Animated Solution for Mathematics - Trigonometry: If $x$ is not an integral multiple of $2\pi$ use mathematical induction to prove that : $\cos x+\cos 2x+....+\cos nx=\cos \frac{n+1}{2}x\sin \frac{nx}{2}\csc \frac{x}{2}$

Welcome, future engineer! Today, we are not just solving a trigonometric series; we are embarking on a journey of logical proof.

Mathematical induction is the domino effect of the mathematical world. If we can knock down the first domino (the base case) and prove that each domino knocks down the next (the inductive step), then the entire infinite chain falls into place.

We aim to prove the following identity:

The Base Case

Every great structure needs a foundation. We start with $n=1$.

On the Left Hand Side (L.H.S.), we have just the first term: $\cos x$.

On the Right Hand Side (R.H.S.), we substitute $n=1$ into our formula:

The $\sin \left(\frac{x}{2}\right)$ terms cancel out, leaving us with $\cos x$. Since L.H.S. $=$ R.H.S., our base case is solid.

The Inductive Hypothesis

Now, we take a leap of faith. We assume the statement $P(k)$ is true for some arbitrary positive integer $k$.

This is our 'Golden Key.' We write:

We are not proving this; we are assuming it is true so that we can use it to unlock the next step. This is the heart of induction.

The Inductive Step

This is where the real work begins. We need to prove $P(k+1)$.

The L.H.S. for $n=k+1$ is the sum of the first $k$ terms plus the $(k+1)$-th term:

Using our Golden Key, we replace the sum of the first $k$ terms with our hypothesis:

Now, we find a common denominator to combine these terms:

The Algebraic Dance

To simplify the numerator, we multiply the numerator and denominator by $2$.

Our numerator becomes:

We apply the identity $2\cos A\sin B=\sin (A+B)-\sin (A-B)$.

For the first term, $A=\frac{k+1}{2}x$ and $B=\frac{kx}{2}$, yielding:

For the second term, $A=(k+1)x$ and $B=\frac{x}{2}$, yielding:

The Triumph of Cancellation

Look closely at the numerator now:

The term $\sin \left(\frac{2k+1}{2}x\right)$ appears with a positive and a negative sign and vanishes. We are left with:

Using the sum-to-product identity $\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$, we transform this into:

When we divide by the $2\sin \left(\frac{x}{2}\right)$ in the denominator, the $2$s cancel. We arrive exactly at the form for $P(k+1)$, completing the proof. The logic is complete.