Q 114 - Precalculus Enhanced with Graphing Utilities

Question

Use the periodic and even–odd properties. 

If $f\left(\theta \right)=\cos \theta$ and $f\left(a\right)=\frac{1}{4}$, find the exact value of :

(a) $f(-a)$

(b) $f\left(a\right)+f(a+2\pi )+f(a-2\pi )$.

Step-by-Step Solution

Answer

(a) The value of $f(-a)$ is $\frac{1}{4}$.

(b) The value of $f\left(a\right)+f(a+2\pi )+f(a-2\pi )$ is $\frac{3}{4}$.

Step 1. Given Information

We have given that following function :- 

$f\left(\theta \right)=\cos \theta$ and $f\left(a\right)=\frac{1}{4}$.

We have to find the value of $f(-a)$ and value of $f\left(a\right)+f(a+2\pi )+f(a-2\pi )$.

To find value of $f(-a)$ we will use even-odd properties and to find the value of $f\left(a\right)+f(a+2\pi )+f(a-2\pi )$ we will use periodic properties.

Step 2. Part (a). To find value of f ( - a )

We have given that :- 

$f\left(\theta \right)=\cos \theta$.

Put $\theta =-a$, then we have :-

$f(-a)=\cos \left(-a\right)$.

We know that cosine function is an even function, that is : 

$\cos \left(-\theta \right)=\cos \theta$.

So that :-

$\cos \left(-a\right)=\cos \left(a\right)$.

This gives us :-

$$\begin{gathered} f\left(-a\right)=\cos \left(a\right) \\ or f(-a)=f\left(a\right) \end{gathered}$$

We have given that :-

$f\left(a\right)=\frac{1}{4}$.

By putting this value we have :- 

$f(-a)=\frac{1}{4}$.

This is the required value. 

Step 3. Part (b) To find value of f ( a ) + f ( a + 2 π ) + f ( ( a - 2 π )

Given that :- 

$f\left(\theta \right)=\cos \theta$.

We know that cosine function is periodic for period $2\pi$.

That is :-

$\cos (\theta +2\pi k)=\cos \theta$ for any integer $k$.

So that we have :-

$$\begin{gathered} f\left(\theta \right)+f(\theta +2\pi )+f(\theta -2\pi )=\cos \theta +\cos (\theta +2\pi )+\cos (\theta -2\pi ) \\ \Rightarrow f\left(\theta \right)+f(\theta +2\pi )+f(\theta -2\pi )=\cos \theta +\cos \theta +\cos \theta \\ \Rightarrow f\left(\theta \right)+f(\theta +2\pi )+f(\theta -2\pi )=3\cos \theta \\ \Rightarrow f\left(\theta \right)+f(\theta +2\pi )+f(\theta -2\pi )=3f\left(\theta \right). \end{gathered}$$

Now take $a$ in place of $\theta$, then we have :-

$f\left(a\right)+f(a+2\pi )+f(a-2\pi )=3f\left(a\right)$.

We have :-

$f\left(a\right)=\frac{1}{4}$.

By putting this value we have :-  

$$\begin{gathered} f\left(a\right)+f(a+2\pi )+f(a-2\pi )=3\times \frac{1}{4} \\ \Rightarrow f\left(a\right)+f(a+2\pi )+f(a-2\pi )=\frac{3}{4} \end{gathered}$$

This is the required value.