What is the value of sinh(2+i) ?

The value of sinh(2+i) is (-cos(1)+e^4cos(1))/(2e^2)+i(sin(1)+e^4sin(1))/(2e^2)

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sinh(2+i)

Solution

sinh (2 + i)

Solution

\frac{- cos ( 1 ) + e ^{4} cos ( 1 )}{2 e ^{2}} + i \frac{sin ( 1 ) + e ^{4} sin ( 1 )}{2 e ^{2}}

Solution steps

sinh (2 + i)

Use the Hyperbolic identity:

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

= \frac{e ^{2 + i} - e ^{- (2 + i)}}{2}

Simplify

\frac{e ^{2 + i} - e ^{- (2 + i)}}{2} : \frac{- cos ( - 1 ) + e ^{4} cos ( 1 )}{2 e ^{2}} + i \frac{- sin ( - 1 ) + e ^{4} sin ( 1 )}{2 e ^{2}}

\frac{e ^{2 + i} - e ^{- (2 + i)}}{2}

e^{2 + i} - e^{- (2 + i)} = e^{2} (cos (1) + i sin (1)) - e^{- 2} (cos (- 1) + i sin (- 1))

e^{2 + i} - e^{- (2 + i)}

Apply imaginary number rule:

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

= e^{2} (cos (1) + i sin (1)) - e^{- (2 + i)}

Apply imaginary number rule:

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

= e^{2} (cos (1) + i sin (1)) - e^{- 2} (cos (- 1) + i sin (- 1))

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

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

Remove parentheses:

(a) = a

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

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

= \frac{e ^{2} ( cos ( 1 ) + i sin ( 1 ) ) - \frac{c o s ( - 1 ) + i s i n ( - 1 )}{e ^{2}}}{2}

Join

e^{2} (cos (1) + sin (1) i) - \frac{cos ( - 1 ) + sin ( - 1 ) i}{e ^{2}} : \frac{e ^{4} cos ( 1 ) + e ^{4} i sin ( 1 ) - cos ( - 1 ) - i sin ( - 1 )}{e ^{2}}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

e^{2} (cos (1) + sin (1) i) e^{2} - (cos (- 1) + sin (- 1) i) = e^{4} (cos (1) + i sin (1)) - (cos (- 1) + i sin (- 1))

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

e^{2} (cos (1) + sin (1) i) e^{2} = e^{4} (cos (1) + i sin (1))

e^{2} (cos (1) + sin (1) i) e^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e^{2} e^{2} = e^{2 + 2}

= (cos (1) + sin (1) i) e^{2 + 2}

Add the numbers:

2 + 2 = 4

= (cos (1) + sin (1) i) e^{4}

= e^{4} (cos (1) + i sin (1)) - (cos (- 1) + i sin (- 1))

= \frac{e ^{4} ( cos ( 1 ) + i sin ( 1 ) ) - ( cos ( - 1 ) + i sin ( - 1 ) )}{e ^{2}}

Expand

(cos (1) + sin (1) i) e^{4} - (cos (- 1) + sin (- 1) i) : e^{4} cos (1) + e^{4} i sin (1) - cos (- 1) - sin (- 1) i

(cos (1) + sin (1) i) e^{4} - (cos (- 1) + sin (- 1) i)

= e^{4} (cos (1) + i sin (1)) - (cos (- 1) + i sin (- 1))

Expand

e^{4} (cos (1) + sin (1) i) : e^{4} cos (1) + e^{4} i sin (1)

e^{4} (cos (1) + sin (1) i)

Apply the distributive law:

a (b + c) = ab + a c

a = e^{4}, b = cos (1), c = sin (1) i

= e^{4} cos (1) + e^{4} sin (1) i

= e^{4} cos (1) + e^{4} i sin (1)

= e^{4} cos (1) + e^{4} i sin (1) - (cos (- 1) + sin (- 1) i)

- (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

= e^{4} cos (1) + e^{4} i sin (1) - cos (- 1) - sin (- 1) i

= \frac{e ^{4} cos ( 1 ) + e ^{4} i sin ( 1 ) - cos ( - 1 ) - i sin ( - 1 )}{e ^{2}}

