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

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 2

Is that correct

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!