Quandaries and Queries


Given that the maximum value of [sin(3y-2)]2 -[cos(3y-2)]2
is k. If y>7, Find the minimum value of y for which
[Sin(3y-2)]2 - [cos(3y-2)]2 =k.
Thank you.

Student K12



For any x,

cos(2x) = cos2x - sin22


[sin(3y-2)]2 -[cos(3y-2)]2 = -cos(6y-4)

The maximum value of -cos(t) occurs at t = + 2n where n is an integer. What is the smallest value of y, y > 7, so that + 2n = 6y - 4 for some integer n?



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