Prove that sin5x cos2x+cos6x sin3x=sin8x cosx
সমাধানঃ
sin 5x cos2x + cos6x sin3x
= $\frac{1}{2}$ (2 sin5x cos2x + 2cos6x sin3x)
= $\frac{1}{2}$ {sin (5x+2x) + sin(5x-2x) + sin(6x+3x)- sin(6x-3x)}
= $\frac{1}{2}$ (sin7x + sin 3x+ sin 9x– sin 3x}
= $\frac{1}{2}$ (sin9x + sin7x)
= $\frac{1}{2}$×2 sin $\frac{9x + 7x}{2}$ cos $\frac{9x – 7x}{2}$
=sin $\frac{16x}{2}$ cos $\frac{2x}{2}$
= sin8x cosx