Question

The value of \[{\left( {\dfrac{{1 + \sin \dfrac{{2\pi }}{9} + i\cos \dfrac{{2\pi }}{9}}}{{1 + \sin \dfrac{{2\pi }}{9} - i\cos \dfrac{{2\pi }}{9}}}} \right)^3}\] is
1. \[\dfrac{{ - 1}}{2}\left( {1 - i\sqrt 3 } \right)\]
2. \[\dfrac{1}{2}\left( {1 - i\sqrt 3 } \right)\]
3. \[\dfrac{{ - 1}}{2}\left( {\sqrt 3 - i} \right)\]
4. \[\dfrac{1}{2}\left( {\sqrt 3 - i} \right)\]

Answer 1

Hint:
In this question, we have to find the value of the given function \[{\left( {\dfrac{{1 + \sin \dfrac{{2\pi }}{9} + i\cos \dfrac{{2\pi }}{9}}}{{1 + \sin \dfrac{{2\pi }}{9} - i\cos \dfrac{{2\pi }}{9}}}} \right)^3}\]. We solve this using the trigonometric functions and trigonometric identities. Method and convert \[\sin \] function into \[\cos \] function Here, we will use the De Moivre’s theorem to solve this question

Formula used:
The basic trigonometric formulas are given as
1. \[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \]
2. \[\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \]
3. \[\cos 2\theta = 2{\cos ^2}\theta - 1\]
4. \[\sin 2\theta = 2\sin \theta \cos \theta \]
5. \[{\left[ {r\left( {\cos \theta + i\sin \theta } \right)} \right]^n} = {r^n}\left( {\cos n\theta + i\sin n\theta } \right)\]
6. \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\]

