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受欢迎的三角函数 >

sinh(ln(2)+pi/2 i)

解答

sinh (ln (2) + \frac{π}{2} i)

解答

\frac{3 cos ( \frac{π}{2} )}{4} + i \frac{5 sin ( \frac{π}{2} )}{4}

求解步骤

sinh (ln (2) + \frac{π}{2} i)

化简

sinh (ln (2) + \frac{π}{2} i) : sinh (\frac{2 ln ( 2 ) + πi}{2})

sinh (ln (2) + \frac{π}{2} i)

乘

\frac{π}{2} i : \frac{πi}{2}

\frac{π}{2} i

分式相乘:

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

= \frac{πi}{2}

= sinh (ln (2) + \frac{πi}{2})

化简

ln (2) + \frac{πi}{2} : \frac{2 ln ( 2 ) + πi}{2}

ln (2) + \frac{πi}{2}

将项转换为分式:

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

= \frac{ln ( 2 ) \cdot 2}{2} + \frac{πi}{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{ln ( 2 ) \cdot 2 + πi}{2}

= sinh (\frac{ln ( 2 ) \cdot 2 + πi}{2})

= sinh (\frac{2 ln ( 2 ) + πi}{2})

= sinh (\frac{2 ln ( 2 ) + πi}{2})

使用三角恒等式改写:

\frac{3 cos ( \frac{π}{2} )}{4} + i \frac{5 sin ( \frac{π}{2} )}{4}

sinh (\frac{2 ln ( 2 ) + πi}{2})

使用双曲函数恒等式:

sinh (x) = \frac{e ^{x} - e ^{- x}}{2}

= \frac{e ^{\frac{2 l n ( 2 ) + πi}{2}} - e ^{- \frac{2 l n ( 2 ) + πi}{2}}}{2}

化简

\frac{e ^{\frac{2 l n ( 2 ) + πi}{2}} - e ^{- \frac{2 l n ( 2 ) + πi}{2}}}{2} : \frac{- cos ( - \frac{π}{2} ) + 4 cos ( \frac{π}{2} )}{4} + i \frac{- sin ( - \frac{π}{2} ) + 4 sin ( \frac{π}{2} )}{4}

\frac{e ^{\frac{2 l n ( 2 ) + πi}{2}} - e ^{- \frac{2 l n ( 2 ) + πi}{2}}}{2}

e^{\frac{2 l n ( 2 ) + πi}{2}} - e^{- \frac{2 l n ( 2 ) + πi}{2}} = e^{l n (2)} (cos (\frac{π}{2}) + i sin (\frac{π}{2})) - e^{- l n (2)} (cos (- \frac{π}{2}) + i sin (- \frac{π}{2}))

e^{\frac{2 l n ( 2 ) + πi}{2}} - e^{- \frac{2 l n ( 2 ) + πi}{2}}

使用虚数运算法则:

e^{a + ib} = e^{a} (cos (b) + i sin (b))

= e^{l n (2)} (cos (\frac{π}{2}) + i sin (\frac{π}{2})) - e^{- \frac{2 l n ( 2 ) + πi}{2}}

使用虚数运算法则:

e^{a + ib} = e^{a} (cos (b) + i sin (b))

= e^{l n (2)} (cos (\frac{π}{2}) + i sin (\frac{π}{2})) - e^{- l n (2)} (cos (- \frac{π}{2}) + i sin (- \frac{π}{2}))

= \frac{e ^{l n (2)} ( cos ( \frac{π}{2} ) + i sin ( \frac{π}{2} ) ) - e ^{- l n (2)} ( cos ( - \frac{π}{2} ) + i sin ( - \frac{π}{2} ) )}{2}

e^{l n (2)} (cos (\frac{π}{2}) + sin (\frac{π}{2}) i) = 2 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))

e^{l n (2)} (cos (\frac{π}{2}) + sin (\frac{π}{2}) i)

e^{l n (2)} = 2

e^{l n (2)}

使用对数计算法则:

a^{l o g_{a} (b)} = b

= 2

= 2 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))

e^{- l n (2)} (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i) = \frac{cos ( - \frac{π}{2} ) + i sin ( - \frac{π}{2} )}{2}

e^{- l n (2)} (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i)

