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受欢迎的三角函数 >

sin(arccos(-1)-arccos(1/2))

解答

sin (arccos (- 1) - arccos (\frac{1}{2}))

解答

\frac{3}{2}

+1

十进制

0.86602 \dots

求解步骤

sin (arccos (- 1) - arccos (\frac{1}{2}))

使用三角恒等式改写:

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

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

使用和差化积恒等式:

arccos (s) - arccos (t) = arccos (s t + (1 - s^{2}) (1 - t^{2}))

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

化简:

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

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

去除括号:

(- a) = - a

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

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

1 \cdot \frac{1}{2}

乘以:

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

= \frac{1}{2}

(1 - (- 1)^{2}) (1 - (\frac{1}{2})^{2}) = 0

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

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

(\frac{1}{2})^{2} = \frac{1}{4}

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

使用法则

1^{a} = 1

1^{2} = 1

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

2^{2} = 4

= \frac{1}{4}

= (1 - 1) (- \frac{1}{4} + 1)

数字相减:

1 - 1 = 0

= 0 \cdot (- \frac{1}{4} + 1)

化简

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

1 - \frac{1}{4}

将项转换为分式:

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

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

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 4 - 1}{4}

1 \cdot 4 - 1 = 3

1 \cdot 4 - 1

数字相乘:

1 \cdot 4 = 4

= 4 - 1

数字相减:

4 - 1 = 3

= 3

= \frac{3}{4}

= 0 \cdot \frac{3}{4}

使用法则

0 \cdot a = 0

= 0

使用法则

0 = 0

= 0

= - \frac{1}{2} + 0

- \frac{1}{2} + 0 = - \frac{1}{2}

= - \frac{1}{2}

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

= sin (arccos (- \frac{1}{2}))

使用三角恒等式改写:

sin (arccos (- \frac{1}{2})) = 1 - (- \frac{1}{2})^{2}

利用以下特性:

sin (arccos (x)) = 1 - x^{2}

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

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

化简

= \frac{3}{2}

流行的例子

ln sin (\frac{π}{2})

sin((5pi)/(12))sin(pi/(12))

sin (\frac{5 π}{12}) sin (\frac{π}{12})

1 + tan^{2} (0)

tan (\frac{6}{4})

arctan(sqrt((1-cos(19))/(1+cos(19))))

arctan (\frac{1 - cos ( 1 9 ^{\circ} )}{1 + cos ( 1 9 ^{\circ} )})