# Question #dea46

Aug 2, 2017

$\frac{\pi}{2} \le \left(\theta\right) < \frac{5 \pi}{4}$

#### Explanation:

For this question, it really helps to be familiar with the unit circle. We will travel around the circle anticlockwise, starting from $\frac{\pi}{2}$ and finishing at $\frac{3 \pi}{2}$.

In the second quadrant, at $\theta = \frac{\pi}{2}$, 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 $\frac{3 \pi}{2}$, 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 $\frac{3 \pi}{2}$, where cos is 0 and sin is -1.

We need to find this crossover point to see when $\cos \left(\theta\right)$ ceases to be less than $\sin \left(\theta\right)$. Do this by equating the two terms:

$\sin \left(\theta\right) = \cos \left(\theta\right)$

Rearrange this equation:

$\sin \frac{\theta}{\cos} \left(\theta\right) = 1 = \tan \left(\theta\right)$

Solve $\tan \left(\theta\right) = 1$

$\theta = \frac{\pi}{4} \pm n \pi , n \in \mathbb{Z}$

$\theta = \frac{5 \pi}{4}$
Now we know that $\cos \theta$$<$$\sin \left(\theta\right)$ for $\frac{\pi}{2} \le \left(\theta\right) < \frac{5 \pi}{4}$