Asked by Anonymous
Given that cos2x=7/12 and "270 equal or < 2x equal or < 360", find sinx.
Please help and Thank you
Please help and Thank you
Answers
Answered by
Reiny
In these type of quesstions it usually asks for the "exact" value of ....
cos 2x = 7/12
cos 2x = 1 - 2sin^2 x
2 sin^2 x = 1 - cos 2x = 1 - 7/12 = 5/12
sin^2 x = 5/24
sin x = ± √(5/24)
but 270 < 2x ≤ 360
135 ≤ x ≤ 180 ---> x in in I or II, so
sinx = +√5/√24 = √30/12
cos 2x = 7/12
cos 2x = 1 - 2sin^2 x
2 sin^2 x = 1 - cos 2x = 1 - 7/12 = 5/12
sin^2 x = 5/24
sin x = ± √(5/24)
but 270 < 2x ≤ 360
135 ≤ x ≤ 180 ---> x in in I or II, so
sinx = +√5/√24 = √30/12
Answered by
Anonymous
how did you get the ã30/12?
Answered by
Anonymous
sqrt(5)/2sqrt(6)
when you rationalize the denominator you times both top and bottom by sqrt(6) only not the 2 right?
when you rationalize the denominator you times both top and bottom by sqrt(6) only not the 2 right?
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