Question

If we have $x = {e^{i\theta }}$ and $y = {e^{i\phi }}$ then prove that ${x^m}{y^n} + \dfrac{1}{{{x^m}{y^n}}} = 2\cos \left( {m\theta + n\phi } \right)$.

Answer 1

Hint – In this particular type of question use the concept that in complex system ${e^{ix}} = \cos x + i\sin x$ and ${e^{ - ix}} = \cos \left( { - x} \right) + i\sin \left( { - x} \right)$, use the concept that cos (A + B) = cos A cos B – sin A sin B so use these properties to reach the solution of the question.


Complete step-by-step solution -
Proof –
As we all know in complex system,
${e^{ix}} = \cos x + i\sin x$ and ${e^{ - ix}} = \cos \left( { - x} \right) + i\sin \left( { - x} \right)$
Now it is given that, $x = {e^{i\theta }}$
Now take the ${m^{th}}$ power in the both sides we have,
Therefore, ${x^m} = {\left( {{e^{i\theta }}} \right)^m} = {e^{im\theta }}$............. (1)
Therefore, ${x^m} = \cos \left( {m\theta } \right) + i\sin \left( {m\theta } \right)$.................(2)
And it is also given that, $y = {e^{i\phi }}$
Now take the ${n^{th}}$ power in the both sides we have,
Therefore, ${y^n} = {\left( {{e^{i\phi }}} \right)^n} = {e^{in\phi }}$............... (3)
Therefore, ${y^n} = \cos \left( {n\phi } \right) + i\sin \left( {n\phi } \right)$................... (4)
Now take the L.H.S of the given equation we have,
$ \Rightarrow {x^m}{y^n} + \dfrac{1}{{{x^m}{y^n}}}$
This equation is also written as,
$ \Rightarrow {x^m}{y^n} + {x^{ - m}}{y^{ - n}}$....................... (5)
Now from equation (1) we have,
${x^m} = {\left( {{e^{i\theta }}} \right)^m} = {e^{im\theta }}$
Therefore, ${x^{ - m}} = {\left( {{e^{i\theta }}} \right)^{ - m}} = {e^{ - im\theta }}$
${x^{ - m}} = \cos \left( { - m\theta } \right) + i\sin \left( { - m\theta } \right)$
Now as we know that cos (-x) = cos x and sin (-x) = -sin x so use this property in the above equation we have,
${x^{ - m}} = \cos \left( {m\theta } \right) - i\sin \left( {m\theta } \right)$............... (6)
Similarly from equation (3) we have,
${y^{ - n}} = \cos \left( {n\phi } \right) - i\sin \left( {n\phi } \right)$.................. (7)
Now substitute the value from equation (2), (4), (6) and (7) in equation (5) we have,
\[ \Rightarrow {x^m}{y^n} + {x^{ - m}}{y^{ - n}} = \left( {\cos \left( {m\theta } \right) + i\sin \left( {m\theta } \right)} \right)\left( {\cos \left( {n\phi } \right) + i\sin \left( {n\phi } \right)} \right) + \left( {\cos \left( {m\theta } \right) - i\sin \left( {m\theta } \right)} \right)\left( {\cos \left( {n\phi } \right) - i\sin \left( {n\phi } \right)} \right)\]Now simplify this we have,
\[
   \Rightarrow {x^m}{y^n} + {x^{ - m}}{y^{ - n}} = \left( {\cos m\theta \cos n\phi + {i^2}\sin m\theta \sin n\phi + i\left( {\cos m\theta \sin n\phi + \sin m\theta \cos n\phi } \right)} \right) + \\
  \left( {\cos m\theta \cos n\phi + {i^2}\sin m\theta \sin n\phi - i\left( {\cos m\theta \sin n\phi + \sin m\theta \cos n\phi } \right)} \right) \\
\]
\[ \Rightarrow {x^m}{y^n} + {x^{ - m}}{y^{ - n}} = 2\left( {\cos m\theta \cos n\phi + {i^2}\sin m\theta \sin n\phi } \right)\]
Now as we know in complex, $i = \sqrt { - 1} \Rightarrow {i^2} = - 1$ so use this in the above equation we have,
\[ \Rightarrow {x^m}{y^n} + {x^{ - m}}{y^{ - n}} = 2\left( {\cos m\theta \cos n\phi - \sin m\theta \sin n\phi } \right)\]
Now as we know that cos (A + B) = cos A cos B – sin A sin B so use this in the above equation we have,
\[ \Rightarrow {x^m}{y^n} + {x^{ - m}}{y^{ - n}} = 2\cos \left( {m\theta + n\phi } \right)\]
= R.H.S
Hence Proved.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall the properties which are stated above then first consider the L.H.S of the given equation and simplify using these properties as above we will get the required result i.e. equal to R.H.S.