Determine if $x(t)$ is periodic. If periodic calculate its period. Determine if $$x(t) = \cos(8t) + 4 \sin(8t)$$ is periodic. If so, calculate its period.
 A: You need to find any common period (typically the smallest period) of the individual periods of the sinusoidal terms $\cos(8t)$ and $4\sin(8t)$. A function $f$ is periodic with period $T>0$ if $f(t)=f(t+T)$ for all $t$.
We can write a sinusoidal term as $\cos\left(2\pi\frac{t}{T}\right)$, where T is the period of the sinusoid.
So we can rewrite the first term $\cos(8t) = \left(\cos(2\pi\frac{t}{\frac{\pi}{4}}\right)$. Then $T=\frac{\pi}{4}$.
The second term $4\sin(8t) = 4\sin\left(2\pi\frac{t}{\frac{\pi}{4}}\right)$. Then $T=\frac{\pi}{4}$.
The smallest common period is $T=\frac{\pi}{4}$.
A: To check if a function is periodic, you need to know that $x(t) = x(t+c)$ for some fixed constant $c$. Let us figure out what $c > 0$ is (if it just so happens to exist). Notice that if $x(t) = x(t+c)$, then
$$
\cos(8t) + 4\sin(8t) = \cos(8(t+c)) + \sin(8(t+c)) = \cos(8t + 8c) + 4\sin(8t+8c).
$$
for all $t$. This is definition of what it means for $x$ to periodic with period $c$. So let's ask ourselves if this can happen and what is the smallest value of $c$ that it does? We know that both sine and cosine are periodic with period $2\pi$. Therefore, since $\cos(8t) = \cos(8t + 8c)$ for all values of $t$, we know that $8t = 8t + 8c - 2\pi$ and so $c = \pi/4$. This value also works for the sine. It is the smallest such $c$ for otherwise sine and cosine would need to have a period less than $2\pi$, which isn't the case. Therefore, $x$ is periodic with period $\pi/4$.
A: Some hints


*

*A function is periodic with period $P$ if $f(t)=f(t+P)$ for all $t$.

*These trigonometric identities can be used to simplify the expression:
$$
\begin{align}
\cos(a+b)&=\cos a\cos b-\sin a\sin b,\\
\sin(a+b)&=\sin a\cos b+\sin b\cos a.
\end{align}
$$
A: Let $1=r\cos A,4=r\sin A$ where $r>0$
So, $1^2+4^2=r^2(\cos^2A+\sin^2A)\implies r^2=17\implies r=\sqrt{17}$
$\cos A=\frac1{\sqrt{17}},\sin A= \frac4{\sqrt{17}},\tan A=4$
$\cos8t+4\sin8t=\sqrt{17}(\cos8t\cos A+\sin8t\sin A)=\sqrt{17}\cos(8t-\arctan 4)$
Now, if $f(t)=\sqrt{17}\cos(8t-\arctan 4), f(t+T)=\sqrt{17}\cos(8(t+T)-\arctan 4)$
The period is the minimum positive constant value of $T$ such that $f(t)=f(T+t)$
$$\sqrt{17}\cos(8t-\arctan 4)=\sqrt{17}\cos(8(t+T)-\arctan 4)$$
iff $$\cos(8t-\arctan 4)=\cos(8(t+T)-\arctan 4)$$
iff $$8t-\arctan 4=2n\pi\pm(8(t+T)-\arctan 4)$$ where $n$ is any integer
Taking the '+' sign, $T=-\frac{2n\pi}8=-n\frac\pi4$
As the period is the minimum constant positive number, it will be $\frac\pi4$ putting $n=-1$
Taking the '-' sign, $8t-\arctan 4=2n\pi-(8(t+T)+\arctan 4)\implies T$ is dependent on $t$ hence not constant 
A: The above signal is  periodic.
Simple hint: If $\pi$ is in the denominator, the signal is non-periodic, otherwise the signal is periodic. As an example, consider
$$W=2\pi f_0 \\
f_0=1/T \\
T=2\pi/w$$
so in the above question, the fundamental frequency is $8$, and
the answer is $\pi/4$ which is periodic, and same for the other.
