For this question, it really helps to be familiar with the unit circle. We will travel around the circle anticlockwise, starting from #pi/2# and finishing at #(3pi)/2#.
In the second quadrant, at #theta=pi/2#, we start with cos = 0 and sin = 1. Moving anticlockwise, cos becomes more negative and sin decreases from 1 to 0. At #theta=pi#, cos = -1 and sin = 0. Therefore, cos is always less than sin.
In the third quadrant, at #theta=pi#, cos starts at -1 and increases towards 0 at #(3pi)/2#, so it is getting larger. Sin starts at 0 and decreases towards -1. Therefore, at some point sin becomes less than cos and remains smaller up until #(3pi)/2#, where cos is 0 and sin is -1.
We need to find this crossover point to see when #cos(theta)# ceases to be less than #sin(theta)#. Do this by equating the two terms:
#sin(theta)=cos(theta)#
Rearrange this equation:
#sin(theta)/cos(theta)=1=tan(theta)#
Solve #tan(theta)=1#
#theta=pi/4+-npi, ninZZ#
The answer we want is in the third quadrant, so
#theta=(5pi)/4#
Now we know that #costheta##<##sin(theta)# for #pi/2<=(theta)<(5pi)/4#
