periodicity of 4sin(6x-pi) — step-by-step solution

Let's find the smallest positive period of

4 \cdot \sin (6 x - π)

The period of

\sin (x)

2 π

; for

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

, we obtain

T = 2 \cdot \frac{π}{| b |}

. The nonzero factor

a

and the shifts

c

and

d

do not change the period.

The coefficient of

x

6 x - π

6

ⓘ Shifting

x

by T changes the angle by one period of the trigonometric function:

b T = 6 \cdot \frac{2 π}{6} = 2 π

. Thus

f (x + T) = f (x)

T = \frac{2 π}{6}

ⓘ A fraction is simplified by dividing the numerator and the denominator by the same number:

\frac{2 π}{6}

\frac{π}{3}

T = \frac{π}{3}

▶Period:

\frac{π}{3}

The numeric value of:

\frac{π}{3}

is:

\frac{\approx 3.1416}{3}

▷

\approx 1.0472