= \frac{\frac{e ^{4} c o s ( 1 ) + e ^{4} i s i n ( 1 ) - c o s ( - 1 ) - i s i n ( - 1 )}{e ^{2}}}{2}

Apply the fraction rule:

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

= \frac{e ^{4} cos ( 1 ) + e ^{4} i sin ( 1 ) - cos ( - 1 ) - sin ( - 1 ) i}{e ^{2} \cdot 2}

Rewrite

\frac{e ^{4} cos ( 1 ) + e ^{4} i sin ( 1 ) - cos ( - 1 ) - sin ( - 1 ) i}{e ^{2} \cdot 2}

in standard complex form:

\frac{e ^{4} cos ( 1 ) - cos ( - 1 )}{2 e ^{2}} + \frac{e ^{4} sin ( 1 ) - sin ( - 1 )}{2 e ^{2}} i

\frac{e ^{4} cos ( 1 ) + e ^{4} i sin ( 1 ) - cos ( - 1 ) - sin ( - 1 ) i}{e ^{2} \cdot 2}

Apply the fraction rule:

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

\frac{e ^{4} cos ( 1 ) + e ^{4} i sin ( 1 ) - cos ( - 1 ) - sin ( - 1 ) i}{e ^{2} \cdot 2} = \frac{e ^{4} cos ( 1 )}{e ^{2} \cdot 2} + \frac{e ^{4} i sin ( 1 )}{e ^{2} \cdot 2} - \frac{cos ( - 1 )}{e ^{2} \cdot 2} - \frac{sin ( - 1 ) i}{e ^{2} \cdot 2}

= \frac{e ^{4} cos ( 1 )}{2 e ^{2}} + \frac{e ^{4} i sin ( 1 )}{2 e ^{2}} - \frac{cos ( - 1 )}{2 e ^{2}} - \frac{i sin ( - 1 )}{2 e ^{2}}

Group like terms

= - \frac{cos ( - 1 )}{2 e ^{2}} - \frac{i sin ( - 1 )}{2 e ^{2}} + \frac{e ^{4} cos ( 1 )}{2 e ^{2}} + \frac{e ^{4} i sin ( 1 )}{2 e ^{2}}

Cancel

\frac{e ^{4} cos ( 1 )}{2 e ^{2}} : \frac{e ^{2} cos ( 1 )}{2}

\frac{e ^{4} cos ( 1 )}{2 e ^{2}}

Cancel

\frac{e ^{4} cos ( 1 )}{2 e ^{2}} : \frac{e ^{2} cos ( 1 )}{2}

\frac{e ^{4} cos ( 1 )}{2 e ^{2}}

Apply exponent rule:

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

\frac{e ^{4}}{e ^{2}} = e^{4 - 2}

= \frac{e ^{4 - 2} cos ( 1 )}{2}

Subtract the numbers:

4 - 2 = 2

= \frac{e ^{2} cos ( 1 )}{2}

= \frac{e ^{2} cos ( 1 )}{2}

= - \frac{cos ( - 1 )}{2 e ^{2}} - \frac{i sin ( - 1 )}{2 e ^{2}} + \frac{e ^{2} cos ( 1 )}{2} + \frac{e ^{4} i sin ( 1 )}{2 e ^{2}}

Cancel

\frac{e ^{4} i sin ( 1 )}{2 e ^{2}} : \frac{e ^{2} i sin ( 1 )}{2}

\frac{e ^{4} i sin ( 1 )}{2 e ^{2}}

Cancel

\frac{e ^{4} i sin ( 1 )}{2 e ^{2}} : \frac{e ^{2} i sin ( 1 )}{2}

\frac{e ^{4} i sin ( 1 )}{2 e ^{2}}

Apply exponent rule:

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

\frac{e ^{4}}{e ^{2}} = e^{4 - 2}

= \frac{i e ^{4 - 2} sin ( 1 )}{2}

Subtract the numbers:

