Question

answer the following. (a) for each function below, indicate whether it is odd, even, or neither. $f(x)=\sec x$ odd even neither $g(x)=\cos x$ odd even neither (b) $\sec (-85^{\circ})=$ (choose one) (c) $\cos (-\frac{2\pi}{5})=$ (choose one)

Explanation:

Step1: Recall the definitions of even and odd functions

An even function satisfies \(f(-x)=f(x)\) for all \(x\) in the domain. An odd function satisfies \(f(-x)=-f(x)\) for all \(x\) in the domain.
For \(y = \sec x=\frac{1}{\cos x}\), we know that \(\cos(-x)=\cos x\). Then \(\sec(-x)=\frac{1}{\cos(-x)}=\frac{1}{\cos x}=\sec x\).
For \(y=\cos x\), \(\cos(-x)=\cos x\).

Step2: Use the property of even - valued trigonometric functions

Since \(\sec x\) and \(\cos x\) satisfy \(f(-x) = f(x)\) (because \(\sec(-x)=\sec x\) and \(\cos(-x)=\cos x\)), they are even functions.
Also, for any trigonometric function \(y = f(x)\) which is even (\(f(-x)=f(x)\)), we have \(\sec(-85^{\circ})=\sec(85^{\circ})\) and \(\cos(-\frac{2\pi}{5})=\cos(\frac{2\pi}{5})\)

Answer:

• \(f(x)=\sec x\): Even
• \(g(x)=\cos x\): Even
• \(\sec(-85^{\circ})=\sec(85^{\circ})\)
• \(\cos(-\frac{2\pi}{5})=\cos(\frac{2\pi}{5})\)