Question

Determine whether the function is even, odd, or neither.

$f\left(x\right)=\sin x+\cos x$

Short answer

The$f\left(x\right)=\sin x+\cos x$is neither an even function nor an odd function.

Step-by-step solution

Step 1. State the definitions of even and odd functions.

A function $f\left(x\right)$ is an even function if $f\left(x\right)=f\left(-x\right)$.

A function$f\left(x\right)$ is an odd function if $f\left(x\right)=f\left(-x\right)$.

Step 2. Calculate f−x. 

Substitute $-x$ for x into the function $f\left(x\right)=\sin x+\cos x$and simplify.

$$\begin{gathered} f\left(x\right)=\sin x+\cos x \\ f\left(-x\right)=\sin \left(-x\right)+\cos \left(-x\right) \end{gathered}$$

Step 3. Simplify.

Note: $\sin \left(-t\right)=-\sin t$, that is., sine function, in general, is an odd function and $\cos \left(-t\right)=\cos t$, that is., cosine function, in general, is an even function.

$f\left(-x\right)=\sin \left(-x\right)+\cos \left(-x\right)$

$$\begin{gathered} =\left(-\sin x\right)+\cos x \\ =-s\mathrm{in}x+\cos x \end{gathered}$$

Since, neither $f\left(-x\right)\ne f\left(x\right)$nor $f\left(-x\right)\ne -f\left(x\right)$, therefore, $f\left(x\right)=\sin x+\cos x$ is neither an even function nor an odd function.