fiziksfun Messages 77 Reaction score 0 Thread starter Apr 12, 2009 #1 can someone please show me why sinxcos2x = (sin3x - sinx)*0.5 I've been working on it for thirty minutes
can someone please show me why sinxcos2x = (sin3x - sinx)*0.5 I've been working on it for thirty minutes
praharmitra Messages 308 Reaction score 1 Apr 12, 2009 #2 can u expand sin(3x) ?? and cos(2x) just simplify both sides..and u'll get it..
qntty Messages 289 Reaction score 4 Apr 12, 2009 #3 You should use the identities [tex]\cos{2\theta} = 1-2\sin^2{\theta}[/tex] [tex]\sin{3\theta} = 3\sin{\theta} - 4\sin^3{\theta}[/tex] Perhaps you were putting these identities in terms of cosines. When possible, stick with as few trig functions as possible and replace the [tex]\cos^2{\theta}[/tex] with [tex]1-\sin^2{\theta}[/tex] or vice versa. Last edited: Apr 12, 2009
You should use the identities [tex]\cos{2\theta} = 1-2\sin^2{\theta}[/tex] [tex]\sin{3\theta} = 3\sin{\theta} - 4\sin^3{\theta}[/tex] Perhaps you were putting these identities in terms of cosines. When possible, stick with as few trig functions as possible and replace the [tex]\cos^2{\theta}[/tex] with [tex]1-\sin^2{\theta}[/tex] or vice versa.