Proof by induction in trigonometry.

Prove that $\cos x +\cos 2x + \cos 3x + ...+ \cos nx =\cos \left(\dfrac{n+1}{2}x\right) \sin \left(\dfrac{nx}{2}\right)\csc \dfrac{x}{2}$

Attempt:

Clearly, $P(1)$ is true.

Assume $P(m)$ is true.

Thus, $P(m+1) = (\cos x +\cos 2x + \cos 3x + ...+ \cos mx)+ \cos((m+1)x)$

$= \cos \left(\dfrac{m+1}{2}x\right) \sin \left(\dfrac{mx}{2}\right)\csc \dfrac{x}{2} + \cos((m+1)x) \\= \csc (\dfrac x 2)\left(\cos \left(\dfrac{m+1}{2}x\right) \sin \left(\dfrac{mx}{2}\right)+ (\cos(m+1)x)\sin (\dfrac x 2)\right)$

What do I do next?

Answer

Formula to be used:

$\sin A- \sin B = \cos\left(\dfrac{A+B}{2}\right)\sin \left(\dfrac{A-B}{2}\right)$

Thus,

$\csc \left(\dfrac x 2 \right)\left(\cos \left(\dfrac{m+1}{2}x\right) \sin \left(\dfrac{mx}{2}\right)+ (\cos(m+1)x)\sin (\dfrac x 2)\right)$

$= \csc \left(\dfrac x 2 \right)\left(\dfrac 1 2 \left(\sin \dfrac{2mx+x}{2} - \sin \dfrac x 2 \right)+ \dfrac 1 2 \left(\sin \dfrac{2mx+3x}{2} - \sin \dfrac{2mx + x}{2 } \right) \right)$

Now, again use the formula on the left out terms.

$= \csc \left(\dfrac x 2 \right)\left(\dfrac 1 2 \left(\sin \dfrac{2mx+3x}{2} - \sin \dfrac x 2 \right) \right)$

$= \cos \left(\dfrac{m+2}{2}x\right) \sin \left(\dfrac{(m+1)x}{2}\right)\csc \dfrac{x}{2}$

Thus, $P(m+1)$ is also true.

Q.E.D.