The graph of f(x)=sin(x+c) is shifted π/3 units left when compared to the graph of g(x)=sin(x). What is the value of c?

1 answer

To determine the value of \( c \) such that the graph of \( f(x) = \sin(x + c) \) is shifted \( \frac{\pi}{3} \) units to the left compared to the graph of \( g(x) = \sin(x) \), we can use the properties of the sine function and the concept of horizontal shifts.

A horizontal shift of a function \( g(x) = \sin(x) \) to the left by \( a \) units can be represented as \( g(x + a) \). In this case, since we want to shift \( g(x) \) left by \( \frac{\pi}{3} \), we get:

\[ h(x) = g\left(x + \frac{\pi}{3}\right) = \sin\left(x + \frac{\pi}{3}\right) \]

We want \( f(x) \) to match this shifted function, such that:

\[ f(x) = \sin(x + c) = \sin\left(x + \frac{\pi}{3}\right) \]

From this equality, we can deduce that the expressions inside the sine functions must be equal. Therefore, we have:

\[ x + c = x + \frac{\pi}{3} \]

To find \( c \), we can simply solve:

\[ c = \frac{\pi}{3} \]

Thus, the value of \( c \) that causes the graph of \( f(x) \) to shift \( \frac{\pi}{3} \) units to the left is:

\[ \boxed{\frac{\pi}{3}} \]