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受欢迎的三角函数 >

csc((37pi)/(12))

解答

csc (\frac{37 π}{12})

解答

- 2 (1 + 3)

+1

十进制

- 3.86370 \dots

求解步骤

csc (\frac{37 π}{12})

csc (\frac{37 π}{12}) = csc (\frac{13 π}{12})

csc (\frac{37 π}{12})

将

\frac{37 π}{12}

改写为

2 π + \frac{13 π}{12}

= csc (2 π + \frac{13 π}{12})

使用周期

csc

:

csc (x + 2 π) = csc (x)

csc (2 π + \frac{13 π}{12}) = csc (\frac{13 π}{12})

= csc (\frac{13 π}{12})

= csc (\frac{13 π}{12})

使用三角恒等式改写:

\frac{1}{sin ( \frac{13 π}{12} )}

csc (\frac{13 π}{12})

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{sin ( \frac{13 π}{12} )}

= \frac{1}{sin ( \frac{13 π}{12} )}

使用三角恒等式改写:

sin (\frac{13 π}{12}) = \frac{2 ( 1 - 3 )}{4}

sin (\frac{13 π}{12})

使用三角恒等式改写:

sin (\frac{5 π}{6}) cos (\frac{π}{4}) + cos (\frac{5 π}{6}) sin (\frac{π}{4})

sin (\frac{13 π}{12})

将

sin (\frac{13 π}{12})

写为

sin (\frac{5 π}{6} + \frac{π}{4})

= sin (\frac{5 π}{6} + \frac{π}{4})

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

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

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

使用以下普通恒等式:

sin (\frac{5 π}{6}) = \frac{1}{2}

sin (\frac{5 π}{6})

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

使用以下普通恒等式:

cos (\frac{5 π}{6}) = - \frac{3}{2}

cos (\frac{5 π}{6})

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{3}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

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

化简

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

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

去除括号:

(- a) = - a

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

因式分解出通项

\frac{2}{2}

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

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

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

使用法则

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

= \frac{1 - 3}{2}

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

分式相乘:

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

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

数字相乘:

2 \cdot 2 = 4

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

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

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

化简

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

\frac{1}{\frac{2 ( 1 - 3 )}{4}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

分解

4 : 2^{2}

因式分解

4 = 2^{2}

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

消掉

\frac{2 ^{2}}{2 ( 1 - 3 )} : \frac{2 ^{\frac{3}{2}}}{1 - 3}

\frac{2 ^{2}}{2 ( 1 - 3 )}

使用根式运算法则:

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

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

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

使用指数法则:

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

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

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

数字相减:

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

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

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

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{2 2}{1 - 3}

\frac{2 2}{1 - 3}

有理化:

- 2 (1 + 3)

\frac{2 2}{1 - 3}

乘以共轭根式

\frac{1 + 3}{1 + 3}

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

(1 - 3) (1 + 3) = - 2

(1 - 3) (1 + 3)

使用平方差公式:

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

a = 1, b = 3

= 1^{2} - (3)^{2}

化简

1^{2} - (3)^{2} : - 2

1^{2} - (3)^{2}

使用法则

1^{a} = 1

1^{2} = 1

= 1 - (3)^{2}

(3)^{2} = 3

(3)^{2}

使用根式运算法则:

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

= (3^{\frac{1}{2}})^{2}

使用指数法则:

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

= 3^{\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

= 3

= 1 - 3

数字相减:

1 - 3 = - 2

= - 2

= - 2

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

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

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

数字相除:

\frac{2}{2} = 1

= - 2 (1 + 3)

= - 2 (1 + 3)

= - 2 (1 + 3)

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