which of the following is the graph of $y = \\cos(2(x+pi))$?

Answer:

First, use the trigonometric identity $\cos(A + B)=\cos A\cos B-\sin A\sin B$. For $y = \cos(2(x+\pi))=\cos(2x + 2\pi)$. Since $\cos(A + 2k\pi)=\cos A$ for any real - number $A$ and integer $k$, then $\cos(2x + 2\pi)=\cos(2x)$. The general form of a cosine function is $y = A\cos(Bx - C)+D$, where for $y=\cos(2x)$, $A = 1$, $B = 2$, $C = 0$, and $D = 0$. The period of the cosine function $y = A\cos(Bx - C)+D$ is given by $T=\frac{2\pi}{|B|}$. Here, $B = 2$, so $T=\frac{2\pi}{2}=\pi$. The amplitude of the function $y=\cos(2x)$ is $|A| = 1$. The graph of $y=\cos(2x)$ is a cosine - type wave with an amplitude of 1 and a period of $\pi$. It oscillates between $y=-1$ and $y = 1$.

Explanation:

Step1: Simplify the function

Use the periodicity of cosine. $\cos(2(x+\pi))=\cos(2x + 2\pi)=\cos(2x)$

Step2: Identify the period

Use the period formula $T=\frac{2\pi}{|B|}$, with $B = 2$, so $T=\pi$

Step3: Identify the amplitude

For $y=\cos(2x)$, $A = 1$, so amplitude $|A| = 1$