Question

show that sinix=isinhx​

Answer 1

Answer:

Sin(ix) = iSinhx

Step-by-step explanation:

We have to prove

Sin(ix) = iSinhx

L.H.S = sin(ix)

Since we know that

Sin(x) =  [ e^(i(x)) - e^(-i(x)) ] / (2i)

By putting ix in place of x we get

Sin(ix) =  [ e^(i(ix)) - e^(-i(ix)) ] / (2i)

          =  [ e(-x) - e^(x) ] / (2i)

by taking -1 common we get

L.H.S =  - [ e^(x) - e^(-x) ] / (2i)    

Multiplying and dividing by i      

L.H.S =  -i [ e^(x) - e^(-x) ] / (2i²)    

          = i [ e^(x) - e^(-x) ] / (2)       ∵ i² = -1

          =  iSinhx                             ∵  [ e^(x) - e^(-x) ] / (2) = Sinhx

          = R.H.S

Hence proved

L.H.S = R.H.S

That is

Sin(ix) = iSinhx

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

FORMULA : 1

Trigonometric function

$\displaystyle \: \sin x = \frac{ {e}^{ix} - {e}^{ - ix} }{2i} \: \: \: ....(1)$

${Where \: \: i = Imaginary \: Number }$

FORMULA : 2

Hyperbolic function

$\displaystyle \: \sin hx = \frac{ {e}^{x} - {e}^{ - x} }{2} \: \: \: ....(2)$

TO PROVE

$\sin ix = i \sin hx$

PROOF

LHS

$= \sin ix$

$\displaystyle \: = \frac{ {e}^{i \times ix} - {e}^{ - i \times ix} }{2i} \: \: (using \:formula \: 1)$

$\displaystyle \: = \frac{ {e}^{ {i}^{2} x} - {e}^{ - {i}^{2} x} }{2i}$

$\displaystyle \: = \frac{ {e}^{ - x} - {e}^{ x} }{2i} \: \: \: ( \because \: {i}^{2} = - 1)$

$\displaystyle \: = i \times \frac{ {e}^{ - x} - {e}^{ x} }{2 {i}^{2} } \: \: \:$

$\displaystyle \: = i \times \frac{ {e}^{ - x} - {e}^{ x} }{ - 2} \: \: \: ( \because \: {i}^{2} = - 1)$

$\displaystyle \: = i \times \frac{ {e}^{ x} - {e}^{ - x} }{2} \: \:$

$= i \times \sin hx \: \: (using \: formula \: 2)$

$= i \sin hx \:$

= RHS

Hence proved