Hi All,
Could someone tell me how to plot the graph and generate the table for y = 9 sin (Ɵ - 15) over 360 degree.
The coordinates need to be a 15 degree intervals
regards
Bungouk
Hi All,
Could someone tell me how to plot the graph and generate the table for y = 9 sin (Ɵ - 15) over 360 degree.
The coordinates need to be a 15 degree intervals
regards
Bungouk
Here is more information on the Taylor series:
Trigonometric functions - Wikipedia, the free encyclopedia
So, what, exactly is the question? If it is, as you initially said, to "graph and generate the table" with "15 degree intervals", then you just need to use your calculator (make sure it is set to "degree" measure, not radian) to to take $\displaystyle \theta= 0, 15, 30, 45, ... 345, 360$ so that $\displaystyle \theta- 15= -15, 0, 15, 30, ... 330, 345$ and so find 9sin(-15), 9 sin(0), 9 sin(15), 9 sin(30), 9 sin(45), ..., 9 sin(330), 9 sin(450) and plot those points.
If, however, the problem is to use the "standard" graph, y= sin(x), and alter it to give y= 9 sin(x- 15), then, yes, "9" is the amplitude: sin(x) is always between-1 and 1 so that, no matter what $\displaystyle \theta$ is, 9 sin(x- 15) is always between -9 and 9. One thing you could do is draw horizontal lines at y= -9 and y= 9 as guides. You know, I presume, that sin(x) is periodic with period 360 degrees, that it is 0 when x= 0, goes up to 1 when x= 90, down to 0 at x= 180, down to -1 at x= 270, back up to 0 at x= 360 and then repeats. Okay, when $\displaystyle \theta- 15= 0$, that is, when $\displaystyle \theta= 15$, $\displaystyle 9sin(\theta- 15)$ is 0. When $\displaystyle \theta- 15= 90$, that is, when $\displaystyle \theta= 105$, $\displaystyle 9sin(\theta - 15)$ is 9, when $\displaystyle \theta- 15= 180$, that is, when $\displaystyle \theta= 195$, [tex]9sin(\theta-15)= 0[k/tex], etc.
Plot on a single sheet of graph paper the two following functions over 360o. Ensure both your axis are clearly labelled and that the graph is appropriately titled. Produce tables to generate all of the coordinates for both functions at approximately 15o intervals.
a) Y = 9 sin (θ – 15)
b) Y = 5 cos (θ + 10)
please see full question above
taking the sine of θ - 15, is what's known as a "phase shift".
since you are going to find sine in 15 degree intervals anyway: do this:
sin(0°) = sin((15-15)°) <---this is your value for θ = 15°
sin(15°) = sin((30-15)°) <--this is your value for θ = 30°
etc.
remember to multiply everything by 9 when you're done. your answers will be numbers between -9 and 9 (if you get something of bigger absolute value than this, you've made a mistake, somewhere).
some points are "easy" to find (all angles are in degrees):
9sin(45-15) = 9sin(30) = 9/2 = 4.5
9sin(60-15) = 9sin(45) = 9√2/2 ~ 6.364
9sin(75-15) = 9sin(60) = 9√3/2 ~ 7.794
9sin(105-15) = 9sin(90) = 9
and so on. the "hard ones" to do are for θ = 30 and θ = 90 (which require finding sin(15) and sin(75)). after that, there is a pattern:
A,B,C,D,E,F,G (for the first 7 15-degree increments, starting with θ = 15)
F,E,D,C,B,A (for the next 6)
-B,-C,-D,-E,-F,-G (for the next 6)
-F,-E,-D,-C,-B (the next one would be A = 9sin(375-15) = 9sin(360) = 9sin(0) = 9sin(15-15), you've come "full circle").
Just draw the curve for sine function.
Now you have to make adjustments as explained:
For the function f(x) = A sin(Bx + C ) + D
A gives the amplitude, B the period, C the horizontal shift and D the vertical shift. I am sure this would help you.