Answer
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Hint: Here we will use the formula for hyperbolic function and the formula of sine function expressed in exponential form. Since the standard formula is expressed in terms of “x”, we will convert the form in “ix” form. Then will simplify the equation for the resultant answer.
Complete step-by-step answer:
Using the identity for hyperbolic function –
$ \sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2} $
Here, the given function is with respect to therefore convert “x” to “ix” form
Therefore, the above equation can be expressed as –
$ \Rightarrow \sinh (ix) = \dfrac{{{e^{ix}} - {e^{ - ix}}}}{2} $ .... (A)
Also, $ {e^{ix}} = \cos x + i\sin x $
Place the above value in equation (A)
$ \Rightarrow \sinh (ix) = \dfrac{{(\cos x + i\sin x) - (\cos ( - x) + i\sin ( - x))}}{2} $
Now, use the properties of sines and cosines and apply accordingly. The odd and even trigonometric functions states that -
$
\sin ( - \theta ) = - \sin \theta \\
\cos ( - \theta ) = \cos \theta \;
$
$ \Rightarrow \sinh (ix) = \dfrac{{(\cos x + i\sin x) - (\cos x - i\sin x)}}{2} $
Now, open the brackets and simplify. Remember when there is a negative sign outside the bracket, all the terms inside the bracket changes. Positive term changes to negative and negative becomes positive.
$ \Rightarrow \sinh (ix) = \dfrac{{\cos x + i\sin x - \cos x + i\sin x}}{2} $
Make the pair of like terms and simplify –
$ \Rightarrow \sinh (ix) = \dfrac{{\underline {\cos x - \cos x} + \underline {i\sin x + i\sin x} }}{2} $
Like terms with equal value and opposite signs cancel each other. And add like terms with positive terms.
$ \Rightarrow \sinh (ix) = \dfrac{{2i\sin x}}{2} $
Common multiple from the numerator and the denominator cancel each other. Therefore remove from the numerator and the denominator.
$ \Rightarrow \sinh (ix) = i\sin x $
So, the correct answer is “Option A”.
Note: Always remember the standard hyperbolic formula for the trigonometric function and the exponential form of the trigonometric function. Remember the standard formula properly as the further solution depends on it. Be clear with the odd and even trigonometric function. Cosine is an even function whereas sine is an odd function.
Complete step-by-step answer:
Using the identity for hyperbolic function –
$ \sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2} $
Here, the given function is with respect to therefore convert “x” to “ix” form
Therefore, the above equation can be expressed as –
$ \Rightarrow \sinh (ix) = \dfrac{{{e^{ix}} - {e^{ - ix}}}}{2} $ .... (A)
Also, $ {e^{ix}} = \cos x + i\sin x $
Place the above value in equation (A)
$ \Rightarrow \sinh (ix) = \dfrac{{(\cos x + i\sin x) - (\cos ( - x) + i\sin ( - x))}}{2} $
Now, use the properties of sines and cosines and apply accordingly. The odd and even trigonometric functions states that -
$
\sin ( - \theta ) = - \sin \theta \\
\cos ( - \theta ) = \cos \theta \;
$
$ \Rightarrow \sinh (ix) = \dfrac{{(\cos x + i\sin x) - (\cos x - i\sin x)}}{2} $
Now, open the brackets and simplify. Remember when there is a negative sign outside the bracket, all the terms inside the bracket changes. Positive term changes to negative and negative becomes positive.
$ \Rightarrow \sinh (ix) = \dfrac{{\cos x + i\sin x - \cos x + i\sin x}}{2} $
Make the pair of like terms and simplify –
$ \Rightarrow \sinh (ix) = \dfrac{{\underline {\cos x - \cos x} + \underline {i\sin x + i\sin x} }}{2} $
Like terms with equal value and opposite signs cancel each other. And add like terms with positive terms.
$ \Rightarrow \sinh (ix) = \dfrac{{2i\sin x}}{2} $
Common multiple from the numerator and the denominator cancel each other. Therefore remove from the numerator and the denominator.
$ \Rightarrow \sinh (ix) = i\sin x $
So, the correct answer is “Option A”.
Note: Always remember the standard hyperbolic formula for the trigonometric function and the exponential form of the trigonometric function. Remember the standard formula properly as the further solution depends on it. Be clear with the odd and even trigonometric function. Cosine is an even function whereas sine is an odd function.
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