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受欢迎的三角函数 >

sin(-315)sin(-15)

解答

sin (- 31 5^{\circ}) sin (- 1 5^{\circ})

解答

- \frac{3 - 1}{4}

+1

十进制

- 0.18301 \dots

求解步骤

sin (- 31 5^{\circ}) sin (- 1 5^{\circ})

利用以下特性:

sin (- x) = - sin (x)

sin (- 1 5^{\circ}) = - sin (1 5^{\circ})

= sin (- 31 5^{\circ}) (- sin (1 5^{\circ}))

利用以下特性:

sin (- x) = - sin (x)

sin (- 31 5^{\circ}) = - sin (31 5^{\circ})

= (- sin (31 5^{\circ})) (- sin (1 5^{\circ}))

化简

= sin (31 5^{\circ}) sin (1 5^{\circ})

使用三角恒等式改写:

sin (31 5^{\circ}) = - \frac{2}{2}

sin (31 5^{\circ})

使用三角恒等式改写:

sin (18 0^{\circ}) cos (13 5^{\circ}) + cos (18 0^{\circ}) sin (13 5^{\circ})

sin (31 5^{\circ})

将

sin (31 5^{\circ})

写为

sin (18 0^{\circ} + 13 5^{\circ})

= sin (18 0^{\circ} + 13 5^{\circ})

使用角和恒等式:

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

= sin (18 0^{\circ}) cos (13 5^{\circ}) + cos (18 0^{\circ}) sin (13 5^{\circ})

= sin (18 0^{\circ}) cos (13 5^{\circ}) + cos (18 0^{\circ}) sin (13 5^{\circ})

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:

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

cos (13 5^{\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{2}{2}

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:

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}

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

化简

= - \frac{2}{2}

使用三角恒等式改写:

sin (1 5^{\circ}) = \frac{6 - 2}{4}

sin (1 5^{\circ})

使用三角恒等式改写:

sin (4 5^{\circ}) cos (3 0^{\circ}) - cos (4 5^{\circ}) sin (3 0^{\circ})

sin (1 5^{\circ})

将

sin (1 5^{\circ})

写为

sin (4 5^{\circ} - 3 0^{\circ})

= sin (4 5^{\circ} - 3 0^{\circ})

使用角差恒等式:

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

= sin (4 5^{\circ}) cos (3 0^{\circ}) - cos (4 5^{\circ}) sin (3 0^{\circ})

= sin (4 5^{\circ}) cos (3 0^{\circ}) - cos (4 5^{\circ}) sin (3 0^{\circ})

使用以下普通恒等式:

sin (4 5^{\circ}) = \frac{2}{2}

sin (4 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:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\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{2}{2}

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}

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

化简

\frac{2}{2} \cdot \frac{3}{2} - \frac{2}{2} \cdot \frac{1}{2} : \frac{6 - 2}{4}

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

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

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

分式相乘:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 3}{4}

化简

23 : 6

23

使用根式运算法则:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

数字相乘:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

\frac{2}{2} \cdot \frac{1}{2} = \frac{2}{4}

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

分式相乘:

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

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

乘以:

2 \cdot 1 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{6}{4} - \frac{2}{4}

使用法则

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

= \frac{6 - 2}{4}

= \frac{6 - 2}{4}

= (- \frac{2}{2}) \frac{6 - 2}{4}

化简

(- \frac{2}{2}) \frac{6 - 2}{4} : - \frac{3 - 1}{4}

(- \frac{2}{2}) \frac{6 - 2}{4}

去除括号:

(- a) = - a

= - \frac{2}{2} \cdot \frac{6 - 2}{4}

分式相乘:

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

= - \frac{2 ( 6 - 2 )}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= - \frac{2 ( 6 - 2 )}{8}

消掉

\frac{2 ( 6 - 2 )}{8} : \frac{6 - 2}{2 ^{2} 2}

\frac{2 ( 6 - 2 )}{8}

分解

8 : 2^{3}

因式分解

8 = 2^{3}

= \frac{2 ( 6 - 2 )}{2 ^{3}}

消掉

\frac{2 ( 6 - 2 )}{2 ^{3}} : \frac{6 - 2}{2 ^{\frac{5}{2}}}

\frac{2 ( 6 - 2 )}{2 ^{3}}

使用根式运算法则:

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

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

= \frac{2 ^{\frac{1}{2}} ( 6 - 2 )}{2 ^{3}}

使用指数法则:

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

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

= \frac{6 - 2}{2 ^{3 - \frac{1}{2}}}

数字相减:

3 - \frac{1}{2} = \frac{5}{2}

= \frac{6 - 2}{2 ^{\frac{5}{2}}}

= \frac{6 - 2}{2 ^{\frac{5}{2}}}

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

2^{\frac{5}{2}}

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

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

使用指数法则:

x^{a + b} = x^{a} x^{b}

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

整理后得

= 2^{2} 2

= \frac{6 - 2}{2 ^{2} 2}

= - \frac{6 - 2}{2 ^{2} 2}

2^{2} = 4

= - \frac{6 - 2}{4 2}

- \frac{6 - 2}{4 2}

有理化:

- \frac{3 - 1}{4}

- \frac{6 - 2}{4 2}

乘以共轭根式

\frac{2}{2}

= - \frac{( 6 - 2 ) 2}{4 2 2}

(6 - 2) 2 = 23 - 2

(6 - 2) 2

= 2 (6 - 2)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 6, c = 2

= 26 - 22

化简

26 - 22 : 23 - 2

26 - 22

26 = 23

26

分解整数

6 = 2 \cdot 3

= 2 2 \cdot 3

使用根式运算法则:

n ab = n a n b

2 \cdot 3 = 23

= 223

使用根式运算法则:

a a = a

22 = 2

= 23

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= 23 - 2

= 23 - 2

422 = 8

422

使用根式运算法则:

a a = a

22 = 2

= 4 \cdot 2

数字相乘:

4 \cdot 2 = 8

= 8

= - \frac{2 3 - 2}{8}

消掉

\frac{2 3 - 2}{8} : \frac{3 - 1}{4}

\frac{2 3 - 2}{8}

分解

23 - 2 : 2 (3 - 1)

23 - 2

改写为

= 23 - 2 \cdot 1

因式分解出通项

2

= 2 (3 - 1)

= \frac{2 ( 3 - 1 )}{8}

约分:

2

= \frac{3 - 1}{4}

= - \frac{3 - 1}{4}

= - \frac{3 - 1}{4}

= - \frac{3 - 1}{4}

流行的例子

csc (- \frac{13 π}{6})

tan (12. 5^{\circ})

1/2 arctan(sqrt(3))

\frac{1}{2} arctan (3)

cos^2(pi/4)-sin((7pi)/6)

cos^{2} (\frac{π}{4}) - sin (\frac{7 π}{6})

cot^{2} (6 0^{\circ})