zaddyzad said:
A) 0 correct ?
B/C) Iv never seen a Sin wave until I just had google graph it for me. Max sin = 1 and Min. sin = -1. And Sinθ=0 when θ=0,180. However how does this help me solve the overall b/c w.o technology.
If you've never seen a sine wave before today, what are you doing answering these questions? And maybe the title should have a been a little different, because it suggests you already know trig but you were just put off by a harder problem.
Anyway, yes, the sine wave oscillates between 1 and -1. This means that if we had the graph y=\sin(x) the max would be 1 and the min would be -1. Thus the graph y=2\sin(x) will oscillate between 2 and -2 because whatever the value of sin(x) is, the y value will be doubled.
In general, y=A\sin(x) will oscillate between A and -A.
The period of a sine wave means how far along the x-axis (or time in your case) the sine wave takes to finish a complete cycle. If we're at the maximum value of 1, then for y=\sin(x) it'll take another 360
o to get back to 1 again. So sin(x)=1 occurs at x = 180
o, 540
o, 900
o, -180
o etc. going on forever. The general formula is x=180^o+n\cdot 360^o for any integer n.
You need to be careful though! Just because the period of the sine wave is 360
o doesn't mean when you solve \sin(x)=0 it'll happen every 360
o because if the wave starts at 1 and by the time it gets back to 1 again, it would have crossed the y-axis twice (went down to -1 then back up again). The general formula for \sin(x)=0 is x=n\cdot 180^o
Finally, similarly to how you multiply the sine wave by a constant and the max/min changes from 1/-1, if you multiply the argument of the sine by a constant, such as y=\sin(bx) then the period of the sine wave will change. If b=2, then the period is halved which means the graph will oscillate twice as fast (it'll only take 180
o to get from 1 back to 1 again).