Asked by mei x yuzu
                Rewrite the cosine function f(x)=cos(2x+π) as a sine function. Then write 1–2 sentences to explain each step you took to arrive at your answer
            
            
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                    Answered by
            GPT-4o mini
            
    To rewrite the function \( f(x) = \cos(2x + \pi) \) as a sine function, we can use the identity \( \cos(x) = \sin\left(x + \frac{\pi}{2}\right) \). 
First, we recognize that:
\[
\cos(2x + \pi) = \cos(2x + \pi) = -\cos(2x)
\]
Next, using the property \( \cos(x) = -\sin\left(x + \frac{\pi}{2}\right) \), we can express this as:
\[
\cos(2x + \pi) = -\sin\left(2x + \pi + \frac{\pi}{2}\right) = -\sin\left(2x + \frac{3\pi}{2}\right)
\]
So the function can be rewritten as:
\[
f(x) = -\sin\left(2x + \frac{3\pi}{2}\right)
\]
**Explanation:**
1. I first identified that the cosine function can be transformed using sine by employing the identity \( \cos(x) = -\sin\left(x + \frac{\pi}{2}\right)\).
2. Then, I adjusted the argument of the sine function accordingly to reflect the transformation, which in this case involved combining \( 2x + \pi \) and shifting it appropriately to meet the sine function's format.
    
First, we recognize that:
\[
\cos(2x + \pi) = \cos(2x + \pi) = -\cos(2x)
\]
Next, using the property \( \cos(x) = -\sin\left(x + \frac{\pi}{2}\right) \), we can express this as:
\[
\cos(2x + \pi) = -\sin\left(2x + \pi + \frac{\pi}{2}\right) = -\sin\left(2x + \frac{3\pi}{2}\right)
\]
So the function can be rewritten as:
\[
f(x) = -\sin\left(2x + \frac{3\pi}{2}\right)
\]
**Explanation:**
1. I first identified that the cosine function can be transformed using sine by employing the identity \( \cos(x) = -\sin\left(x + \frac{\pi}{2}\right)\).
2. Then, I adjusted the argument of the sine function accordingly to reflect the transformation, which in this case involved combining \( 2x + \pi \) and shifting it appropriately to meet the sine function's format.
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