Complete step-by-step solution:
The given function is \[{\left( {\dfrac{{1 + \sin \dfrac{{2\pi }}{9} + i\cos \dfrac{{2\pi }}{9}}}{{1 + \sin \dfrac{{2\pi }}{9} - i\cos \dfrac{{2\pi }}{9}}}} \right)^3}\].
Let us assume \[I = {\left( {\dfrac{{1 + \sin \dfrac{{2\pi }}{9} + i\cos \dfrac{{2\pi }}{9}}}{{1 + \sin \dfrac{{2\pi }}{9} - i\cos \dfrac{{2\pi }}{9}}}} \right)^3}\]
Firstly, we will apply the formula \[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \] and
\[\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta \] in the given function, we get
\[I = {\left( {\dfrac{{1 + \cos \left( {\dfrac{\pi }{2} - \dfrac{{2\pi }}{9}} \right) + i\sin \left( {\dfrac{\pi }{2} - \dfrac{{2\pi }}{9}} \right)}}{{1 + \cos \left( {\dfrac{\pi }{2} - \dfrac{{2\pi }}{9}} \right) - i\sin \left( {\dfrac{\pi }{2} - \dfrac{{2\pi }}{9}} \right)}}} \right)^3}\]
Now, we will simplify the sine and cosine function in the above expression, we get
\[I = {\left( {\dfrac{{1 + \cos \dfrac{{5\pi }}{{18}} + i\sin \dfrac{{5\pi }}{{18}}}}{{1 + \cos \dfrac{{5\pi }}{{18}} - i\sin \dfrac{{5\pi }}{{18}}}}} \right)^3}\]
Further, we use the trigonometry identities \[\cos 2\theta = 2{\cos ^2}\theta - 1\] and expressed \[1 + \cos \theta \] as \[2{\cos ^2}\dfrac{\theta }{2}\] and also use \[\sin 2\theta = 2\sin \theta \cos \theta \] and expressed as \[\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}\], we get
\[I = {\left( {\dfrac{{2{{\cos }^2}\dfrac{{5\pi }}{{36}} + 2i\sin \dfrac{{5\pi }}{{36}}\cos \dfrac{{5\pi }}{{36}}}}{{2{{\cos }^2}\dfrac{{5\pi }}{{36}} - 2i\sin \dfrac{{5\pi }}{{36}}\cos \dfrac{{5\pi }}{{36}}}}} \right)^3}\]
Furthermore, we will taken out \[2\cos \dfrac{{5\pi }}{{36}}\] as common from both numerator and denominator and cancel out, we get
\[\begin{array}{l}I = {\left( {\dfrac{{2\cos \dfrac{{5\pi }}{{36}}\left( {\cos \dfrac{{5\pi }}{{36}} + i\sin \dfrac{{5\pi }}{{36}}} \right)}}{{2\cos \dfrac{{5\pi }}{{36}}\left( {\cos \dfrac{{5\pi }}{{36}} - i\sin \dfrac{{5\pi }}{{36}}} \right)}}} \right)^3}\\I = {\left( {\dfrac{{\cos \dfrac{{5\pi }}{{36}} + i\sin \dfrac{{5\pi }}{{36}}}}{{\cos \dfrac{{5\pi }}{{36}} - i\sin \dfrac{{5\pi }}{{36}}}}} \right)^3}\end{array}\]
Now, we will multiply and divide both numerator and denominator by \[\cos \dfrac{\theta }{2} + i\sin \dfrac{\theta }{2}\], we get
\[I = {\left( {\dfrac{{\cos \dfrac{{5\pi }}{{36}} + i\sin \dfrac{{5\pi }}{{36}}}}{{\cos \dfrac{{5\pi }}{{36}} - i\sin \dfrac{{5\pi }}{{36}}}} \times \dfrac{{\cos \dfrac{{5\pi }}{{36}} + i\sin \dfrac{{5\pi }}{{36}}}}{{\cos \dfrac{{5\pi }}{{36}} + i\sin \dfrac{{5\pi }}{{36}}}}} \right)^3}\]
Further, we will use the formula \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\], we get
\[I = {\left( {\dfrac{{{{\left( {\cos \dfrac{{5\pi }}{{36}} + i\sin \dfrac{{5\pi }}{{36}}} \right)}^2}}}{{{{\cos }^2}\dfrac{{5\pi }}{{36}} - {i^2}{{\sin }^2}\dfrac{{5\pi }}{{36}}}}} \right)^3}\]
Furthermore, we will use the formula \[{i^2} = - 1\]in the above expression, we get
\[I = {\left( {\dfrac{{{{\left( {\cos \dfrac{{5\pi }}{{36}} + i\sin \dfrac{{5\pi }}{{36}}} \right)}^2}}}{{{{\cos }^2}\dfrac{{5\pi }}{{36}} + {{\sin }^2}\dfrac{{5\pi }}{{36}}}}} \right)^3}\]
As we know that \[{\cos ^2}\theta + {\sin ^2}\theta = 1\], we get
\[I = {\left( {\cos \dfrac{{5\pi }}{{36}} + i\sin \dfrac{{5\pi }}{{36}}} \right)^6}\]
Now, we will apply the demoivre’s theorem \[{\left[ {r\left( {\cos \theta + i\sin \theta } \right)} \right]^n} = {r^n}\left( {\cos n\theta + i\sin n\theta } \right)\], we get
\[\begin{array}{l}I = \cos 6 \times \dfrac{{5\pi }}{{36}} + i\sin 6 \times \dfrac{{5\pi }}{{36}}\\I = \cos \dfrac{{5\pi }}{6} + i\sin \dfrac{{5\pi }}{6}\end{array}\]
Further, we will write \[\dfrac{{5\pi }}{6} = \dfrac{\pi }{2} + \dfrac{\pi }{3}\] to simplify the above equation, we get
\[I = \cos \left( {\dfrac{\pi }{2} + \dfrac{\pi }{3}} \right) + i\sin \left( {\dfrac{\pi }{2} + \dfrac{\pi }{3}} \right)\]
Furthermore, we will apply the formula \[\cos \left( {\dfrac{\pi }{2} + \theta } \right) = - \sin \theta \] and
\[\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta \], we get
\[I = - \sin \dfrac{\pi }{3} + i\cos \dfrac{\pi }{3}\]
As we know that \[\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}\] and
\[\cos \dfrac{\pi }{3} = \dfrac{1}{2}\], we get
\[I = - \dfrac{1}{2}\left( {\sqrt 3 - i} \right)\]
Hence, the option (3) is correct

Note
In these types of questions, we should remember the formulas of trigonometry formulas and their value and their conversion also. We can also solve this question using exponential form that is \[{e^{i\theta }} = \cos \theta + i\sin \theta \]