Question

suppose a function ( f(x) ) is defined when ( 0 leq x leq 2pi ). this function can be extended into a periodic function over all real numbers by interpreting ( x ) as the radian measure of an angle traversed counterclockwise around the unit circle. for example, ( fleft(\frac{5pi}{2}
ight) ) may be defined to be equal to ( fleft(\frac{pi}{2}
ight) ).

if ( f(pi)=4 ), what is the value of ( f(7pi) ) when ( f(x) ) is extended in this way?

Explanation:

Step1: Find the equivalent angle within \(0\leq x\leq2\pi\)

We know that the function \(f(x)\) is periodic with period \(2\pi\). To find an equivalent angle for \(7\pi\) within the interval \(0\leq x\leq2\pi\), we use the formula \(7\pi = 2\pi\times3+\pi\).

Step2: Use the periodic property of the function

Since \(f(x)\) is periodic with period \(2\pi\), \(f(7\pi)=f(2\pi\times3 + \pi)\). For a periodic function \(y = f(x)\) with period \(T\) (here \(T = 2\pi\)), \(f(x + nT)=f(x)\) where \(n\in\mathbb{Z}\). So \(f(7\pi)=f(\pi)\)

Answer:

\(4\)