periodicity of f(x)=2sin(pi/2 x) — step-by-step solution

We save the definition:

f (x) = 2 \cdot \sin (\frac{π}{2} \cdot x)

Let's find the smallest positive period of

2 \cdot \sin (\frac{π}{2} \cdot x)

The period of

\sin (x)

2 π

; for

a \cdot \sin (b x + c) + d

, we obtain

T = 2 \cdot \frac{π}{abs (b)}

. The nonzero factor

a

and the shifts

c

and

d

do not change the period.

The coefficient of

x

\frac{π x}{2}

\frac{π}{2}

T = \frac{2 π}{\frac{π}{2}}

T = 2 π \cdot \frac{2}{π}

T = \frac{4 π}{π}

T = 4

▶Period:

4