使用指数法则:

a^{- b} = \frac{1}{a ^{b}}

e^{- l n (2)} = \frac{1}{e ^{l n (2)}}

= \frac{1}{e ^{l n (2)}} (cos (- \frac{π}{2}) + i sin (- \frac{π}{2}))

分式相乘:

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

= \frac{1 \cdot ( cos ( - \frac{π}{2} ) + sin ( - \frac{π}{2} ) i )}{e ^{l n (2)}}

1 \cdot (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i) = cos (- \frac{π}{2}) + i sin (- \frac{π}{2})

1 \cdot (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i)

乘以:

1 \cdot (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i) = (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i)

= (cos (- \frac{π}{2}) + i sin (- \frac{π}{2}))

去除括号:

(a) = a

= cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i

= \frac{cos ( - \frac{π}{2} ) + i sin ( - \frac{π}{2} )}{e ^{l n (2)}}

e^{l n (2)} = 2

e^{l n (2)}

使用对数计算法则:

a^{l o g_{a} (b)} = b

= 2

= \frac{cos ( - \frac{π}{2} ) + i sin ( - \frac{π}{2} )}{2}

= \frac{2 ( cos ( \frac{π}{2} ) + i sin ( \frac{π}{2} ) ) - \frac{c o s ( - \frac{π}{2} ) + i s i n ( - \frac{π}{2} )}{2}}{2}

化简

2 (cos (\frac{π}{2}) + sin (\frac{π}{2}) i) - \frac{cos ( - \frac{π}{2} ) + sin ( - \frac{π}{2} ) i}{2} : \frac{4 cos ( \frac{π}{2} ) + 4 i sin ( \frac{π}{2} ) - cos ( - \frac{π}{2} ) - i sin ( - \frac{π}{2} )}{2}

2 (cos (\frac{π}{2}) + sin (\frac{π}{2}) i) - \frac{cos ( - \frac{π}{2} ) + sin ( - \frac{π}{2} ) i}{2}

将项转换为分式:

2 (cos (\frac{π}{2}) + i sin (\frac{π}{2})) = \frac{2 ( cos ( \frac{π}{2} ) + sin ( \frac{π}{2} ) i ) 2}{2}

= \frac{2 ( cos ( \frac{π}{2} ) + sin ( \frac{π}{2} ) i ) \cdot 2}{2} - \frac{cos ( - \frac{π}{2} ) + sin ( - \frac{π}{2} ) i}{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{2 ( cos ( \frac{π}{2} ) + sin ( \frac{π}{2} ) i ) \cdot 2 - ( cos ( - \frac{π}{2} ) + sin ( - \frac{π}{2} ) i )}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 ( cos ( \frac{π}{2} ) + i sin ( \frac{π}{2} ) ) - ( cos ( - \frac{π}{2} ) + i sin ( - \frac{π}{2} ) )}{2}

乘开

4 (cos (\frac{π}{2}) + sin (\frac{π}{2}) i) - (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i) : 4 cos (\frac{π}{2}) + 4 i sin (\frac{π}{2}) - cos (- \frac{π}{2}) - sin (- \frac{π}{2}) i

4 (cos (\frac{π}{2}) + sin (\frac{π}{2}) i) - (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i)

= 4 (cos (\frac{π}{2}) + i sin (\frac{π}{2})) - (cos (- \frac{π}{2}) + i sin (- \frac{π}{2}))

乘开

4 (cos (\frac{π}{2}) + sin (\frac{π}{2}) i) : 4 cos (\frac{π}{2}) + 4 i sin (\frac{π}{2})

