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受欢迎的三角函数 >

2sin(1.1pi)

解答

2 sin (1.1 π)

解答

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

+1

十进制

- 0.61803 \dots

求解步骤

2 sin (1.1 π)

= 2 sin (\frac{11}{10} π)

化简:

\frac{11}{10} π = \frac{11 π}{10}

\frac{11}{10} π

分式相乘:

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

= \frac{11 π}{10}

= 2 sin (\frac{11 π}{10})

使用三角恒等式改写:

sin (\frac{11 π}{10}) = - \frac{1 - cos ( \frac{π}{5} )}{2}

sin (\frac{11 π}{10})

将

sin (\frac{11 π}{10})

写为

sin (\frac{\frac{11 π}{5}}{2})

= sin (\frac{\frac{11 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{11 π}{5} )}{2}

cos (\frac{11 π}{5}) = cos (\frac{π}{5})

cos (\frac{11 π}{5})

将

\frac{11 π}{5}

改写为

2 π + \frac{π}{5}

= cos (2 π + \frac{π}{5})

使用周期

cos

:

cos (x + 2 π) = cos (x)

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

= cos (\frac{π}{5})

= - \frac{1 - cos ( \frac{π}{5} )}{2}

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

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

2 - \frac{1 - cos ( \frac{π}{5} )}{2}

去除括号:

(- a) = - a

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

使用指数法则:

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

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

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

乘

2^{2} \cdot \frac{1 - cos ( \frac{π}{5} )}{2} : 2 (- cos (\frac{π}{5}) + 1)

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

分式相乘:

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

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

约分:

2

= 2 (- cos (\frac{π}{5}) + 1)

= - 2 (- cos (\frac{π}{5}) + 1)

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

= - 2 - cos (\frac{π}{5}) + 1

= - 2 - cos (\frac{π}{5}) + 1

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

流行的例子

161cos(14)+(555sin(14))/8-(21cos(14))/4

161 cos (1 4^{\circ}) + \frac{555 sin ( 1 4 ^{\circ} )}{8} - \frac{21 cos ( 1 4 ^{\circ} )}{4}

-(1^2sin(1)+2(1cos(1)-sin(1)))/(1^3)

- \frac{1 ^{2} sin ( 1 ) + 2 ( 1 cos ( 1 ) - sin ( 1 ) )}{1 ^{3}}

sin (27. 5^{\circ})

3 sin (9)

arctan((-2)/(-4))

arctan (\frac{- 2}{- 4})