y=1747.29+1636.95sin(0.52x-1.71). What is the period for this function? and if you know the y value how do you figure out the x value? For the first question I know that to get the period you do...
y=1747.29+1636.95sin(0.52x-1.71). What is the period for this function? and if you know the y value how do you figure out the x value?
For the first question I know that to get the period you do 2pi/k but that does not give me a whole number such as 10 or 5, instead i get a number multiplied by pi which is not what i want. For the second question I figured out that at x=3.6 y=2011.32 and now I am asked to figure out more x values with the y value equal to 2011.32 using this equation, so i plugged in 2011.32 in for y and now i don't know what to do.
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We are given `y=1747.29+1636.95sin(.52x-1.71)`
(1) You are correct about the period. The period will be `p=(2pi)/.52=(50pi)/13~~12.083` This is what it is -- we might like to have "nice" answers but that rarely happens in the real world. The problem is what value to approximate to.
(2) If y=2011.32:
`2011.32=1747.29+1636.95sin(.52x-1.71)`
`264.03=1636.95sin(.52x-1.71)`
`.1612938697=sin(.52x-1.71)`
(a) `"Sin"^(-1)(sin(.52x-1.71))="Sin"^(-1)(.1612938697)`
`.52x-1.71=.1620015487`
`.52x=1.872001549`
`x=3.600002978` thus your answer of x=3.6
(b) When using the arcsin (inverse sin) there is the possibility that the angle is in another quadrant, so you also have:
`"Sin"^(-1)(sin(.52x-1.71))=pi-"Sin"^(-1)(.1612938697)`
`.52x-1.71=2.979591105`
`.52x=4.689591105`
`x=9.018444432`
Thus to a good approximation we have two points where the y value is 2011.32: x=3.6 and x=9.02.
To find more, we add the period to the two possible answers:
`y=2011.32==>x~~3.6+-12.083` or `x~~9.018+-12.083`
The graph:
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