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 >

-4cos(pi/3)(1+ln(7sin(pi/3)))

解答

- 4 cos (\frac{π}{3}) (1 + ln (7 sin (\frac{π}{3})))

解答

- 2 (1 + ln (\frac{7 3}{2}))

+1

十进制

- 5.60413 \dots

求解步骤

- 4 cos (\frac{π}{3}) (1 + ln (7 sin (\frac{π}{3})))

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

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{1}{2}

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}

= - 4 \cdot \frac{1}{2} (1 + ln (7 \cdot \frac{3}{2}))

化简

- 4 \cdot \frac{1}{2} (1 + ln (7 \cdot \frac{3}{2})) : - 2 (1 + ln (\frac{7 3}{2}))

- 4 \cdot \frac{1}{2} (1 + ln (7 \cdot \frac{3}{2}))

分式相乘:

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

= - \frac{1 \cdot 4}{2} (ln (7 \cdot \frac{3}{2}) + 1)

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

\frac{1 \cdot 4}{2}

数字相乘:

1 \cdot 4 = 4

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

= - 2 (ln (7 \cdot \frac{3}{2}) + 1)

乘

7 \cdot \frac{3}{2} : \frac{7 3}{2}

7 \cdot \frac{3}{2}

分式相乘:

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

= \frac{3 \cdot 7}{2}

= - 2 (ln (\frac{7 3}{2}) + 1)

= - 2 (1 + ln (\frac{7 3}{2}))

流行的例子

(tan(40)+tan(20))/(1-tan(40)tan(20))

\frac{tan ( 4 0 ^{\circ} ) + tan ( 2 0 ^{\circ} )}{1 - tan ( 4 0 ^{\circ} ) tan ( 2 0 ^{\circ} )}

4 \cdot tan (6 0^{\circ})

sin((3pi)/(11))

sin (\frac{3 π}{11})

8 cos (2 0^{\circ})

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