## JenniferSmart1 Group Title graphing $y=sin(2x-\pi)$ did I graph this correctly? one year ago one year ago

1. JenniferSmart1 Group Title

|dw:1360883013269:dw|

2. JenniferSmart1 Group Title

I guess I can get as far as doing the phase shift but I don't know what to do with the 2 in front of the x $y=sin(2(x-\pi/2))$|dw:1360883459233:dw|

3. JenniferSmart1 Group Title

@TuringTest

4. JenniferSmart1 Group Title

the phase shift is pi/2 isn't it?

5. TuringTest Group Title

$\sin(-\pi)=0$so the graph should start at x=0, y=0

6. JenniferSmart1 Group Title

what does the phase shift do then?

7. JenniferSmart1 Group Title

oh I see

8. JenniferSmart1 Group Title

no not really

9. JenniferSmart1 Group Title
10. TuringTest Group Title

a phase shift of -pi will shift the graph a half-period to the right, which for the sin is equivalent to turning it upside-down

11. JenniferSmart1 Group Title

Oh i see. so the phase shift is pi and not pi/2

12. TuringTest Group Title

right, you could also see that by applying$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$

13. JenniferSmart1 Group Title

makes sense. Thanks!

14. TuringTest Group Title

welcome :D