contestada

[tex]in cos(3x)+1, why isn't The "Period" =pi/3?
Shift should be "0", I hope.
What's the range?[/tex]

Respuesta :

LammettHash
Let the period be [tex]p\neq0[/tex]. Then by definition, a [tex]p[/tex]-periodic function [tex]f(x)[/tex] satisfies

[tex]f(x)=f(x+p)[/tex]

where here [tex]f(x)=\cos3x[/tex].

[tex]\cos3x=\cos3(x+p)=\cos(3x+3p)[/tex]

We know that [tex]\cos x[/tex] is [tex]2\pi[/tex]-periodic, i.e.

[tex]\cos x=\cos(x+2\pi)[/tex]

which reveals that we should also have

[tex]\cos3x=\cos(3x+2\pi)[/tex]

and so [tex]p[/tex] is such that [tex]3p=2\pi\implies p=\dfrac{2\pi}3[/tex].

In general, the period of [tex]\cos nx[/tex] is [tex]\dfrac{2\pi}n[/tex]. (The same applies for [tex]\sin nx[/tex].)

The range can be found by recalling that [tex]-1\le\cos3x\le1[/tex], which means [tex]0\le\cos3x+1\le2[/tex].

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