ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: /0 on April 09, 2008, 09:36:45 pm

Title: Finding the right angle in an equation
Post by: /0 on April 09, 2008, 09:36:45 pm: $\mbox{For}\ \pi < x < \frac{3\pi}{2}\ \mbox{with}\ \sin{x}=-\sin{\frac{\pi}{6}},\ \mbox{find the value of x} \left(\mbox{do not evaulate}\sin{\frac{\pi}{6}}\right)$

Every time I keep getting $\frac{11\pi}{6}$ ... this isn't in the domain but I think it is an equivalent angle. I don't know how to make it so that it is within the domains though. My teacher used a pictorial method with the unit circle but I'm lazy and I would rather use identities.
Title: Re: Finding the right angle in an equation
Post by: AppleXY on April 09, 2008, 09:43:22 pm: -sin(pi/6) = sin(-pi/6)

sin is neg @ $\pi +\theta$

$\therefore \frac{\pi}{6}+\pi$

= $\frac{7\pi}{6}$
Title: Re: Finding the right angle in an equation
Post by: /0 on April 09, 2008, 09:48:55 pm: Thanks, ApppleXY, I was a bit confused about using the symmetry properties in reverse.