Which of these functions is an even function? f(x) = sin(-3 pi x), f(x) = tan(3 pi x), f(x) = cos(5 pi x / 4), f(x) = csc(-pi x / 2).

Step-by-step solution

1. 1

   State the test. A function is even if $f(-x) = f(x)$ for every $x$ in its domain (graph symmetric about the $y$-axis) and odd if $f(-x) = -f(x)$ (symmetric about the origin). So the whole question is: substitute $-x$ and see which sign you get back.

2. 2

   Recall the parity of the basic trig functions. $\cos$ is even: $\cos(-\theta) = \cos\theta$. Its reciprocal $\sec$ is even too. Meanwhile $\sin$, $\tan$, $\cot$, $\csc$ are all odd: $\sin(-\theta) = -\sin\theta$, and so on. A linear inside argument $kx$ does not disturb this, because replacing $x$ by $-x$ just negates $kx$.

3. 3

   Test the three odd candidates.

   $\sin(-3\pi(-x)) = \sin(3\pi x) = -\sin(-3\pi x)$

   $\tan(3\pi(-x)) = \tan(-3\pi x) = -\tan(3\pi x)$

   $\csc\!\left(-\tfrac{\pi}{2}(-x)\right) = \csc\!\left(\tfrac{\pi}{2}x\right) = -\csc\!\left(-\tfrac{\pi}{2}x\right)$

   Each returns the negative of the original, so all three are odd, not even.

4. 4

   Test the cosine candidate. Because cosine is even,

   $\cos\!\left(\tfrac{5\pi}{4}(-x)\right) = \cos\!\left(-\tfrac{5\pi}{4}x\right) = \cos\!\left(\tfrac{5\pi}{4}x\right) = f(x)$

   So $f(-x) = f(x)$ and this one is even.

5. 5

   Spot-check numerically. Take $x = 0.3$: $\cos\!\bigl(\tfrac{5\pi}{4}(0.3)\bigr) = \cos(1.1781) = 0.3827$ and $\cos\!\bigl(\tfrac{5\pi}{4}(-0.3)\bigr) = \cos(-1.1781) = 0.3827$ ✓ — equal. For the tangent option at the same $x$: $\tan(0.9\pi) = -0.3249$ versus $\tan(-0.9\pi) = +0.3249$ — opposite, confirming odd. Answer: $f(x) = \cos\!\left(\tfrac{5\pi}{4}x\right)$.