Question

Which of the following correctly rewrites the sine function f(x)=sin (2x+π )-1 as a translated cosine function? (1 point) f(x)=cos (x+π )-1 f(x)=cos (2(x+ π /2 -1)) f(x)=cos (2x+ π /2 )-1 f(x)=cos (2x+π )-1

Answer 1

Is that correct

Answer 2

To rewrite the function \( f(x) = \sin(2x + \pi) - 1 \) as a cosine function, we start by using the relationship between sine and cosine:

\[
\sin(\theta) = \cos\left(\theta - \frac{\pi}{2}\right)
\]

In this case, we have:

\[
\sin(2x + \pi) = \cos\left((2x + \pi) - \frac{\pi}{2}\right) = \cos\left(2x + \frac{\pi}{2}\right)
\]

Now substituting this back into the function:

\[
f(x) = \sin(2x + \pi) - 1 = \cos\left(2x + \frac{\pi}{2}\right) - 1
\]

Thus, we find that the function can be rewritten as:

\[
f(x) = \cos(2x + \frac{\pi}{2}) - 1
\]

Therefore, the correct choice from the options provided is:

\[
\boxed{f(x) = \cos(2x + \frac{\pi}{2}) - 1}
\]

Answer 3

Let's carefully verify the process of transforming \( f(x) = \sin(2x + \pi) - 1 \) into a cosine function.

1. **Using the angle addition formula for sine:**
\[
\sin(2x + \pi) = \sin(2x) \cos(\pi) + \cos(2x) \sin(\pi)
\]
Since \(\cos(\pi) = -1\) and \(\sin(\pi) = 0\), this simplifies to:
\[
\sin(2x + \pi) = -\sin(2x)
\]

2. **Replacing in the original function:**
\[
f(x) = -\sin(2x) - 1
\]

3. **Expressing \(-\sin(2x)\) using cosine:**
We know that:
\[
-\sin(2x) = \cos\left(2x + \frac{\pi}{2}\right)
\]
This is due to the identity \( \sin(\theta) = \cos\left(\theta - \frac{\pi}{2}\right) \), thus:
\[
-\sin(2x) = \cos\left(2x + \frac{\pi}{2}\right)
\]

4. **Substituting back to get \( f(x) \):**
\[
f(x) = \cos(2x + \frac{\pi}{2}) - 1
\]

This confirms that the rewritten form of the function is:

\[
f(x) = \cos(2x + \frac{\pi}{2}) - 1
\]

So, the previous response where I stated:

\[
\boxed{f(x) = \cos(2x + \frac{\pi}{2}) - 1}
\]

is indeed correct. Thank you for your patience!