Question

Prove that $\cos x + \cos 3x + ... + \cos \left( {2n - 1} \right)x = \dfrac{{\sin 2nx}}{{2\sin x}}$, $x \ne K$, $K \in I$ and then deduce that $\sin x + 3\sin 3x + ... + \left( {2n - 1} \right)\sin \left( {2n - 1} \right)x = \dfrac{{\left[ {\left( {2n + 1} \right)\sin \left( {2n - 1} \right)x - \left( {2n - 1} \right)\sin \left( {2n + 1} \right)x} \right] }}{{4{{\sin }^2}x}}$.

Answer 1

Hint: We have to prove that the given trigonometric series is equal to the expression given in the RHS. We can multiply the numerator and denominator in the LHS by $2\sin x$ and then use trigonometric identities to prove this. Further we have to deduce another identity from this series. For this we can simply differentiate the given series.

Complete step by step solution:
We have to prove that $\cos x + \cos 3x + ... + \cos \left( {2n - 1} \right)x = \dfrac{{\sin 2nx}}{{2\sin x}}$. Here $x$ is not an integer.
We can start with the LHS and simplify it such that we arrive at the expression similar to that given in the RHS.
We can multiply the numerator and denominator by $2\sin x$ in the LHS to get,
$
  \dfrac{{2\sin x \times \left[ {\cos x + \cos 3x + ... + \cos \left( {2n - 1} \right)x} \right] }}{{2\sin x}} \\
   = \dfrac{{2\sin x\cos x + 2\sin x\cos 3x + ... + 2\sin x\cos \left( {2n - 1} \right)x}}{{2\sin x}} \;
 $
We can use the trigonometric identity $2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)$
Therefore we get,
$
  \dfrac{{\left( {\sin \left( {x + x} \right) + \sin \left( {x - x} \right)} \right) + \left( {\sin \left( {x + 3x} \right) + \sin \left( {x - 3x} \right)} \right) + ... + \left( {\sin \left( {x + \left( {2n - 1} \right)x} \right) + \sin \left( {x - \left( {2n - 1} \right)x} \right)} \right)}}{{2\sin x}} \\
   = \dfrac{{\left( {\sin 2x - 0} \right) + \left( {\sin 4x - \sin 2x} \right) + \left( {\sin 6x - \sin 4x} \right)... + \left( {\sin 2nx - \sin \left( {2n - 2} \right)x} \right)}}{{2\sin x}} \;
 $
We can observe that the first term in each bracket is cancelled out by the second term of the consecutive bracket in the numerator. When we cancel all such terms we will be left with,
$ = \dfrac{{\sin 2nx}}{{2\sin x}}$
This is equal to the required RHS.
Hence, we proved $\cos x + \cos 3x + ... + \cos \left( {2n - 1} \right)x = \dfrac{{\sin 2nx}}{{2\sin x}}$.
Now for the second part of the question we will differentiate both the sides with respect to $x$.
 \[
   \Rightarrow \dfrac{{d\left[ {\cos x + \cos 3x + ... + \cos \left( {2n - 1} \right)x} \right] }}{{dx}} = \dfrac{{d\left[ {\dfrac{{\sin 2nx}}{{2\sin x}}} \right] }}{{dx}} \\
   \Rightarrow \dfrac{{d\left( {\cos x} \right)}}{{dx}} + \dfrac{{d\left( {\cos 3x} \right)}}{{dx}} + ... + \dfrac{{d\left( {\cos \left( {2n - 1} \right)x} \right)}}{{dx}} = \dfrac{{2\sin x\dfrac{{d\left( {\sin 2nx} \right)}}{{dx}} - \sin 2nx\dfrac{{d\left( {2\sin x} \right)}}{{dx}}}}{{{{\left( {2\sin x} \right)}^2}}} \\
   \Rightarrow - \sin x - 3\sin 3x - ... - \left( {2n - 1} \right)\sin \left( {2n - 1} \right)x = \dfrac{{4n\sin x\cos 2nx - 2\sin 2nx\cos x}}{{4{{\sin }^2}x}} \\
   \Rightarrow \sin x + 3\sin 3x + ... + \left( {2n - 1} \right)\sin \left( {2n - 1} \right)x = \dfrac{{2\sin 2nx\cos x - 4n\sin x\cos 2nx}}{{4{{\sin }^2}x}} \\
   \Rightarrow \sin x + 3\sin 3x + ... + \left( {2n - 1} \right)\sin \left( {2n - 1} \right)x = \dfrac{{\left( {\sin \left( {2nx + x} \right) + \sin \left( {2nx - x} \right)} \right) - 2n\left( {\sin \left( {x + 2nx} \right) + \sin \left( {x - 2nx} \right)} \right)}}{{4{{\sin }^2}x}} \\
   \Rightarrow \sin x + 3\sin 3x + ... + \left( {2n - 1} \right)\sin \left( {2n - 1} \right)x = \dfrac{{\sin \left( {2n + 1} \right)x + \sin \left( {2n - 1} \right)x - 2n\sin \left( {2n + 1} \right)x + 2n\sin \left( {2n - 1} \right)x}}{{4{{\sin }^2}x}} \\
   \Rightarrow \sin x + 3\sin 3x + ... + \left( {2n - 1} \right)\sin \left( {2n - 1} \right)x = \dfrac{{\left( {2n + 1} \right)\sin \left( {2n - 1} \right)x - \left( {2n - 1} \right)\sin \left( {2n + 1} \right)x}}{{4{{\sin }^2}x}} \;
 \]
Here we can see that LHS and RHS are both the same as given in the question.
Hence we got the required identity.

Note: There may be other methods to solve this problem like using exponential form of the complex number. Using complex numbers for such questions is a common way to solve the problem but this problem was easily solved by using trigonometric identity. We have to practice more and more problems in solving series to be able to have the aptitude to guess the right approach.