Is this the correct period? What is the period for the following:
$$  y = 10 \sin\Bigl(\frac{2\pi}{365}(x-50)\Bigr) $$
Is the period $$ \frac{2\pi}{\frac{2\pi}{365}} $$  which would be $365$?
 A: In general, divide the period of any function $f(x)$ by the coefficient of $x$ i.e. if $T$ is the period of any function say $f(x)$ then the period of function $f(\alpha x+\beta)$ is $$=\frac{T}{\alpha}$$ 
Hence, for the given function:$y = 10 \sin\left(\frac{2\pi}{365}(x-50)\right)=10 \sin\left(\frac{2\pi}{365}x-\frac{200\pi }{365}\right)$  Hence the period of given function  $$=\frac{\text{period of}\ \sin x}{\text{coefficient of} \ x}=\frac{2\pi}{\frac{2\pi }{365}}=365$$ 
A: Yes. You are correct, the function $$a \sin (bx + c)$$ has period $\frac{2\pi}{b}$. In your case $b = \frac{2\pi}{365}$, so the period is (as you rightly said) $$\bbox[10px, border:blue 1px solid]{\text{period} = \frac{2\pi}{\frac{2\pi}{365}} = 365}$$

The same thing applies to the cosine function, $a \cos (bx +c)$ has period $2\pi/b$, whilst the tangent function $a \tan (bx +c)$ has period $\pi/b$ because the tangent function is $\pi$-periodic.
A: For what value of $\Delta x$ will cause an increase of $2\pi$ in the argument of $\sin$.
$$\frac{2\pi}{365}(x-50) + \color{blue}{2\pi} = \frac{2\pi}{365}((x+\color{purple}{\Delta x})-50)$$
Try doing some manipulations to the LHS to get the RHS. What does $\Delta x$ equal?
Now try to find the period of $\sin(a(x-c))$
$$a(x-c) + \color{blue}{2\pi} = a\bigg((x+\color{purple}{\Delta x})-c\bigg)$$
What does $\Delta x$ equal now?
