Question

Periodic Function Flexibility Practice
Complete this assessment to review what you’ve learned. It will not count toward your grade.
2 of 52 of 5 Items








Question
Rewrite the sine function f(x)=−sin(2x+π2)−1
as a cosine function.

Option #1: f(x)=−cos(2x)−1

Option #2: f(x)=−cos(2(x+π2))−1

(1 point)
Option #
is correct.
Remaining Attempts : 3
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Answer 1

To rewrite the sine function \( f(x) = -\sin(2x + \frac{\pi}{2}) - 1 \) as a cosine function, we can use the identity \( \sin(x) = -\cos\left(x - \frac{\pi}{2}\right) \).

First, we can rewrite \( \sin(2x + \frac{\pi}{2}) \) as:

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

Thus, we have:

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

Now, substituting this into the original function gives us:

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

Based on the options:

Option #1: \( f(x) = -\cos(2x) - 1 \)

Option #2: \( f(x) = -\cos(2(x+\frac{\pi}{2})) - 1 \)

The correct answer is **Option #1**: \( f(x) = -\cos(2x) - 1 \).

So, you should select Option #1 as the correct answer.