Sinusoid, amplitude, period and symmetry

Problem

The function $y=-1+2\cos\!\left(\dfrac{2\pi}{3}x-\dfrac{\pi}{3}\right)$ is given, written in the form $y=a+b\cos(cx+d)$. a) Find $a,b,c,d$ and from them the midline (mean value), the amplitude and the period. b) Show that the function $y=1-2\sin\!\left(\dfrac{5\pi}{6}+\dfrac{2\pi}{3}x\right)$ is obtained from the previous one by a symmetry about the origin.

Solution

a) Comparing: $a=-1$ (midline $y=-1$), $b=2$ (amplitude $2$), $c=\dfrac{2\pi}{3}$, $d=-\dfrac{\pi}{3}$. The period is $T=\dfrac{2\pi}{c}=\dfrac{2\pi}{2\pi/3}=3$. b) The symmetry about the origin maps $(x,y)\to(-x,-y)$: the symmetric curve has equation $-y=-1+2\cos\!\left(-\dfrac{2\pi}{3}x-\dfrac{\pi}{3}\right)$, i.e., since cosine is even, $y=1-2\cos\!\left(\dfrac{2\pi}{3}x+\dfrac{\pi}{3}\right)$. Since $\sin\!\left(\dfrac{2\pi}{3}x+\dfrac{5\pi}{6}\right)=\cos\!\left(\dfrac{2\pi}{3}x+\dfrac{5\pi}{6}-\dfrac{\pi}{2}\right)=\cos\!\left(\dfrac{2\pi}{3}x+\dfrac{\pi}{3}\right)$, we obtain exactly $y=1-2\sin\!\left(\dfrac{5\pi}{6}+\dfrac{2\pi}{3}x\right)$. $\boxed{\,a=-1,\ b=2,\ c=\tfrac{2\pi}{3},\ d=-\tfrac{\pi}{3}\,;\quad T=3\,}$