problem 3 should be graphed according to the same guidelines as above, except: (1) sketch both functions on…

problem 3 should be graphed according to the same guidelines as above, except: (1) sketch both functions on the same set of axes, and (2) graph 1 period of the larger function and 2 periods of the smaller function. label each function. 3. y=-3 sin 1/2 x + 1 and y = 2 cos 1/4 x - 2

problem 3 should be graphed according to the same guidelines as above, except: (1) sketch both functions on the same set of axes, and (2) graph 1 period of the larger function and 2 periods of the smaller function. label each function. 3. y=-3 sin 1/2 x + 1 and y = 2 cos 1/4 x - 2

Answer

Explanation:

Step1: Recall period - formula for trigonometric functions

For $y = A\sin(Bx - C)+D$, the period $T_{sin}=\frac{2\pi}{|B|}$, and for $y = A\cos(Bx - C)+D$, the period $T_{cos}=\frac{2\pi}{|B|}$. For $y=-3\sin(\frac{1}{2}x)+1$, $B = \frac{1}{2}$, so $T_{1}=\frac{2\pi}{\frac{1}{2}} = 4\pi$. For $y = 2\cos(\frac{1}{4}x)-2$, $B=\frac{1}{4}$, so $T_{2}=\frac{2\pi}{\frac{1}{4}}=8\pi$. Since $8\pi>4\pi$, $y = 2\cos(\frac{1}{4}x)-2$ is the larger - function and $y=-3\sin(\frac{1}{2}x)+1$ is the smaller - function.

Step2: Identify key - points for $y=-3\sin(\frac{1}{2}x)+1$

The general form of a sine function is $y = A\sin(Bx - C)+D$. Here, $A=-3$, $B=\frac{1}{2}$, $C = 0$, $D = 1$. The amplitude is $|A| = 3$. For one - period ($x$ from $0$ to $4\pi$): When $x = 0$, $y=-3\sin(0)+1=1$. When $x=\pi$, $y=-3\sin(\frac{\pi}{2})+1=-3 + 1=-2$. When $x = 2\pi$, $y=-3\sin(\pi)+1=1$. When $x = 3\pi$, $y=-3\sin(\frac{3\pi}{2})+1=3 + 1=4$. When $x = 4\pi$, $y=-3\sin(2\pi)+1=1$. For the second period ($x$ from $4\pi$ to $8\pi$), we can repeat the pattern.

Step3: Identify key - points for $y = 2\cos(\frac{1}{4}x)-2$

The general form of a cosine function is $y = A\cos(Bx - C)+D$. Here, $A = 2$, $B=\frac{1}{4}$, $C = 0$, $D=-2$. The amplitude is $|A| = 2$. For one - period ($x$ from $0$ to $8\pi$): When $x = 0$, $y=2\cos(0)-2=2 - 2=0$. When $x = 2\pi$, $y=2\cos(\frac{\pi}{2})-2=-2$. When $x = 4\pi$, $y=2\cos(\pi)-2=-2 - 2=-4$. When $x = 6\pi$, $y=2\cos(\frac{3\pi}{2})-2=-2$. When $x = 8\pi$, $y=2\cos(2\pi)-2=2 - 2=0$.

Step4: Sketch the graphs

On the same set of axes, plot the key - points for $y=-3\sin(\frac{1}{2}x)+1$ for two periods ($x$ from $0$ to $8\pi$) and for $y = 2\cos(\frac{1}{4}x)-2$ for one period ($x$ from $0$ to $8\pi$). Label the function $y=-3\sin(\frac{1}{2}x)+1$ and $y = 2\cos(\frac{1}{4}x)-2$.

Answer:

The graphs of $y=-3\sin(\frac{1}{2}x)+1$ (two periods) and $y = 2\cos(\frac{1}{4}x)-2$ (one period) are sketched on the same set of axes with key - points calculated as above and each function is labeled.