For this question, it really helps to be familiar with the unit circle. We will travel around the circle anticlockwise, starting from pi/2π2 and finishing at (3pi)/23π2.
In the second quadrant, at theta=pi/2θ=π2, we start with cos = 0 and sin = 1. Moving anticlockwise, cos becomes more negative and sin decreases from 1 to 0. At theta=piθ=π, cos = -1 and sin = 0. Therefore, cos is always less than sin.
In the third quadrant, at theta=piθ=π, cos starts at -1 and increases towards 0 at (3pi)/23π2, 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)/23π2, where cos is 0 and sin is -1.
We need to find this crossover point to see when cos(theta)cos(θ) ceases to be less than sin(theta)sin(θ). Do this by equating the two terms:
sin(theta)=cos(theta)sin(θ)=cos(θ)
Rearrange this equation:
sin(theta)/cos(theta)=1=tan(theta)sin(θ)cos(θ)=1=tan(θ)
Solve tan(theta)=1tan(θ)=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
