Question

Find \[\sinh \left( {\dfrac{\pi }{6}\iota } \right) = \]
(1) \[\dfrac{{ - \iota }}{2}\]
(2) \[\dfrac{\iota }{2}\]
(3) \[\dfrac{{\iota \sqrt 3 }}{2}\]
(4) \[\dfrac{{ - \iota \sqrt 3 }}{2}\]

Answer 1

Hint: In this question, we have to find the value of \[\sinh \left( {\dfrac{\pi }{6}\iota } \right)\] .As we don’t know the direct value of hyperbolic functions. So, in order to solve this question, we will first explore the relation between complex hyperbolic and complex trigonometric functions, i.e., \[\sinh \left( {\iota x} \right) = \iota \sin x\] .After that we will substitute the value of \[x\] as \[\dfrac{\pi }{6}\] in the above result to get the required value.

Complete step-by-step answer:
First of all, we will first explore the relation between complex hyperbolic and complex trigonometric functions.
Now, as we know that
Exponential form of \[\sin x\] is \[\dfrac{{{e^{\iota x}} - {e^{ - \iota x}}}}{{2\iota }}{\text{ }} - - - \left( 1 \right)\]
And exponential form of \[\sinh x\] is \[\dfrac{{{e^x} - {e^{ - x}}}}{2}{\text{ }} - - - \left( 2 \right)\]
Now, if we substitute \[\iota x\] in the place of \[x\] in equation \[\left( 2 \right)\] ,we get
\[\sinh \left( {\iota x} \right) = \dfrac{{{e^{\iota x}} - {e^{ - \iota x}}}}{2}{\text{ }} - - - \left( 3 \right)\]
On multiplying and dividing by \[\iota \] on the right-hand side of the equation \[\left( 3 \right)\] ,we get
\[\sinh \left( {\iota x} \right) = \iota \left( {\dfrac{{{e^{\iota x}} - {e^{ - \iota x}}}}{{2\iota }}} \right){\text{ }} - - - \left( 4 \right)\]
Now, using equation \[\left( 1 \right)\] we can write equation \[\left( 4 \right)\] as,
\[\sinh \left( {\iota x} \right) = \iota \sin x\]
which is the relation we required.
Now, we have to find the value of \[\sinh \left( {\dfrac{\pi }{6}\iota } \right)\]
So, here \[x = \dfrac{\pi }{6}\]
So, on substituting the value of \[x\] , we get
\[\sinh \left( {\dfrac{\pi }{6}\iota } \right) = \iota \sin \left( {\dfrac{\pi }{6}} \right)\]
As we know that the value of \[\sin \left( {\dfrac{\pi }{6}} \right)\] is equal to \[\dfrac{1}{2}\]
Therefore, we get
\[\sinh \left( {\dfrac{\pi }{6}\iota } \right) = \iota \left( {\dfrac{1}{2}} \right)\]
\[ \Rightarrow \sinh \left( {\dfrac{\pi }{6}\iota } \right) = \dfrac{\iota }{2}\]
Thus, the value of \[\sinh \left( {\dfrac{\pi }{6}\iota } \right)\] is equal to \[\dfrac{\iota }{2}\]
So, the correct answer is “Option 2”.

Note: The first mistake students can make while solving these types of questions is to write the formulas for \[\sin x\] and \[\sinh x\] in their exponential form. So, don’t get confused between the two. And another chance of mistake can happen while finding the relation, so try not to do the same.
Also, there are some more important relations between complex hyperbolic and complex trigonometric functions, that are:
(i) \[\sin \left( x \right) = - \iota \sinh \left( {\iota x} \right)\]
(ii) \[\sin \left( {\iota x} \right) = \iota \sinh \left( x \right)\]
(iii) \[\sinh \left( x \right) = - \iota \cos \left( {\iota x} \right)\]
(iv) \[\cosh \left( x \right) = \cos \left( {\iota x} \right)\]
(v) \[\cos \left( x \right) = \cosh \left( {\iota x} \right)\]