sinx+sin2x+sin3x+ +sin nx=sin((n+1)/2)x ((sin nx)/2) / sin(x/2) - Maths - Principle of Mathematical Induction

Question

sinx+sin2x+sin3x+ +sin nx=sin((n+1)/2)x ((sin nx)/2) /
sin(x/2)

Ajanta Trivedi · Accepted Answer

TPT : $\sin x + \sin 2 x + \sin 3 x + + \sin n x = \frac{\sin (\frac{n + 1}{2}) x \cdot \sin \frac{n x}{2}}{\sin \frac{x}{2}}$ step I : for n = 1 LHS = $\sin x$ RHS = $\frac{\sin (\frac{1 + 1}{2}) x \cdot \sin \frac{x}{2}}{\sin \frac{x}{2}} = \sin x$ thus the result is true for n = 1 Step II: let it is true for n = k therefore $\sin x + \sin 2 x + \sin 3 x + + \sin k x = \frac{\sin (\frac{k + 1}{2}) x \cdot \sin \frac{k x}{2}}{\sin \frac{x}{2}}$ (1) now in the third step , we will try to prove that it is true for n = (k+1) step III: $\sin x + \sin 2 x + \sin 3 x + + \sin k x + \sin (k + 1) x$ $= \frac{\sin (\frac{k + 1}{2}) x \cdot \sin \frac{k x}{2}}{\sin \frac{x}{2}} + \sin (k + 1) x = \frac{\sin (\frac{k + 1}{2}) x \cdot \sin \frac{k x}{2}}{\sin \frac{x}{2}} + 2 \sin (\frac{k + 1}{2}) x \cdot \cos (\frac{k + 1}{2}) x$ $= \frac{\sin (\frac{k + 1}{2}) x}{\sin \frac{x}{2}} [\sin \frac{k x}{2} + 2 \cos (\frac{k + 1}{2}) x \cdot \sin \frac{x}{2}] = \frac{\sin (\frac{k + 1}{2}) x}{\sin \frac{x}{2}} [\sin \frac{k x}{2} + \sin {(\frac{k + 1}{2}) x + \frac{x}{2}} - \sin {(\frac{k + 1}{2}) x - \frac{x}{2}}]$ $[∵ 2 \cos A \sin B = \sin (A + B) - \sin (A - B)]$ $= \frac{\sin (\frac{k + 1}{2}) x}{\sin \frac{x}{2}} [\sin \frac{k x}{2} + \sin (\frac{k + 2}{2}) x - \sin \frac{k}{2}] = \frac{\sin (\frac{k + 1}{2}) x}{\sin \frac{x}{2}} \cdot \sin \frac{(k + 1) + 1}{2} x$ thus it is true for n = (k+1) therefore it will be true for all values of n belongs to N hope this helps you cheers!!

sinx+sin2x+sin3x+...+sin nx=sin((n+1)/2)x .((sin nx)/2) / sin(x/2)