Question

prove that sinix= isinhx​

Answer 1

Answer:

Step-by-step explanation:

Hope it helps you....

Answer 2

Answer:

To prove:

• $\sin ix=i\sin hx$

Proof:

We know that sine of an angle can be written in terms of the exponential as follows:

$\sin t=\dfrac{e^{it}-e^{-it}}{2i}$.

The sine hyperbolic of an angle can be written in terms of the exponential as

$\sin ht=\dfrac{e^{t}-e^{-t}}{2}$.

Therefore,

$\sin (ix)=\dfrac{e^{i(ix)}-e^{-i(ix)}}{2i}\\=\dfrac{e^{i^2x}-e^{-i^2x}}{2i}\\=\dfrac{e^{-1\cdot x}-e^{+1\cdot x}}{2i}\ \ \ \ \ \ \ \ \because i^2=-1\\=\dfrac{e^{- x}-e^{x}}{2i}\\=-\dfrac{e^{x}-e^{-x}}{2i}\\=\dfrac{-1}{i}\dfrac{e^x-e^{-x}}{2}\\=\dfrac{-1}{i}\sin h(x)\\=\dfrac{-1}{i}\cdot \dfrac ii \ \sin h(x)\\=\dfrac{-i}{i^2}\sin h(x)\\=\dfrac{-i}{-1}\sin h(x)\\=i\sin h(x)$

Hence, LHS = RHS, proved.