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受欢迎的三角函数 >

10picos(2pi0.8)+e^{30.8}*3

解答

10 π cos (2 π \cdot 0.8) + e^{3 \cdot 0.8} \cdot 3

解答

10 π (1 - \frac{5 - 5}{4}) + 3 e^{\frac{12}{5}}

+1

十进制

42.77758 \dots

求解步骤

10 π cos (2 π 0.8) + e^{3 \cdot 0.8} \cdot 3

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

使用三角恒等式改写:

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

cos (2 π \frac{4}{5})

使用倍角公式:

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

= 1 - 2 sin^{2} (π \frac{4}{5})

化简:

π \frac{4}{5} = \frac{4 π}{5}

π \frac{4}{5}

分式相乘:

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

= \frac{4 π}{5}

= 1 - 2 sin^{2} (\frac{4 π}{5})

= 10 π (1 - 2 sin^{2} (\frac{4 π}{5})) + e^{3 \cdot \frac{4}{5}} \cdot 3

10 π (1 - 2 sin^{2} (\frac{4 π}{5})) + e^{3 \cdot \frac{4}{5}} \cdot 3 = 10 π (1 - 2 sin^{2} (\frac{4 π}{5})) + 3 e^{\frac{12}{5}}

10 π (1 - 2 sin^{2} (\frac{4 π}{5})) + e^{3 \cdot \frac{4}{5}} \cdot 3

e^{3 \cdot \frac{4}{5}} = e^{\frac{12}{5}}

e^{3 \cdot \frac{4}{5}}

乘

3 \cdot \frac{4}{5} : \frac{12}{5}

3 \cdot \frac{4}{5}

分式相乘:

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

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

数字相乘:

4 \cdot 3 = 12

= \frac{12}{5}

= e^{\frac{12}{5}}

= 10 π (- 2 sin^{2} (\frac{4 π}{5}) + 1) + 3 e^{\frac{12}{5}}

= 10 π (1 - 2 sin^{2} (\frac{4 π}{5})) + 3 e^{\frac{12}{5}}

使用三角恒等式改写:

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

sin (\frac{4 π}{5})

使用三角恒等式改写:

sin (\frac{π}{5})

sin (\frac{4 π}{5})

使用基本三角恒等式:

sin (x) = sin (π - x)

= sin (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}

= sin (\frac{π}{5})

= sin (\frac{π}{5})

使用三角恒等式改写:

\frac{2 5 - 5}{4}

sin (\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}

两边进行平方

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

利用以下特性:

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

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

代入

cos (\frac{π}{5}) = \frac{5 + 1}{4}

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

整理后得

sin^{2} (\frac{π}{5}) = \frac{5 - 5}{8}

在两侧开平方

sin (\frac{π}{5}) = \pm \frac{5 - 5}{8}

sin (\frac{π}{5})

不能为负

sin (\frac{π}{5}) = \frac{5 - 5}{8}

整理后得

sin (\frac{π}{5}) = \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}

= 10 π 1 - 2 (\frac{2 5 - 5}{4})^{2} + 3 e^{\frac{12}{5}}

化简

10 π 1 - 2 (\frac{2 5 - 5}{4})^{2} + 3 e^{\frac{12}{5}} : 10 π (1 - \frac{5 - 5}{4}) + 3 e^{\frac{12}{5}}

10 π 1 - 2 (\frac{2 5 - 5}{4})^{2} + 3 e^{\frac{12}{5}}

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

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

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

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

使用指数法则:

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

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

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

(2 5 - 5)^{2} = (2)^{2} (5 - 5)^{2}

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

(2)^{2} : 2

使用根式运算法则:

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

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

使用指数法则:

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

= 2^{\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

= 2

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

(5 - 5)^{2} : 5 - 5

使用根式运算法则:

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

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

使用指数法则:

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

= (5 - 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 - 5

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

约分:

2

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

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

分式相乘:

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

= \frac{( 5 - 5 ) \cdot 2}{2 ^{3}}

约分:

2

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

2^{2} = 4

= \frac{5 - 5}{4}

= 10 π (- \frac{5 - 5}{4} + 1) + 3 e^{\frac{12}{5}}

= 10 π (1 - \frac{5 - 5}{4}) + 3 e^{\frac{12}{5}}

流行的例子

arcsec (\frac{2}{3})

2 tan (\frac{π}{3})

7 cos (16 0^{\circ})

sin (2 + i)

tan((11pi)/6-pi/4)

tan (\frac{11 π}{6} - \frac{π}{4})