- Thread starter
- #1

#### sweatingbear

##### Member

- May 3, 2013

- 91

The conventional(?) approach is to first solve \(\displaystyle \cos(x) = \sin(x)\) and then draw the graphs for each function in order to find the correct interval. However, I would love to know if there is a way to do it strictly algebraically.

I attempted writing \(\displaystyle \sin (90^\circ - x) < \sin (x)\) which yields \(\displaystyle x > 45^\circ + n \cdot 180^\circ\) but I am unable to find an upper bound for \(\displaystyle x\).