What is the period and amplitude for $y = cos 9 x$ $y=cos9x$ ?

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Answer 1 · 2018-07-01T11:48:27

The period is $= \frac{2}{9} π$ $=2/9pi$ and the amplitude is $= 1$ $=1$

The period $T$ $T$ of a periodic function $f (x)$ $f(x)$ is such that

$f (x) = f (x + T)$ $f(x)=f(x+T)$

Here,

$f (x) = cos 9 x$ $f(x)=cos9x$

Therefore,

$f (x + T) = cos 9 (x + T)$ $f(x+T)=cos9(x+T)$

$= cos (9 x + 9 T)$ $=cos(9x+9T)$

$= cos 9 x cos 9 T + sin 9 x sin 9 T$ $=cos9xcos9T+sin9xsin9T$

Comparing $f (x)$ $f(x)$ and $f (x + T)$ $f(x+T)$

${(cos9T=1),(sin9tT=0):}$

$=>$ , $9T=2pi$

$=>$ , $T=(2pi)/9$

The amplitude is $=1$ as

$-1<=cosx<=1$

graph{cos(9x) [-1.914, 3.56, -0.897, 1.84]}

What is the period and amplitude for y=cos9xy=cos9xy=cos9x?