Question

the graph pf sin(x+c) is shifted pie/5 units to the right compared to sin x what is the value of c

Answer 1

If the graph of \( \sin(x + c) \) is shifted \( \frac{\pi}{5} \) units to the right compared to \( \sin(x) \), we can express this relationship mathematically.

A function \( f(x) \) shifted to the right by \( a \) units is represented as \( f(x - a) \). Therefore, if \( \sin(x + c) \) is equal to \( \sin(x) \) shifted to the right by \( \frac{\pi}{5} \), we can write:

\[
\sin(x + c) = \sin\left(x - \frac{\pi}{5}\right)
\]

For the sine function:

\[
\sin(x + c) = \sin(x - \frac{\pi}{5}) \implies x + c = x - \frac{\pi}{5} + 2k\pi \text{ or } x + c = \pi - (x - \frac{\pi}{5}) + 2k\pi \text{ for some integer } k.
\]

From the first equation:

\[
c = -\frac{\pi}{5} + 2k\pi
\]

From the second equation, we have:

\[
x + c = \pi - x + \frac{\pi}{5} + 2k\pi
\]
\[
2x + c = \pi + \frac{\pi}{5} + 2k\pi
\]
\[
c = \pi + \frac{\pi}{5} - 2x + 2k\pi
\]

However, the first equation \( c = -\frac{\pi}{5} + 2k\pi \) directly gives us a valid solution for \( c \).

For \( k = 0 \):

\[
c = -\frac{\pi}{5}
\]

Thus, the value of \( c \) is:

\[
\boxed{-\frac{\pi}{5}}
\]

This is consistent with the observed shift to the right by \( \frac{\pi}{5} \) units.