Please help me with this problem?

Given:

#h_J(t) = 42+sin((pit)/12)#

#h_P(t) = 42-cos((pit)/12)#

We want to know when:

#h_P(t) > h_J(t)#

That is:

#42-cos((pit)/12) > 42+sin((pit)/12)#

Subtract #42-cos((pit)/12)# from both sides to get:

#0 > sin((pit)/12) + cos((pit)/12)#

In order for the right hand side to be zero, we would require #sin# and #cos# to be the same size but of opposite signs.

The angles at which #sin theta# and #cos theta# are of equal size are the midpoints of the four quadrants, i.e. #pi/4#, #(3pi)/4#, #(5pi)/4#, #(7pi)/4#.

Of these, the quadrants in which #sin# and #cos# are of opposite signs are Q2 and Q4, with midpoints #(3pi)/4# and #(7pi)/4#

Hence the sign of #sin theta + cos theta# changes at these two points in #[0, 2pi]#.

Note that #sin(0)+cos(0) = 0+1 = 1 > 0#. So the inequality is not satisfied at #t=0#. This is what we would expect, since the entrance to the Ferris wheel is at the lowest point.

Hence the interval for which #h_P(t) > h_J(t)# is:

#(3pi)/4 < (pit)/12 < (7pi)/4#

Multiply through by #12/pi# to get:

#9 < t < 21#