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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 references.

Question is y= 3sin(8(x+4))+5

find the period.

my answer =8 book answer = π/4

Anyhelp is good help

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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

2 Answers 2

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.

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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

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