which function is the same as $y = 3cosleft(2left(x+\frac{pi}{2}\right)\right)-2$?\n$y = 3sinleft(2left(x+\fr…

Explanation:

Step1: Use the cosine - sine identity

Recall the identity $\cos(A + \frac{\pi}{2})=-\sin(A)$. For the function $y = 3\cos\left(2\left(x+\frac{\pi}{2}\right)\right)-2$, let $u = 2x$. Then we have $y=3\cos\left(u + \pi\right)-2$. We know that $\cos(A + \pi)=-\cos(A)$ and also $\cos\left(2\left(x+\frac{\pi}{2}\right)\right)=\cos(2x+\pi)$. Another way is to use the identity $\cos\left(\alpha+\frac{\pi}{2}\right)=-\sin\alpha$. Here $\alpha = 2\left(x+\frac{\pi}{4}\right)$. [ \begin{align*} y&=3\cos\left(2\left(x+\frac{\pi}{2}\right)\right)-2\ &=3\cos\left(2x+\pi\right)-2\ &=- 3\cos(2x)-2 \end{align*} ] Also, using the co - function identity $\cos\left(\theta+\frac{\pi}{2}\right)=-\sin\theta$. If we start with $y = 3\cos\left(2\left(x+\frac{\pi}{2}\right)\right)-2$, and rewrite it as $y = 3\cos\left(2x+\pi\right)-2$. We know that $\cos(A + B)=\cos A\cos B-\sin A\sin B$. For $\cos\left(2x+\pi\right)=\cos2x\cos\pi-\sin2x\sin\pi=-\cos2x$. If we use the identity $\cos\left(\alpha+\frac{\pi}{2}\right)=-\sin\alpha$ with $\alpha = 2\left(x + \frac{\pi}{4}\right)$: [ \begin{align*} y&=3\cos\left(2\left(x+\frac{\pi}{2}\right)\right)-2\ &=3\cos\left(2x+\pi\right)-2\ &=-3\sin\left(2\left(x+\frac{\pi}{4}\right)\right)-2 \end{align*} ]

Step2: Compare with options

We compare the simplified form $y=-3\sin\left(2\left(x+\frac{\pi}{4}\right)\right)-2$ with the given options.

Answer:

$y = - 3\sin\left(2\left(x+\frac{\pi}{4}\right)\right)-2$ (the second option)