The conclusion you seem to be making is not correct; just knowing that $f(1.1)>f(1)$, for example, does not tell you that $f(1.1)$ is a local maximum value. To show that $f(a)$ is a local maximum value, you'd have to prove that $f(x)\le f(a)$ for every $x$ in the domain of $f$ sufficiently close to $a$. Here, you cannot do that; $f$ has no local minimum or maximum values by your implied argument.
Note that the definition you give is for relative maximum and minimum values. Whereas you seem to be referring to global maximum and minimum values in the sequel.
A local maximum value $f(x_0)$ is a the largest value the function attains in a neighborhood of $x_0$. Note, the definition said that $f(x_0)$ is greater than or equal to the values of $f(x)$ at all nearby points $x$.
A global maximum value $f(x_0)$ must satisfy $f(x_0)$ is greater than or equal to the values of $f(x)$ for all $x$.
There are similar definitions for local and global minimum values.
These concepts also take into account the domain of the function.
For instance, if you restrict $f(x)=x$ to have domain $[0,1.1]$, say, then $f(0)=0$ will be both a local and a global minimum value and $f(1.1)=1.1$ will be both a local and a global maximum value. To be precise, we would say $f$ has both a local and global maximum value of $f(1.1)=1.1$ over $[0,1.1]$; and, $f$ has both a local and a global minimum value of $f(0)=0$ over $[0,1.1]$.
If you consider $f(x)=x$ to have domain $\Bbb R$, as I assumed in the first paragraph, then it would have no local or global maximum values by your observations. It would also have no local or global minimum values.
By the cosine addition formula $f(t)\sin(t)$ is a telescopic sum:
$$ f(t)\sin(t) = \frac{\cos(4t)-\cos(14t)}{2}$$
and $f(t)$ is a $2\pi$-periodic, odd function, hence in order to find $\max/\min f$ it is enough to find $\max f$.
Since
$$ f'(t) = \frac{5\cos(3t)-3\cos(5t)-15\cos(13t)+13\cos(15t)}{4\sin^2 t} $$
by using Chebyshev polynomials of the first kind we get that $f(t)$ attains its maximum value when $\cos(t)$ is a root of $105 - 3100 x^2 + 27488 x^4 - 103552 x^6 + 187648 x^8 - 161792 x^{10} +
53248 x^{12}$. Numerically
$$ \max f \approx 4.4648, \qquad \min f \approx -4.4648. $$
We may notice this is just a bit tighter than the trivial bound $|f|\leq 5$.
Best Answer
Now that you know $f’(x)$ has a real root which is $x=0$, you can use it to draw the sign scheme of $f’(x)$. More specifically, $f’(x)>0$ is when $x>0$ and $f’(x)<0$ is when $x<0$. This means $f(x)$ is a monotone increasing function on $(0,\infty)$ and monotone decreasing on $(-\infty,0)$. This means that in the given interval of $x$, the minimum can occur at $x=0$, and maximum either at $x=\ln(7)$ or $x=-\ln(2)$. Checking both these possibilities we conclude that $$\min_{x=0} f(x)=1$$ $$\max_{x=\ln 7} f(x)=\frac{25}{7}$$