# Jo

posted by Jo

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

1. Jo

sorry

2. Reiny

To solve 200 = 300sin(5pi/11)(t-4) + 500
we get -2/3 = sin(5pi/11)(t-4)

So I know that (5pi/11)(t-4) 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)(t-4) = pi + .72972
solving this gave me 6.7 (which you had)

For the angle in IV
(5pi/11)(t-4) = 2pi - .72972
solving this gave the the 7.9

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