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So confused for this trig question:

Guest Feb 6, 2018
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 #1
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2.

 

T  =  11cos ( pi * t  / 12 )  + 71

 

In this case....t  = 2  [ 2 hours after 2 PM ].......so we have

 

T  =  11 cos (pi * 2 /12)  + 71  =

 

11cos (pi/6)  +  71  =

 

11 [ √3/ 2 ]  +  71  =

 

(11/2)√3   + 71   =

 

[ 11√3  + 142 ]  / 2   ≈  80.53°

 

 

Note that this temperature  will occur again at  -pi/6   because  cos (pi/6)  = cos (-pi/6)

 

So......solving this for t...we have 

 

pi * t /12  =  -pi / 6

 

t    / 12  =  - 1 / 6       multiply  both sides by 12

 

t =  -12 / 6   =    - 2....i.e., 2 hours before 2 PM   =  12 noon

 

This makes sense.......the temp  would increase until 2PM.....and at equal times from 2PM, namely at  12 noon    and at 2 PM.....the temps should be equal

 

 

cool cool cool

CPhill  Feb 6, 2018
edited by CPhill  Feb 6, 2018

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