Question

Determine whether the functions are even, odd, or neither. a. \(y=2 \sin x\) b. \(y=-\frac{\cos ^{2} x}{x}\) c. \(y=-\csc x\)

Short answer

a. The function \(y=2 \sin x\) is odd. b. The function \(y = -\frac{\cos^2 x}{x}\) is neither. c. The function \(y=-\csc x\) is neither.

Step-by-step solution

a. Determine the parity of \(y=2 \sin x\)

To determine if the function \(y=2 \sin x\) is even, odd, or neither, we will check whether the function satisfies the conditions for being even or odd. Evaluate \(y(-x)\): $$ y(-x)=2 \sin (-x) $$ Since the sine function has the property \(\sin (-x)=-\sin x\), we can write: $$ y(-x)=2(-\sin x)=-2 \sin x $$ Comparing the expressions for \(y(x)=2 \sin x\) and \(y(-x)=-2 \sin x\), we can conclude that \(y(-x)=-y(x)\). Therefore, the function \(y=2 \sin x\) is odd.

b. Determine the parity of \(y=-\frac{\cos^2 x}{x}\)

To determine if the function \(y = -\frac{\cos^2 x}{x}\) is even, odd, or neither, we will check whether the function satisfies the definitions of even and odd functions. Evaluate \(y(-x)\): $$ y(-x)=-\frac{\cos^2 (-x)}{-x} $$ Since the cosine function has the property \(\cos (-x)=\cos x\), we can write: $$ y(-x)=-\frac{\cos^2 x}{-x}=\frac{\cos^2 x}{x} $$ Comparing the expressions for \(y(x)=-\frac{\cos^2 x}{x}\) and \(y(-x)=\frac{\cos^2 x}{x}\), we can conclude that the function does not satisfy the conditions for being even or odd, so \(y = -\frac{\cos^2 x}{x}\) is neither.

c. Determine the parity of \(y=-\csc x\)

To determine if the function \(y=-\csc x\) is even, odd, or neither, we will check whether the function satisfies the conditions for being even or odd. Evaluate \(y(-x)\): $$ y(-x)=-\csc (-x) $$ Since the cosecant function has the property \(\csc (-x)=-\csc x\), we can write: $$ y(-x)=\csc x $$ Comparing the expressions for \(y(x)=-\csc x\) and \(y(-x)=\csc x\), we can conclude that the function does not satisfy the conditions for being even nor odd, so \(y=-\csc x\) is neither.