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受欢迎的三角函数 >

1-cos(60)

解答

1 - cos (6 0^{\circ})

解答

\frac{1}{2}

+1

十进制

0.5

求解步骤

1 - cos (6 0^{\circ})

使用以下普通恒等式:

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 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{1}{2}

= 1 - \frac{1}{2}

化简

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

数字相减:

2 - 1 = 1

= 1

= \frac{1}{2}

= \frac{1}{2}

流行的例子

sqrt((1-cos(32))/(1+cos(32)))

\frac{1 - cos ( 3 2 ^{\circ} )}{1 + cos ( 3 2 ^{\circ} )}

(2tan((2pi)/3))/(1-tan^2((2pi)/3))

\frac{2 tan ( \frac{2 π}{3} )}{1 - tan ^{2} ( \frac{2 π}{3} )}

8(cos(30)+isin(30))

8 (cos (3 0^{\circ}) + i sin (3 0^{\circ}))

3 \cdot sin (\frac{2 π}{3})

18 sin (7 1^{\circ})