What is the value of cosh((ipi)/2) ?

The value of cosh((ipi)/2) is cos(pi/2)

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cosh((ipi)/2)

cosh (\frac{iπ}{2})

cos (\frac{π}{2})

Decimal Notation

0

Solution steps

cosh (\frac{iπ}{2})

Use the Hyperbolic identity:

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

= \frac{e ^{\frac{πi}{2}} + e ^{- \frac{πi}{2}}}{2}

Simplify

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

\frac{e ^{\frac{πi}{2}} + e ^{- \frac{πi}{2}}}{2}

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

e^{\frac{πi}{2}} + e^{- \frac{πi}{2}}

Apply imaginary number rule:

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

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

Apply imaginary number rule:

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

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

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

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

Use the following property:

sin (- x) = - sin (x)

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

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

Use the following property:

cos (- x) = cos (x)

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

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

Simplify

= cos (\frac{π}{2})

\frac{10}{tan ( 3 0 ^{\circ} )}

4 cos (31 5^{\circ})

- tan (\frac{π}{2})

csc^{2} (4 5^{\circ})

\frac{8}{sin ( 6 0 ^{\circ} )}