# Find the period of $y= 3\sin(8(x+4))+5$

I am doing a math problem for my homework and I know I got the answer wrong by looking at the back of the book. I am just trying to find out how to get that answer for future reference.

The question is to find the period of $y= 3\sin(8(x+4))+5$

My answer is $8$ but the book's answer is $\dfrac\pi4$.

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Well, what is the period of $\sin(x)$? What about $\sin(2x)$ or even $\sin(8x)$? –  M.B. Oct 29 '13 at 18:44

If you've had material on transformation of functions, you would have met "horizontal stretches/compressions" for $\ f(kx) \$ , which takes the function $\ f(x) \$ and "squeezes" it toward the $\ y$-axis if $\ k > 1 \$ and "stretches" it away from the $\ y$-axis if $\ 0 < k < 1 \$ . Another way to think of this is that $\ f(kx) \$ says "plot the function $\ f(x) \ \ k \$ times 'faster' along the $\ x$-axis" , when $\ k > 1 \$ ("squeeze it inward"); when $\ k \$ is between 0 and 1 , write it as a fraction $\ k \ = \ \frac{1}{n} \$ , so "plot $\ f(x) \ \ n \$ times slower" ("stretch it out").
For periodic functions, like the trig functions, this has the effect of changing the period for sine, cosine, secant, and cosecant to $\ T \ = \ \frac{2 \pi}{k}$ , and, for tangent and cotangent, to $\ T \ = \ \frac{ \pi}{k}$ . In your function, the multiplier $\ k \$ is equal to 8 , so the period of this transformed sine function is $\ T \ = \ \frac{2 \pi}{8} \ = \ \frac{\pi}{4}$ .
The phase-shift of -4 in the function $\ \sin( \ 8 \ ( x + 4 ) \ )$ doesn't affect the period at all; it is only that multiplier 8 that matters. Likewise, the "vertical stretch" of 3 and the "vertical shift" of +5 do not alter the new period.
The period is calculated by asking when the argument inside the $\sin$ changes by $2\pi$. So if $p$ is the period, what you want to ask is when is $8([x+p]+4) = 2\pi + 8(x+4)$? Solve this for $p$ and your answer will agree with the book