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

Hint: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.