Section+1.12A

These are the notes from class on Tuesday, September 13th.

Today, we learned more about solving equations with sinusoidal functions and graphing sinusoidal functions. We started off with a problem from Section 1.12, "Check Your Understanding." 4. a) math $ 4\cos2x+3= 5 $ math math $ 4\cos2x=2 $ math math $ \cos2x=\frac{1}{2} $ math math $ 2x=\cos^{-1}\frac{1}{2} $ math math $ 2x=\frac{\pi }{3} $ math math $ x=\frac{\pi }{6} $ math

b) math y=4\cos 2x+3 $ math amplitude = 4 (vertical stretch by 4) period: 2pi/2 = pi vertical shift/displacement = 3 (shift up 3)

c) math $ 4\cos2x+3=5 $ math math $ 4\cos 2x=2 $ math math $ \cos 2x=\frac{1}{2} $ math math $ 2x=\cos^{-1}\frac{1}{2}+2\pi k; 2x=2\pi -\cos^{-1}\frac{1}{2}+2\pi k $ math math $ 2x=\frac{\pi }{3}+k\pi $ math math $ x=\pi-\frac{\pi}{6}+k\pi=\frac{5\pi}{6}+k\pi $ math

After this, Ms. Tyson made up a problem of her own. Ex. math $ 5\sin\frac{\pi}{5}x-3=1 $ math math $ 5\sin\frac{\pi}{5}x=4 $ math math $ \sin\frac{\pi}{5}x=\frac{4}{5} $ math math $ \frac{\pi}{5}x=\sin^{-1}\frac{4}{5} $ math math $ x=\sin^{-1}\frac{4}{5}\ast \frac{5}{\pi} $ math math x=1.48+10k, x=\pi -1.48+10k $ math

math $ y=5\sin \frac{\pi}{5}x-3=1 $ math amplitude = 5 (vertical stretch by 5) period = 2pi * 5/pi = 10 (horizontal stretch = 5/pi) vertical shift/displacement = -3 (shift down 3)