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
Hi,
For any x,
cos(2x) = cos2x - sin22
thus
[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?
Penny