What is the value of sinh(i) ?

The value of sinh(i) is isin(1)

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sinh(i)

sinh (i)

i sin (1)

Solution steps

sinh (i)

Use the Hyperbolic identity:

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

= \frac{e ^{i} - e ^{- i}}{2}

Simplify

\frac{e ^{i} - e ^{- i}}{2} : \frac{- cos ( - 1 ) + cos ( 1 )}{2} + i \frac{- sin ( - 1 ) + sin ( 1 )}{2}

\frac{e ^{i} - e ^{- i}}{2}

e^{i} - e^{- i} = cos (1) + i sin (1) - (cos (- 1) + i sin (- 1))

e^{i} - e^{- i}

Apply imaginary number rule:

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

= cos (1) + i sin (1) - e^{- i}

Apply imaginary number rule:

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

= cos (1) + i sin (1) - (cos (- 1) + i sin (- 1))

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

Expand

cos (1) + sin (1) i - (cos (- 1) + sin (- 1) i) : cos (1) + sin (1) i - cos (- 1) - sin (- 1) i

cos (1) + sin (1) i - (cos (- 1) + sin (- 1) i)

= cos (1) + i sin (1) - (cos (- 1) + i sin (- 1))

- (cos (- 1) + sin (- 1) i) : - cos (- 1) - sin (- 1) i

- (cos (- 1) + sin (- 1) i)

Distribute parentheses

= - (cos (- 1)) - (sin (- 1) i)

Apply minus-plus rules

+ (- a) = - a

= - cos (- 1) - sin (- 1) i

= cos (1) + sin (1) i - cos (- 1) - sin (- 1) i

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

Rewrite

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

in standard complex form:

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

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

Apply the fraction rule:

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

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

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

Group like terms

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

Group the real part and the imaginary part of the complex number

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

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

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

Apply rule

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

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

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

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

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

Apply rule

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

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

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

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

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

Use the following property:

sin (- x) = - sin (x)

sin (- 1) = - sin (1)

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

Use the following property:

cos (- x) = cos (x)

cos (- 1) = cos (1)

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

Simplify

= i sin (1)