Question

Separate ${\sin ^{ - 1}}\left( {\cos \theta + i\sin \theta } \right)$ into real and imaginary parts, where θ is a positive acute angle.

Answer 1

Here, we need to find real and imaginary values separately. We will assume $\cos \theta + i\sin \theta = \sin (x + iy)$. Then, we will use the identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$ and $\sin (A + B) = \sin A\cos B + \cos A\sin B$. And substituting it and then separating the real and imaginary part, we will solve and get the final output.

Complete step by step answer:
Given that, ${\sin ^{ - 1}}\left( {\cos \theta + i\sin \theta } \right)$
Let, $\cos \theta + i\sin \theta = \sin (x + iy)$
We know that, $\sin (A + B) = \sin A\cos B + \cos A\sin B$
Using this formula, we will get,
$\Rightarrow \cos \theta + i\sin \theta = \sin x\cos iy + \cos x\sin iy$
We know that, $\cos iy = \cosh y$ and $\sin iy = i\sinh y$
Substituting these values, we will get,
$\Rightarrow \cos \theta + i\sin \theta = \sin x\cosh y + i\cos x\sinh y$

Now, we will equate the real and imaginary parts separately as below:
Real part: $\cos \theta = \sin x\cosh y$ ----- (1)
Imaginary part: $\sin \theta = \cos x\sinh y$ ---- (2)
Next, squaring and adding (1) and (2), we will get,
$\Rightarrow {\cos ^2}\theta + {\sin ^2}\theta = {\sin ^2}x{\cosh ^2}y + {\cos ^2}x{\sinh ^2}y$
We will use the trigonometric ratios identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$, applying this we will get,
$\Rightarrow 1 = {\sin ^2}x{\cosh ^2}y + {\cos ^2}x{\sinh ^2}y$
Taking ${\cosh ^2}y = 1 + {\sinh ^2}y$ and applying this, we will get,
$\Rightarrow 1 = {\sin ^2}x\left( {1 + {{\sinh }^2}y} \right) + {\cos ^2}x{\sinh ^2}y$

Opening the brackets, we will get,
$\Rightarrow 1 = {\sin ^2}x + {\sin ^2}x{\sinh ^2}y + {\cos ^2}x{\sinh ^2}y$
$\Rightarrow 1 = {\sin ^2}x + {\sinh ^2}y\left( {{{\sin }^2}x + {{\cos }^2}x} \right)$
$\Rightarrow 1 = {\sin ^2}x + {\sinh ^2}y\left( 1 \right)$ $\left( {\because {{\sin }^2}x + {{\cos }^2}x = 1} \right)$
$\Rightarrow 1 = {\sin ^2}x + {\sinh ^2}y$
By using transposing method, we will move the RHS term i.e. ${\sin ^2}x$ to LHS, we will get,
$\Rightarrow 1 - {\sin ^2}x = {\sinh ^2}y$
We know that, $1 - {\sin ^2}x = {\cos ^2}x$ and substituting this value, we will get,
$\Rightarrow {\cos ^2}x = {\sinh ^2}y$ ----- (3)

Now, we will use the equation (2), to get the real part value, i.e.
$\sin \theta = \cos x\sinh y$
$\Rightarrow \dfrac{{\sin \theta }}{{\cos x}} = \sinh y$
$\Rightarrow \sinh y = \dfrac{{\sin \theta }}{{\cos x}}$
Squaring on both the sides, we will get,
$\Rightarrow {\left( {\sinh y} \right)^2} = {\left( {\dfrac{{\sin \theta }}{{\cos x}}} \right)^2}$
$\Rightarrow {\sinh ^2}y = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}x}}$

From equation (3), we ${\cos ^2}x = {\sinh ^2}y$ and so using this in above equation, we will get,
$\Rightarrow {\cos ^2}x = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}x}}$
$\Rightarrow {\cos ^2}x\left( {{{\cos }^2}x} \right) = {\sin ^2}\theta$
$\Rightarrow {\cos ^4}x = {\sin ^2}\theta$
As we are given, $\theta$ being a positive acute angle, $\sin \theta$ is positive.Taking square root on both sides, we will get,
$\Rightarrow {\cos ^2}x = \sin \theta$
As x lies between $- \dfrac{\pi }{2}$ to $\dfrac{\pi }{2}$.
Again, taking square root on both the sides, we will get,
$\Rightarrow \cos x = \sqrt {\sin \theta }$ ------- (4)
$\Rightarrow x = {\cos ^{ - 1}}\left( {\sqrt {\sin \theta } } \right)$

Also, to get the imaginary part value, we will use the relation from equation (2), and we get,
$\sin \theta = \cos x\sinh y$
Using equation (4), we will get,
$\Rightarrow \sin \theta = \sqrt {\sin \theta } \sinh y$
$\Rightarrow \sinh y = \dfrac{{\sin \theta }}{{\sqrt {\sin \theta } }}$
$\Rightarrow \sinh y = \dfrac{{\sqrt {\sin \theta } \sqrt {\sin \theta } }}{{\sqrt {\sin \theta } }}$ $\left( {\because \sin \theta = \sqrt {\sin \theta } \sqrt {\sin \theta } } \right)$
On simplifying this, we will get,
$\Rightarrow \sinh y = \sqrt {\sin \theta }$
$\therefore y = log[sin\theta + (1 + sin\theta )]\;$

Hence, the separate values of real and imaginary parts are $x = {\cos ^{ - 1}}\left( {\sqrt {\sin \theta } } \right)$ and $y = log[sin\theta + (1 + sin\theta )]\;$ respectively, when $\theta$ is a positive acute angle.

Note: In this problem, the important step lies in the determination of the angle $\theta$. As it is given that, $\theta$ is an acute angle so we will consider only that value of $\theta$ which measures less than ${90^ \circ }$. Also, trigonometry is one of those divisions in mathematics that helps in finding the angles and the missing sides of a triangle too, with the help of trigonometric ratios.