- #1
estex198
- 14
- 0
Problem: Solve for t, 20 = 100 sin 2pi(50)t
note: pi = "pie"
I must be doing something wrong here. To solve algebraically, I first divide both sides by 100. Then, I get the inverse cosine of both sides, and set the angle in radians (.2013579208) equal to "2pi(50)t". Lastly I divide the derived angle by 100pi , which gives me .0063258456 for t.
When solving the equation graphically using a calculator, I get a different result for t. To solve graphically on my TI-83 Plus, I hit "y=" and enter for Y1 = 100*sin(2pi*50*X) - 20 and hit "graph". Once the graph has loaded I hit "2nd" + "calc" and choose the calculate zero function. I choose the intersection closest to x=0 to ensure the smallest positive angle, and for y=0, I get X = 6.4094E-4. This equates to 6.4094 * 10^-4, or .00064094, right? So why am I not getting the same answer for both of my solution methods?
Thanks in advance!
note: pi = "pie"
I must be doing something wrong here. To solve algebraically, I first divide both sides by 100. Then, I get the inverse cosine of both sides, and set the angle in radians (.2013579208) equal to "2pi(50)t". Lastly I divide the derived angle by 100pi , which gives me .0063258456 for t.
When solving the equation graphically using a calculator, I get a different result for t. To solve graphically on my TI-83 Plus, I hit "y=" and enter for Y1 = 100*sin(2pi*50*X) - 20 and hit "graph". Once the graph has loaded I hit "2nd" + "calc" and choose the calculate zero function. I choose the intersection closest to x=0 to ensure the smallest positive angle, and for y=0, I get X = 6.4094E-4. This equates to 6.4094 * 10^-4, or .00064094, right? So why am I not getting the same answer for both of my solution methods?
Thanks in advance!