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受欢迎的三角函数 >

(sin(3015)tan(4290))/(cos(2730))

解答

\frac{sin ( 301 5 ^{\circ} ) tan ( 429 0 ^{\circ} )}{cos ( 273 0 ^{\circ} )}

解答

\frac{2}{3}

+1

十进制

0.47140 \dots

求解步骤

\frac{sin ( 301 5 ^{\circ} ) tan ( 429 0 ^{\circ} )}{cos ( 273 0 ^{\circ} )}

sin (301 5^{\circ}) = sin (13 5^{\circ})

sin (301 5^{\circ})

将

301 5^{\circ}

改写为

36 0^{\circ} \cdot 8 + 13 5^{\circ}

= sin (36 0^{\circ} 8 + 13 5^{\circ})

使用周期

sin

:

sin (x + 36 0^{\circ} \cdot k) = sin (x)

sin (36 0^{\circ} \cdot 8 + 13 5^{\circ}) = sin (13 5^{\circ})

= sin (13 5^{\circ})

= \frac{sin ( 13 5 ^{\circ} ) tan ( 429 0 ^{\circ} )}{cos ( 273 0 ^{\circ} )}

tan (429 0^{\circ}) = tan (15 0^{\circ})

tan (429 0^{\circ})

将

429 0^{\circ}

改写为

18 0^{\circ} \cdot 23 + 15 0^{\circ}

= tan (18 0^{\circ} 23 + 15 0^{\circ})

使用周期

tan

:

tan (x + 18 0^{\circ} \cdot k) = tan (x)

tan (18 0^{\circ} \cdot 23 + 15 0^{\circ}) = tan (15 0^{\circ})

= tan (15 0^{\circ})

= \frac{sin ( 13 5 ^{\circ} ) tan ( 15 0 ^{\circ} )}{cos ( 273 0 ^{\circ} )}

cos (273 0^{\circ}) = cos (21 0^{\circ})

cos (273 0^{\circ})

将

273 0^{\circ}

改写为

36 0^{\circ} \cdot 7 + 21 0^{\circ}

= cos (36 0^{\circ} 7 + 21 0^{\circ})

使用周期

cos

:

cos (x + 36 0^{\circ} \cdot k) = cos (x)

cos (36 0^{\circ} \cdot 7 + 21 0^{\circ}) = cos (21 0^{\circ})

= cos (21 0^{\circ})

= \frac{sin ( 13 5 ^{\circ} ) tan ( 15 0 ^{\circ} )}{cos ( 21 0 ^{\circ} )}

使用以下普通恒等式:

sin (13 5^{\circ}) = \frac{2}{2}

sin (13 5^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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}

使用三角恒等式改写:

tan (15 0^{\circ}) = - \frac{3}{3}

tan (15 0^{\circ})

使用三角恒等式改写:

\frac{sin ( 15 0 ^{\circ} )}{cos ( 15 0 ^{\circ} )}

tan (15 0^{\circ})

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( 15 0 ^{\circ} )}{cos ( 15 0 ^{\circ} )}

= \frac{sin ( 15 0 ^{\circ} )}{cos ( 15 0 ^{\circ} )}

使用以下普通恒等式:

sin (15 0^{\circ}) = \frac{1}{2}

sin (15 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 (15 0^{\circ}) = - \frac{3}{2}

cos (15 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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}

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

化简

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

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

使用分式法则:

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

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

分式相除:

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

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

整理后得

= - \frac{2}{2 3}

约分:

2

= - \frac{1}{3}

- \frac{1}{3}

有理化:

- \frac{3}{3}

- \frac{1}{3}

乘以共轭根式

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

= - \frac{3}{3}

使用三角恒等式改写:

cos (21 0^{\circ}) = - \frac{3}{2}

cos (21 0^{\circ})

使用三角恒等式改写:

cos (18 0^{\circ}) cos (3 0^{\circ}) - sin (18 0^{\circ}) sin (3 0^{\circ})

cos (21 0^{\circ})

将

cos (21 0^{\circ})

写为

cos (18 0^{\circ} + 3 0^{\circ})

= cos (18 0^{\circ} + 3 0^{\circ})

使用角和恒等式:

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

= cos (18 0^{\circ}) cos (3 0^{\circ}) - sin (18 0^{\circ}) sin (3 0^{\circ})

= cos (18 0^{\circ}) cos (3 0^{\circ}) - sin (18 0^{\circ}) sin (3 0^{\circ})

使用以下普通恒等式:

cos (18 0^{\circ}) = (- 1)

cos (18 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

使用以下普通恒等式:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 (18 0^{\circ}) = 0

sin (18 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

使用以下普通恒等式:

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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}

= (- 1) \frac{3}{2} - 0 \cdot \frac{1}{2}

化简

= - \frac{3}{2}

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

化简

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

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

去除括号:

(- a) = - a

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

使用分式法则:

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

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

使用分式法则:

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

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

乘

\frac{2}{2} \cdot \frac{3}{3} \cdot 2 : \frac{2}{3}

\frac{2}{2} \cdot \frac{3}{3} \cdot 2

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 \cdot 2}{2 \cdot 3}

约分:

2

= \frac{2 3}{3}

使用根式运算法则:

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

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

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

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

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

数字相减:

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

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

使用根式运算法则:

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

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

= \frac{2}{3}

合并相同指数项 :

\frac{x}{y} = \frac{x}{y}

= \frac{2}{3}

= \frac{\frac{2}{3}}{3}

\frac{2}{3} = \frac{2}{3}

\frac{2}{3}

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{2}{3}

= \frac{\frac{2}{3}}{3}

使用分式法则:

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

= \frac{2}{3 3}

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= \frac{2}{3}

= \frac{2}{3}

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sinh(-arccosh(2))

sinh (- arccosh (2))

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sin (\frac{π}{9}) - sin (\frac{2 π}{9}) - sin (\frac{3 π}{9}) + sin (\frac{4 π}{9})

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sin (2) \cdot sin (3)