Question

the black graph is y = cos x. choose the equation for the red graph. y = cos(x) - 1 y = cos(x) + 1 y = cos(x - π/2) y = cos(x + π/2)

Explanation:

Step1: Recall the phase - shift formula

The general form of a cosine function is \(y = A\cos(B(x - C))+D\), where \(C\) is the phase - shift (horizontal shift). If \(C>0\), the graph shifts to the right by \(C\) units; if \(C < 0\), the graph shifts to the left by \(|C|\) units.

Step2: Analyze the key points

For \(y=\cos(x)\), when \(x = 0\), \(y=\cos(0)=1\). For the red graph, when \(y = 1\), \(x=\frac{\pi}{2}\).
We know that if \(y=\cos(x - C)\), when \(y = 1\), \(x - C=2k\pi\) (\(k\in\mathbb{Z}\)). Let \(k = 0\), then \(x=C\) gives \(y = 1\).
Substituting \(x=\frac{\pi}{2}\) (the \(x\) - value of the maximum of the red graph) into \(y=\cos(x - C)\) and setting \(y = 1\) (since \(\cos(0)=1\)), we get \(x - C=0\). When \(x=\frac{\pi}{2}\), \(C=\frac{\pi}{2}\).
So the function of the red graph is \(y=\cos(x-\frac{\pi}{2})\).
We can also use the trigonometric identity \(\cos(A - B)=\cos A\cos B+\sin A\sin B\). Here \(y=\cos(x-\frac{\pi}{2})=\cos x\cos\frac{\pi}{2}+\sin x\sin\frac{\pi}{2}=\sin x\). Another way is to check the transformation:

• For \(y = \cos(x)-1\), the graph of \(y=\cos(x)\) is shifted down by 1 unit. The maximum value of \(y=\cos(x)\) is \(1\), and for \(y=\cos(x)-1\) the maximum value is \(0\).
• For \(y=\cos(x)+1\), the graph of \(y = \cos(x)\) is shifted up by 1 unit. The \(x\) - value of the maximum of \(y=\cos(x)\) is \(x = 2k\pi\), \(k\in\mathbb{Z}\), and for \(y=\cos(x)+1\) the \(x\) - value of the maximum is still \(x = 2k\pi\), \(k\in\mathbb{Z}\).
• For \(y=\cos(x+\frac{\pi}{2})\), using the identity \(\cos(A + B)=\cos A\cos B-\sin A\sin B\), \(y=\cos(x+\frac{\pi}{2})=\cos x\cos\frac{\pi}{2}-\sin x\sin\frac{\pi}{2}=-\sin x\). The maximum of \(y =-\sin x\) occurs at \(x=-\frac{\pi}{2}+2k\pi\), \(k\in\mathbb{Z}\).

Answer:

\(y=\cos(x - \pi/2)\)