Question


Periodic Function Flexibility Quick Check
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Question
Rewrite the sine function f(x)=2sin(x+π2)−1 as a cosine function.(1 point)
Responses

f(x)=2cos(x)−1
f left parenthesis x right parenthesis equals 2 cosine left parenthesis x right parenthesis minus 1

f(x)=2cos(x+π4)−1
f left parenthesis x right parenthesis equals 2 cosine left parenthesis x plus Start Fraction pi over 4 End Fraction right parenthesis minus 1

f(x)=2cos(x−π2)−1
f left parenthesis x right parenthesis equals 2 cosine left parenthesis x minus Start Fraction pi over 2 End Fraction right parenthesis minus 1

f(x)=2cos(x+π)−1
f left parenthesis x right parenthesis equals 2 cosine left parenthesis x plus pi right parenthesis minus 1
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Answers

GPT-4o mini
To rewrite the sine function \( f(x) = 2\sin\left(x + \frac{\pi}{2}\right) - 1 \) as a cosine function, we can use the identity that relates sine and cosine:

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

Using this identity, we can rewrite the function:

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

So the correct response is:

\[
f(x) = 2\cos(x) - 1
\]

Thus, the answer is:

**f(x) = 2cos(x) - 1**

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