What is the value of cosh(i pi/4) ?

The value of cosh(i pi/4) is cos(pi/4)

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cosh(i pi/4)

cosh (i \frac{π}{4})

cos (\frac{π}{4})

Decimal Notation

0.70710 \dots

Solution steps

cosh (i \frac{π}{4})

Simplify:

i \frac{π}{4} = \frac{πi}{4}

i \frac{π}{4}

Multiply fractions:

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

= \frac{πi}{4}

= cosh (\frac{πi}{4})

Rewrite using trig identities:

cos (\frac{π}{4})

cosh (\frac{πi}{4})

Use the Hyperbolic identity:

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

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

Simplify

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

\frac{e ^{\frac{πi}{4}} + e ^{- \frac{πi}{4}}}{2}

e^{\frac{πi}{4}} + e^{- \frac{πi}{4}} = cos (\frac{π}{4}) + i sin (\frac{π}{4}) + cos (- \frac{π}{4}) + i sin (- \frac{π}{4})

e^{\frac{πi}{4}} + e^{- \frac{πi}{4}}

Apply imaginary number rule:

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

= cos (\frac{π}{4}) + i sin (\frac{π}{4}) + e^{- \frac{πi}{4}}

Apply imaginary number rule:

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

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

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

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

Use the following property:

sin (- x) = - sin (x)

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

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

Use the following property:

cos (- x) = cos (x)

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

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

Simplify

= cos (\frac{π}{4})

= cos (\frac{π}{4})

sin (2 0^{\circ}) \cdot 2

arccos (- 0.64)

arctan (1) + π

\frac{7}{sin ( 4 5 ^{\circ} )}

cos (\frac{6}{5})