4 (cos (\frac{π}{2}) + sin (\frac{π}{2}) i)

使用分配律:

a (b + c) = ab + a c

a = 4, b = cos (\frac{π}{2}), c = sin (\frac{π}{2}) i

= 4 cos (\frac{π}{2}) + 4 sin (\frac{π}{2}) i

= 4 cos (\frac{π}{2}) + 4 i sin (\frac{π}{2})

= 4 cos (\frac{π}{2}) + 4 i sin (\frac{π}{2}) - (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i)

- (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i) : - cos (- \frac{π}{2}) - sin (- \frac{π}{2}) i

- (cos (- \frac{π}{2}) + sin (- \frac{π}{2}) i)

打开括号

= - (cos (- \frac{π}{2})) - (sin (- \frac{π}{2}) i)

使用加减运算法则

+ (- a) = - a

= - cos (- \frac{π}{2}) - sin (- \frac{π}{2}) i

= 4 cos (\frac{π}{2}) + 4 i sin (\frac{π}{2}) - cos (- \frac{π}{2}) - sin (- \frac{π}{2}) i

= \frac{4 cos ( \frac{π}{2} ) + 4 i sin ( \frac{π}{2} ) - cos ( - \frac{π}{2} ) - i sin ( - \frac{π}{2} )}{2}

= \frac{\frac{4 c o s ( \frac{π}{2} ) + 4 i s i n ( \frac{π}{2} ) - c o s ( - \frac{π}{2} ) - i s i n ( - \frac{π}{2} )}{2}}{2}

使用分式法则:

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

= \frac{4 cos ( \frac{π}{2} ) + 4 i sin ( \frac{π}{2} ) - cos ( - \frac{π}{2} ) - sin ( - \frac{π}{2} ) i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 cos ( \frac{π}{2} ) + 4 i sin ( \frac{π}{2} ) - cos ( - \frac{π}{2} ) - i sin ( - \frac{π}{2} )}{4}

将

\frac{4 cos ( \frac{π}{2} ) + 4 i sin ( \frac{π}{2} ) - cos ( - \frac{π}{2} ) - sin ( - \frac{π}{2} ) i}{4}

改写成标准复数形式:

\frac{4 cos ( \frac{π}{2} ) - cos ( - \frac{π}{2} )}{4} + \frac{4 sin ( \frac{π}{2} ) - sin ( - \frac{π}{2} )}{4} i

\frac{4 cos ( \frac{π}{2} ) + 4 i sin ( \frac{π}{2} ) - cos ( - \frac{π}{2} ) - sin ( - \frac{π}{2} ) i}{4}

使用分式法则:

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

\frac{4 cos ( \frac{π}{2} ) + 4 i sin ( \frac{π}{2} ) - cos ( - \frac{π}{2} ) - sin ( - \frac{π}{2} ) i}{4} = \frac{4 cos ( \frac{π}{2} )}{4} + \frac{4 i sin ( \frac{π}{2} )}{4} - \frac{cos ( - \frac{π}{2} )}{4} - \frac{sin ( - \frac{π}{2} ) i}{4}

= \frac{4 cos ( \frac{π}{2} )}{4} + \frac{4 i sin ( \frac{π}{2} )}{4} - \frac{cos ( - \frac{π}{2} )}{4} - \frac{i sin ( - \frac{π}{2} )}{4}

对同类项分组

= \frac{4 cos ( \frac{π}{2} )}{4} - \frac{cos ( - \frac{π}{2} )}{4} + \frac{4 i sin ( \frac{π}{2} )}{4} - \frac{i sin ( - \frac{π}{2} )}{4}

数字相除:

\frac{4}{4} = 1

= cos (\frac{π}{2}) - \frac{cos ( - \frac{π}{2} )}{4} + i sin (\frac{π}{2}) - \frac{i sin ( - \frac{π}{2} )}{4}

