Question

function f(x) = -2sin(x-pi)+3 as a cosine function.

Answer 1

To express the function \( f(x) = -2 \sin(x - \pi) + 3 \) as a cosine function, we can use the relationship between sine and cosine functions. We know that:

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

Let's rewrite the sine term in the equation:

1. **Phase Shift:** Using the sine to cosine conversion:
\[
\sin(x - \pi) = \sin\left(x - \frac{2\pi}{2}\right) = -\sin\left(x - \frac{\pi}{2}\right)
\]
Hence:
\[
-2\sin(x - \pi) = 2 \sin\left(x - \frac{\pi}{2}\right)
\]

However, note that \(-\sin\) can instead be represented as a cosine:
\[
-2\sin(x - \pi) = 2\cos\left(x - \frac{\pi}{2} + \pi\right) = 2\cos\left(x + \frac{\pi}{2}\right)
\]

2. **Adding the Constant:**
The transformed function becomes:
\[
f(x) = 2\cos\left(x + \frac{\pi}{2}\right) + 3
\]

Thus, you can express the function \( f(x) \) as:
\[
f(x) = 2\cos\left(x + \frac{\pi}{2}\right) + 3
\]

This function indicates a vertical shift of 3 units upwards, with an amplitude of 2, and a phase shift of \(\frac{\pi}{2}\).