Question

the period of (1/3)^sinx +(1/3)^cosx is

Answer 1

Given : (1/3)^sinx +(1/3)^cosx

$(\dfrac{1}{3})^{\sin x}+(\dfrac{1}{3})^{\cos x}$

To Find : Period :

Solution:

$f(x)=(\dfrac{1}{3})^{\sin x}+(\dfrac{1}{3})^{\cos x}$

$f(x+t)=(\dfrac{1}{3})^{\sin (x+t)}+(\dfrac{1}{3})^{\cos (x+t)}$

f(x) = f(x + t)   where t is the period

Sin (x + t) = Sinxcost + cosxSint

Cos(x + t) = CosxCost - SinxSint

There can be two cases :

First :

Sin(x + t) = Sinx   and  Cos(x + t)   =  Cosx

Sinxcost + cosxSint = Sinx    only possible when t = 0 or 2π

CosxCost - SinxSint =  Cosx   only possible when t = 0 or 2π

Hence period should be  2π

Second case

Sin(x + t) = Cosx   and  Cos(x + t)   =  Sinx

Sinxcost +  cosxSint =  Cosx    => t =  π/2

CosxCost -  SinxSint =  Sinx    =>   t = 3 π/2

Hence no common solution

So period of the function is 2π

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