Question

what are the period and amplitude of this graph: http://www.wolframalpha.com/input/?i=sin%28x%29%2C+0%3Cx%3C4pi

Answer 1

Peak amplitude: 1
Peak to peak amplitude: 2

Answer 2

http://www.lynnabbott.com/members/Body_Wisdom_Wheel/Images/Amplitude.png

Answer 3

period is the standard 2pi

Answer 4

what is the function of this graph: y=sin(θ∘){0∘≤θ≤720∘}

Answer 5

that is same function of graph you posted in wolfram

Answer 6

okay so is it y=3sin2piθ

Answer 7

what no?

Answer 8

it is for the function y =a sin nx, when a = amplitude and n = period

Answer 9

not quite...period = 2pi/n

Answer 10

okay cool! so it is 4sin 2pi/n

Answer 11

for what? you said the function was y = sin(x) ....thats what graph said

Answer 12

y =a sin nx

Answer 13

you have to change it

Answer 14

is the amplitude 4

Answer 15

i have no idea...lol what graph are we talking about that has amplitude of 4???

Answer 16

i am so confused it is 2 right

Answer 17

Well, I tried to help until you got frustrated and left.
Why won't you explain us what are you exactly trying to do before it happens again?

Answer 18

stop asking me, i am as confused as you ....are we talking about the graph of sin(x) ?

Answer 19

@pitamar lol, so i see there is a trend

Answer 20

Descrive the characteristics of the graph.

Answer 21

Ye! he stole my graph and brought it here >=(
And to the characeristics of the graph
amplitude is 1, period is \(2 \pi \) and domain is \(0 \le x \le 4\pi \)

Answer 22

the characteristics of sin(x)
amplitude = 1
period = 2pi

Answer 23

how can i locate the solution for sin x = 0.5 {0 < x < 720)

Answer 24

unit circle or calculator

Answer 25

on the graph

Answer 26

No really need
\[
sin(x) = 0.5 = \frac{1}{2}
\]
And it's known that the value that gives that is \(30^\circ\) for example.
So we say:
\[
sin(x) = sin(180^\circ - x) \\
sin(30^\circ) = sin(180^\circ - 30^\circ) = sin(150^\circ)
\]
And since:
\[
sin(x) = sin(360^\circ \pm x)
\]
we can say that \(sin(x) = 0.5\) when:
\[
x = 30^\circ + 360^\circ K \\
x = 150^\circ + 360^\circ K
\]
The only values of x that actually fit in our domain are between \(0^\circ\) to \(720^\circ\) so
\[
x = 30^\circ \quad or \\
x = 30^\circ + 360^\circ = 390^\circ\quad or \\
x = 150^\circ \quad or \\
x = 150^\circ + 360^\circ = 510^\circ
\]
The graph is in radians so converted to radians this is:
\[
x = \frac{\pi}{6} \quad or \\
x = \frac{\pi}{6} + 2 \pi = \frac{13\pi}{6} \quad or \\
x = \frac{5\pi}{6} \quad or \\
x = \frac{5\pi}{6} + 2 \pi = \frac{17\pi}{6}
\]

Answer 27

thanks

Answer 28

sure np, is that all clear?

Answer 29

yep