ) = cos (- θ)

cos (x) = cos (- x)

= cos (- x)

使用周期

cos

:

cos (2 π + θ) = cos (θ)

cos (- x) = cos (2 π - x)

= cos (2 π - x)

= cos (2 π - \frac{6 π}{5})

化简:

2 π - \frac{6 π}{5} = \frac{4 π}{5}

2 π - \frac{6 π}{5}

将项转换为分式:

2 π = \frac{2 π 5}{5}

= \frac{2 π 5}{5} - \frac{6 π}{5}

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{2 π 5 - 6 π}{5}

2 π 5 - 6 π = 4 π

2 π 5 - 6 π

数字相乘:

2 \cdot 5 = 10

= 10 π - 6 π

同类项相加:

10 π - 6 π = 4 π

= 4 π

= \frac{4 π}{5}

= cos (\frac{4 π}{5})

= cos (\frac{4 π}{5})

使用三角恒等式改写:

- cos (\frac{π}{5})

cos (\frac{4 π}{5})

使用基本三角恒等式:

cos (x) = - cos (π - x)

= - cos (π - \frac{4 π}{5})

化简:

π - \frac{4 π}{5} = \frac{π}{5}

π - \frac{4 π}{5}

将项转换为分式:

π = \frac{π 5}{5}

= \frac{π 5}{5} - \frac{4 π}{5}

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 5 - 4 π}{5}

同类项相加:

5 π - 4 π = π

= \frac{π}{5}

= - cos (\frac{π}{5})

= - cos (\frac{π}{5})

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{5 + 1}{4}

= \frac{π ( - \frac{5 + 1}{4} ) ^{2}}{25}

化简

\frac{π ( - \frac{5 + 1}{4} ) ^{2}}{25} : \frac{π ( 3 + 5 )}{200}

\frac{π ( - \frac{5 + 1}{4} ) ^{2}}{25}

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

π (- \frac{5 + 1}{4})^{2}

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

(- \frac{5 + 1}{4})^{2}

使用指数法则:

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

若

n

是偶数

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

= (\frac{5 + 1}{4})^{2}

= π (\frac{1 + 5}{4})^{2}

= \frac{π ( \frac{1 + 5}{4} ) ^{2}}{25}

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

(\frac{5 + 1}{4})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 5 + 1 ) ^{2}}{4 ^{2}}

(5 + 1)^{2} = 6 + 25

(5 + 1)^{2}

使用完全平方公式:

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

a = 5, b = 1

= (5)^{2} + 25 \cdot 1 + 1^{2}

化简

(5)^{2} + 25 \cdot 1 + 1^{2} : 6 + 25

(5)^{2} + 25 \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (5)^{2} + 2 \cdot 1 \cdot 5 + 1

(5)^{2} = 5

(5)^{2}

使用根式运算法则:

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

= (5^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 5

25 \cdot 1 = 25

25 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 25

= 5 + 25 + 1

数字相加:

5 + 1 = 6

= 6 + 25

= 6 + 25

= \frac{6 + 2 5}{4 ^{2}}

分解

6 + 25 : 2 (3 + 5)

6 + 25

改写为

= 2 \cdot 3 + 25

因式分解出通项

2

= 2 (3 + 5)

= \frac{2 ( 3 + 5 )}{4 ^{2}}

分解

4^{2} : 2^{4}

因式分解

4 = 2^{2}

= (2^{2})^{2}

化简

(2^{2})^{2} : 2^{4}

(2^{2})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= 2^{4}

= 2^{4}

= \frac{2 ( 3 + 5 )}{2 ^{4}}

约分:

2

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

= \frac{π \frac{3 + 5}{2 ^{3}}}{25}

乘

π \frac{3 + 5}{2 ^{3}} : \frac{π ( 3 + 5 )}{8}

π \frac{3 + 5}{2 ^{3}}

分式相乘:

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

= \frac{( 3 + 5 ) π}{2 ^{3}}

2^{3} = 8

= \frac{π ( 3 + 5 )}{8}

= \frac{\frac{π ( 3 + 5 )}{8}}{25}

使用分式法则:

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

= \frac{( 3 + 5 ) π}{8 \cdot 25}

数字相乘:

8 \cdot 25 = 200

= \frac{π ( 3 + 5 )}{200}

= \frac{π ( 3 + 5 )}{200}

流行的例子

arctan (17)

cos (3 0^{\circ} + 12 0^{\circ})

arctan (27 0^{\circ})

50 sin (8 0^{\circ})

tan (21. 7^{\circ})