3. (Maximum matk 8 ) (a) Show that \( \sin 2 x+\cos 2 x-1=2 \sin x(\cos x-\sin x) \). (b) Honco or ohnowise, solve \( \sin 2 x+\cos 2 x-1+\cos x-\sin x=0 \) for \( 0

Ask by Cummings Ramos. in the United Arab Emirates

Jan 19,2025

To show that \( \sin 2x + \cos 2x - 1 = 2 \sin x (\cos x - \sin x) \): Starting with the left side: \[ \sin 2x + \cos 2x - 1 = 2 \sin x \cos x + (1 - 2\sin^2 x) - 1 \] This simplifies to: \[ 2 \sin x \cos x - 2 \sin^2 x = 2\sin x (\cos x - \sin x). \] Thus, we have verified the identity. Now for part (b), we need to solve \( \sin 2x + \cos 2x - 1 + \cos x - \sin x = 0 \). Using the earlier result, we can substitute \( \sin 2x + \cos 2x - 1 \). This gives us: \[ 2 \sin x (\cos x - \sin x) + \cos x - \sin x = 0. \] Factoring out \( \cos x - \sin x \), we get: \[ (\cos x - \sin x)(2 \sin x + 1) = 0. \] Thus, either \( \cos x - \sin x = 0 \) or \( 2 \sin x + 1 = 0 \). 1. From \( \cos x - \sin x = 0 \), we get \( \tan x = 1 \), leading to \( x = \frac{\pi}{4} + n\pi \). Within \( 0 < x < 2\pi \), this gives us \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \). 2. From \( 2 \sin x + 1 = 0 \), we find \( \sin x = -\frac{1}{2} \), leading to \( x = \frac{7\pi}{6} \) and \( x = \frac{11\pi}{6} \). So the solutions are: \[ x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{6}, \frac{11\pi}{6}. \]