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受欢迎的三角函数 >

(sin(150)+cot(315))/(sqrt(3)cos(210))

解答

\frac{sin ( 15 0 ^{\circ} ) + cot ( 31 5 ^{\circ} )}{3 cos ( 21 0 ^{\circ} )}

解答

\frac{1}{3}

+1

十进制

0.33333 \dots

求解步骤

\frac{sin ( 15 0 ^{\circ} ) + cot ( 31 5 ^{\circ} )}{3 cos ( 21 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}

使用三角恒等式改写:

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

cot (31 5^{\circ})

cot (31 5^{\circ}) = cot (13 5^{\circ})

cot (31 5^{\circ})

将

31 5^{\circ}

改写为

18 0^{\circ} + 13 5^{\circ}

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

使用周期

cot

:

cot (x + 18 0^{\circ}) = cot (x)

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

= cot (13 5^{\circ})

= cot (13 5^{\circ})

使用以下普通恒等式:

cot (13 5^{\circ}) = - 1

cot (13 5^{\circ})

cot (x)

周期表（周期为

18 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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

= - 1

= - 1

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

化简

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

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

去除括号:

(- a) = - a

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

使用分式法则:

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

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

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

化简

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

- \frac{1}{2} + 1

将项转换为分式:

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

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

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{- 1 + 1 \cdot 2}{2}

- 1 + 1 \cdot 2 = 1

- 1 + 1 \cdot 2

数字相乘:

1 \cdot 2 = 2

= - 1 + 2

数字相加/相减:

- 1 + 2 = 1

= 1

= \frac{1}{2}

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

使用分式法则:

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

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

乘

23 \frac{3}{2} : 3

23 \frac{3}{2}

分式相乘:

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

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

约分:

2

= 33

使用根式运算法则:

a a = a

33 = 3

= 3

= \frac{1}{3}

= \frac{1}{3}

流行的例子

0.23 \cdot sin (62. 2^{\circ})

tan((3pi)/(16))

tan (\frac{3 π}{16})

19.6 cos (3 0^{\circ})

(sin(pi/6))/(cos^2(pi/6))

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

sinh(log_{10}(2)-log_{10}(5))

sinh (lo g_{10} (2) - lo g_{10} (5))