Jo
posted by Jo on .
That makes sense but were did you get of 7.9 from. Once I know that I under stand subtracting the period... just don't know were that value came from
h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ? i d = 1 2 6 0 2 4 6 6 3 4

sorry

To solve 200 = 300sin(5pi/11)(t4) + 500
we get 2/3 = sin(5pi/11)(t4)
So I know that (5pi/11)(t4) is in either III or IV, since the sine is negative in those quadrants
To find the "angle in standard position" I ignore the negative and take the inverse sine
That got me .72972 (you must have had that)
For the angle in III
(5pi/11)(t4) = pi + .72972
solving this gave me 6.7 (which you had)
For the angle in IV
(5pi/11)(t4) = 2pi  .72972
solving this gave the the 7.9