In the inductive step, it needs to prove that, if\(P\left( k \right)\)is true, then\(P\left( {k + 1} \right)\)is also true.
That is,\(P\left( k \right) \to P\left( {k + 1} \right)\)is true for all positive integers k.
In the inductive hypothesis, it assumes that\(P\left( k \right)\)is true for any arbitrary positive integer\(k\).
That is, it is written as\(\sum\limits_{j = 1}^k {\cos jx} = \frac{{\cos \left( {\left( {k + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)\sin \left( {{{kx} \mathord{\left/
{\vphantom {{kx} 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}{{\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}\).
Now it must have to show that\(P\left( {k + 1} \right)\)is also true.
\(\sum\limits_{j = 1}^{k + 1} {\cos jx} = \sum\limits_{j = 1}^k {\cos jx} + \cos \left( {k + 1} \right)x\)
From inductive hypothesis it assumes\(\sum\limits_{j = 1}^k {\cos jx} = \frac{{\cos \left( {\left( {k + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)\sin \left( {{{kx} \mathord{\left/
{\vphantom {{kx} 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}{{\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}\).
Therefore, it is written as:
\(\begin{array}{c}\sum\limits_{j = 1}^{k + 1} {\cos jx} = \sum\limits_{j = 1}^k {\cos jx} + \cos \left( {\left( {k + 1} \right)x} \right)\\ = \frac{{\cos \left( {\left( {k + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)\sin \left( {{{kx} \mathord{\left/
{\vphantom {{kx} 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}{{\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}} + \cos \left( {\left( {k + 1} \right)x} \right)\\ = \frac{{\cos \left( {\left( {k + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)\sin \left( {{{kx} \mathord{\left/
{\vphantom {{kx} 2}} \right.
\kern-\nulldelimiterspace} 2}} \right) + \cos \left( {\left( {k + 1} \right)x} \right)\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}{{\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}\end{array}\)
Applying the formula of\(\cos a\,\sin b = \frac{{\sin \left( {a + b} \right) - \sin \left( {a - b} \right)}}{2}\), then it is written as:
\(\begin{array}{c}\sum\limits_{j = 1}^{k + 1} {\cos jx} = \frac{{\sin \left( {\left( {k + 1} \right){x \mathord{\left/
{\vphantom {x {2 + {{kx} \mathord{\left/
{\vphantom {{kx} 2}} \right.
\kern-\nulldelimiterspace} 2}}}} \right.
\kern-\nulldelimiterspace} {2 + {{kx} \mathord{\left/
{\vphantom {{kx} 2}} \right.
\kern-\nulldelimiterspace} 2}}}} \right) - \sin \left( {\left( {k + 1} \right){x \mathord{\left/
{\vphantom {x {2 - {{kx} \mathord{\left/
{\vphantom {{kx} 2}} \right.
\kern-\nulldelimiterspace} 2}}}} \right.
\kern-\nulldelimiterspace} {2 - {{kx} \mathord{\left/
{\vphantom {{kx} 2}} \right.
\kern-\nulldelimiterspace} 2}}}} \right) + \sin \left( {\left( {k + 1} \right)x + {x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)\sin \left( {{{\left( {k + 1} \right)x - x} \mathord{\left/
{\vphantom {{\left( {k + 1} \right)x - x} 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}{{2\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}\\ = \frac{{\sin \left( {\left( {2k + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right) - \sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right) + \sin \left( {\left( {2k + 3} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right) - \sin \left( {{{\left( {2k + 1} \right)x} \mathord{\left/
{\vphantom {{\left( {2k + 1} \right)x} 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}{{2\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}\\ = \frac{{\sin \left( {\left( {2k + 3} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right) - \sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}{{2\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}\\\frac{{ = \sin \left( {\left( {k + 2} \right){x \mathord{\left/
{\vphantom {x {2 + \left( {k + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}}}} \right.
\kern-\nulldelimiterspace} {2 + \left( {k + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}}}} \right) - \sin \left( {\left( {k + 2} \right){x \mathord{\left/
{\vphantom {x {2 - \left( {k + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}}}} \right.
\kern-\nulldelimiterspace} {2 - \left( {k + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}}}} \right)}}{{2\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}\end{array}\)
Again, using formula\(\cos a\,\sin b = \frac{{\sin \left( {a + b} \right) - \sin \left( {a - b} \right)}}{2}\), then it is written as:
\(\sum\limits_{j = 1}^{k + 1} {\cos jx} = \frac{{\cos \left( {\left( {k + 2} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)\sin \left( {{{\left( {k + 1} \right)x} \mathord{\left/
{\vphantom {{\left( {k + 1} \right)x} 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}{{\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}\)
From the above, it can see that\(P\left( {k + 1} \right)\)is also true.
Hence,\(P\left( {k + 1} \right)\)is true under the assumption that\(P\left( k \right)\)is true. This
completes the inductive step.
Hence using mathematical induction, it is proved that\({{\sum\limits_{j = 1}^n {\cos jx = \cos \left( {\left( {n + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)} \sin \left( {{{nx} \mathord{\left/
{\vphantom {{nx} 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)} \mathord{\left/
{\vphantom {{\sum\limits_{j = 1}^n {\cos jx = \cos \left( {\left( {n + 1} \right){x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)} \sin \left( {{{nx} \mathord{\left/
{\vphantom {{nx} 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)} {\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}} \right.
\kern-\nulldelimiterspace} {\sin \left( {{x \mathord{\left/
{\vphantom {x 2}} \right.
\kern-\nulldelimiterspace} 2}} \right)}}\).