将复数的实部和虚部分组

= (cos (\frac{π}{2}) - \frac{cos ( - \frac{π}{2} )}{4}) + (sin (\frac{π}{2}) - \frac{sin ( - \frac{π}{2} )}{4}) i

sin (\frac{π}{2}) - \frac{sin ( - \frac{π}{2} )}{4} = \frac{4 sin ( \frac{π}{2} ) - sin ( - \frac{π}{2} )}{4}

sin (\frac{π}{2}) - \frac{sin ( - \frac{π}{2} )}{4}

将项转换为分式:

sin (\frac{π}{2}) = \frac{sin ( \frac{π}{2} ) 4}{4}

= \frac{sin ( \frac{π}{2} ) \cdot 4}{4} - \frac{sin ( - \frac{π}{2} )}{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{sin ( \frac{π}{2} ) \cdot 4 - sin ( - \frac{π}{2} )}{4}

= (cos (\frac{π}{2}) - \frac{cos ( - \frac{π}{2} )}{4}) + \frac{4 sin ( \frac{π}{2} ) - sin ( - \frac{π}{2} )}{4} i

cos (\frac{π}{2}) - \frac{cos ( - \frac{π}{2} )}{4} = \frac{4 cos ( \frac{π}{2} ) - cos ( - \frac{π}{2} )}{4}

cos (\frac{π}{2}) - \frac{cos ( - \frac{π}{2} )}{4}

将项转换为分式:

cos (\frac{π}{2}) = \frac{cos ( \frac{π}{2} ) 4}{4}

= \frac{cos ( \frac{π}{2} ) \cdot 4}{4} - \frac{cos ( - \frac{π}{2} )}{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{cos ( \frac{π}{2} ) \cdot 4 - cos ( - \frac{π}{2} )}{4}

= \frac{4 cos ( \frac{π}{2} ) - cos ( - \frac{π}{2} )}{4} + \frac{4 sin ( \frac{π}{2} ) - sin ( - \frac{π}{2} )}{4} i

= \frac{4 cos ( \frac{π}{2} ) - cos ( - \frac{π}{2} )}{4} + \frac{4 sin ( \frac{π}{2} ) - sin ( - \frac{π}{2} )}{4} i

= \frac{- cos ( - \frac{π}{2} ) + 4 cos ( \frac{π}{2} )}{4} + i \frac{- sin ( - \frac{π}{2} ) + 4 sin ( \frac{π}{2} )}{4}

利用以下特性:

sin (- x) = - sin (x)

sin (- \frac{π}{2}) = - sin (\frac{π}{2})

= \frac{- cos ( - \frac{π}{2} ) + 4 cos ( \frac{π}{2} )}{4} + i \frac{- ( - sin ( \frac{π}{2} ) ) + 4 sin ( \frac{π}{2} )}{4}

利用以下特性:

cos (- x) = cos (x)

cos (- \frac{π}{2}) = cos (\frac{π}{2})

= \frac{- cos ( \frac{π}{2} ) + 4 cos ( \frac{π}{2} )}{4} + i \frac{- ( - sin ( \frac{π}{2} ) ) + 4 sin ( \frac{π}{2} )}{4}

化简

= \frac{3 cos ( \frac{π}{2} )}{4} + i \frac{5 sin ( \frac{π}{2} )}{4}

= \frac{3 cos ( \frac{π}{2} )}{4} + i \frac{5 sin ( \frac{π}{2} )}{4}

流行的例子

6cos(45)+3sec(45)

6 cos (4 5^{\circ}) + 3 sec (4 5^{\circ})

tan(arccos(-(sqrt(3))/2)+pi)

tan (arccos (- \frac{3}{2}) + π)

cosh (0) - 1

2sec^2(pi/2)tan(pi/2)

2 sec^{2} (\frac{π}{2}) tan (\frac{π}{2})

(1-sin(30))/(1+sin(30))

\frac{1 - sin ( 3 0 ^{\circ} )}{1 + sin ( 3 0 ^{\circ} )}