zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

ln(sin(pi/3))-ln(sin(pi/4))

解答

ln (sin (\frac{π}{3})) - ln (sin (\frac{π}{4}))

解答

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

+1

十进制

0.20273 \dots

求解步骤

ln (sin (\frac{π}{3})) - ln (sin (\frac{π}{4}))

使用以下普通恒等式:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{π}{3})

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{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}

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

化简

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

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

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

ln (\frac{3}{2})

使用对数计算法则:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

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

= ln (3) - ln (2)

化简

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

ln (3)

改写为

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

使用对数运算法则:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

假定

x \geq 0

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

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

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

ln (\frac{2}{2})

使用对数计算法则:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

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

= ln (2) - ln (2)

化简

ln (2) : \frac{1}{2} ln (2)

ln (2)

改写为

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

使用对数运算法则:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

假定

x \geq 0

= \frac{1}{2} ln (2)

因式分解出通项

ln (2)

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

\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

数字相减:

1 - 2 = - 1

= - 1

= \frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

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

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

使用法则

- (- a) = a

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

同类项相加:

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

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

因式分解出通项

ln (2)

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

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

- 1 + \frac{1}{2}

将项转换为分式:

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

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

- 1 \cdot 2 + 1 = - 1

- 1 \cdot 2 + 1

数字相乘:

1 \cdot 2 = 2

= - 2 + 1

数字相加/相减:

- 2 + 1 = - 1

= - 1

= \frac{- 1}{2}

使用分式法则:

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

= - \frac{1}{2}

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

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

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

流行的例子

35 sin (4 5^{\circ})

arcsin (0.1915)

(5sin(3)+7cos(87))csc(3)

(5 sin (3^{\circ}) + 7 cos (8 7^{\circ})) csc (3^{\circ})

cos(40)+sin(40)

cos (4 0^{\circ}) + sin (4 0^{\circ})

(6(9.81)-6(9.81)sin(30))/(6+6)

\frac{6 ( 9.81 ) - 6 ( 9.81 ) sin ( 3 0 ^{\circ} )}{6 + 6}