Question

The imaginary part of \[{\sin ^2}(x + iy)\] is,
(1)\[\frac{1}{2}\cosh 2x\cos 2y\]
(2) \[\frac{1}{2}\cos 2x\cosh 2y\]
(3) \[\frac{1}{2}\sinh 2x\sin 2y\]
(4) \[\frac{1}{2}\sin 2x\sinh 2y\]

Answer 1

Hint:This question is from the chapter complex number. First of all, simplify the given expression using the concept of trigonometry. After that, compare the simplified expression with \[A + iB\]. In which one part is real and another part will be imaginary.

Formula Used: 1) \[\begin{array}{*{20}{c}}
  {\cos 2x}& = &{1 - 2{{\sin }^2}x}
\end{array}\]
2) \[\begin{array}{*{20}{c}}
  {2{{\sin }^2}x}& = &{1 - }
\end{array}\cos 2x\]
3) \[\begin{array}{*{20}{c}}
  {\cos (p + q)}& = &{\cos p\cos q - \sin p\sin q}
\end{array}\]

Complete step by step Solution:
Complex numbers are number that contains the real part and the imaginary part. In other words, we can say that complex numbers are the combination of real and imaginary numbers. It is denoted by Z. There are a lot of applications of the complex numbers in daily life such as electric circuits and lots more. Let us consider a complex number that is,
\[\begin{array}{*{20}{c}}
  { \Rightarrow Z}& = &{A + iB}
\end{array}\]
In this complex number, A is the real part and\[iB\] is the imaginary part. \[i\]is an imaginary number. A and b are the real numbers. Imaginary numbers are the square root of -1. Imaginary numbers are denoted by\[i\].
Now, we have given that
\[\begin{array}{*{20}{c}}
  { \Rightarrow Z}& = &{{{\sin }^2}(x + iy)}
\end{array}\]
Divide and multiply by the 2 in the above equation, so we can write.
\[\begin{array}{*{20}{c}}
  { \Rightarrow Z}& = &{\frac{2}{2}{{\sin }^2}(x + iy)}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow Z}& = &{\frac{1}{2}\left[ {2{{\sin }^2}(x + iy)} \right]}
\end{array}\] ……….. (a)
Now we know that
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\cos 2x}& = &{1 - 2{{\sin }^2}x}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  {2{{\sin }^2}x}& = &{1 - }
\end{array}\cos 2x\]
Now from equation (a), we will get.
\[\begin{array}{*{20}{c}}
  { \Rightarrow Z}& = &{\frac{1}{2}\left[ {1 - \cos 2(x + iy)} \right]}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow Z}& = &{\frac{1}{2}\left[ {1 - \cos (2x + 2iy)} \right]}
\end{array}\]
And now, we will use,
\[\begin{array}{*{20}{c}}
  { \Rightarrow \cos (p + q)}& = &{\cos p\cos q - \sin p\sin q}
\end{array}\]
Therefore, we will get it.
\[\begin{array}{*{20}{c}}
  { \Rightarrow Z}& = &{\frac{1}{2}\left[ {1 - \left\{ {\cos 2x\cos i2y - \sin 2x\sin i2y} \right\}} \right]}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow Z}& = &{\frac{1}{2}\left[ {1 - \cos 2x\cos i2y + \sin 2x\sin i2y} \right]}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow Z}& = &{\frac{1}{2}\left[ {1 - \cos 2x\cosh 2y + i\sin 2x\sinh 2y} \right]}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow Z}& = &{\frac{1}{2} - \frac{1}{2}\cos 2x\cosh 2y + i\frac{1}{2}\sin 2x\sinh 2y}
\end{array}\]
So, the imaginary part is \[\frac{1}{2}\sin 2x\sinh 2y\]
Now, the final answer is \[\frac{1}{2}\sin 2x\sinh 2y\].

Hence, the correct option is 4.

Note: Before comparing the expression, simplify the given expression. Once we get the simplified form of the given expression, we will compare the expression. sinh and cosh are mathematical functions. sinh is called hyperbolic sine and cosh is called hyperbolic cosine.