Is $\cos(a(\pi-t))$ an even function?

Question

I've an assignment due in a couple of days, and I'm wondering if my teacher made a mistake in the question below, or if I'm missing something silly.

Let $a \in [0, 1]$ and $f_a$ the $2\pi$-periodic function defined as $f_a(t) = \cos(a(π − t))$ for $t ∈ [0, 2π]$.

Show that $f_a$ is continuous and even.

I've tried proving $f_a(t)=f_a(-t)$ with Euler and trigonometry formula to no avail. Any hint on what I'm doing wrong?

Answer 1

Did you try picking some arbitrary value of $a$ (or even just $a=1/2$) and plotting the function? Sure doesn't look even to me.

A plot is not a proof but you can also use the angle sum formulas, $$\cos\left(\frac{\pi}{2} - \frac{t}{2}\right) = \cos\frac{\pi}{2}\cos\frac{t}{2} + \sin\frac{\pi}{2}\sin\frac{t}{2} = \sin\frac{t}{2},$$ which is clearly not even.

Answer 2

You need to use the periodicity of $f_a$. Previously, I suggested take $t=1, a=\frac 13$. We get $$f_{1/3}(1)=\cos((\pi-1)/3)\approx 0.756\\ f_{1/3}(-1)=\cos((\pi+1)/3)\approx 0.189$$ But this is not correct. $f_a(t)$ is defined by starting with $t \in [0,2\pi]$ (should be $t \in [0,2\pi[$ to avoid a conflict) and then invoking periodicity. So $$f_{1/3}(-1)=f_{1/3}(2\pi-1)=\cos((\pi-(2\pi-1))/3)=\cos((-\pi+1)/3)\approx 0.756$$ You can use the fact that $\cos (\pi-x)$ is symmetric around $\pi$ because $\cos(-x)=\cos(x)$ to get there.

Answer 3

Let $a \in ]0, 1[$ and $f_a$ the 2π-periodic function defined as $f_a(t) = \cos(a(π − t))$ for $t ∈ [0, 2π[$.

Show that $f_a$ is continuous and even.

It is true that (for example)
$\cos(\pi - \theta) = \cos(\pi + \theta).$

So, the assertion is true if $a$ happens to equal $1$, because you end up comparing $\cos(\pi - t)$ versus $\cos(\pi + t)$.

Further, if the assertion was changed to
$f_a = \cos[\pi - (at)]$,
rather than $f_a = \cos[a(\pi - t)]$,
then the assertion would again be true, for the same reason.

What has gone wrong, which makes the assertion false, is that instead of comparing $\cos[\pi - (at)]$ versus $\cos[\pi + (at)]$ you are comparing $\cos[(a\pi) - (at)]$ versus $\cos[(a\pi) + at].$

This means that instead of having the central angle about which the angle $\theta$ is reflected be the angle $\pi$, the central angle is $(a\pi)$.

This makes me wonder whether there is a typo in the problem. That is, I wonder whether the OP (i.e. original poster) has mistranslated the original assertion.