4 - 2 = 2

= \frac{e ^{2} i sin ( 1 )}{2}

= \frac{e ^{2} i sin ( 1 )}{2}

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

Group like terms

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

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

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

\frac{e ^{2} sin ( 1 )}{2} - \frac{sin ( - 1 )}{2 e ^{2}} = \frac{e ^{4} sin ( 1 ) - sin ( - 1 )}{2 e ^{2}}

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

Least Common Multiplier of

2, 2 e^{2} : 2 e^{2}

2, 2 e^{2}

Lowest Common Multiplier (LCM)

Least Common Multiplier of

2, 2 : 2

2, 2

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

2

2

= 2

Multiply the numbers:

2 = 2

= 2

Compute an expression comprised of factors that appear either in

2

2 e^{2}

= 2 e^{2}

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

2 e^{2}

For

\frac{e ^{2} sin ( 1 )}{2} :

multiply the denominator and numerator by

e^{2}

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

= \frac{e ^{4} sin ( 1 )}{2 e ^{2}} - \frac{sin ( - 1 )}{2 e ^{2}}

Since the denominators are equal, combine the fractions:

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

= \frac{e ^{4} sin ( 1 ) - sin ( - 1 )}{2 e ^{2}}

= (\frac{e ^{2} cos ( 1 )}{2} - \frac{cos ( - 1 )}{2 e ^{2}}) + \frac{e ^{4} sin ( 1 ) - sin ( - 1 )}{2 e ^{2}} i

\frac{e ^{2} cos ( 1 )}{2} - \frac{cos ( - 1 )}{2 e ^{2}} = \frac{e ^{4} cos ( 1 ) - cos ( - 1 )}{2 e ^{2}}

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

Least Common Multiplier of

2, 2 e^{2} : 2 e^{2}

2, 2 e^{2}

Lowest Common Multiplier (LCM)

Least Common Multiplier of

2, 2 : 2

2, 2

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

2

2

= 2

Multiply the numbers:

2 = 2

= 2

Compute an expression comprised of factors that appear either in

2

2 e^{2}

= 2 e^{2}

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

2 e^{2}

For

\frac{e ^{2} cos ( 1 )}{2} :

multiply the denominator and numerator by

e^{2}

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

= \frac{e ^{4} cos ( 1 )}{2 e ^{2}} - \frac{cos ( - 1 )}{2 e ^{2}}

Since the denominators are equal, combine the fractions:

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

= \frac{e ^{4} cos ( 1 ) - cos ( - 1 )}{2 e ^{2}}

= \frac{e ^{4} cos ( 1 ) - cos ( - 1 )}{2 e ^{2}} + \frac{e ^{4} sin ( 1 ) - sin ( - 1 )}{2 e ^{2}} i

= \frac{e ^{4} cos ( 1 ) - cos ( - 1 )}{2 e ^{2}} + \frac{e ^{4} sin ( 1 ) - sin ( - 1 )}{2 e ^{2}} i

= \frac{- cos ( - 1 ) + e ^{4} cos ( 1 )}{2 e ^{2}} + i \frac{- sin ( - 1 ) + e ^{4} sin ( 1 )}{2 e ^{2}}

Use the following property:

sin (- x) = - sin (x)

sin (- 1) = - sin (1)

= \frac{- cos ( - 1 ) + e ^{4} cos ( 1 )}{2 e ^{2}} + i \frac{- ( - sin ( 1 ) ) + e ^{4} sin ( 1 )}{2 e ^{2}}

Use the following property:

cos (- x) = cos (x)

cos (- 1) = cos (1)

= \frac{- cos ( 1 ) + e ^{4} cos ( 1 )}{2 e ^{2}} + i \frac{- ( - sin ( 1 ) ) + e ^{4} sin ( 1 )}{2 e ^{2}}

Simplify

= \frac{- cos ( 1 ) + e ^{4} cos ( 1 )}{2 e ^{2}} + i \frac{sin ( 1 ) + e ^{4} sin ( 1 )}{2 e ^{2}}

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Frequently Asked Questions (FAQ)

What is the value of sinh(2+i) ?
The value of sinh(2+i) is (-cos(1)+e^4cos(1))/(2e^2)+i(sin(1)+e^4sin(1))/(2e^2)