Answer: [tex]0 \le \theta < \frac{\pi}{4}[/tex]
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How to get this answer:
Use the unit circle to note that [tex]\sin\theta = \cos\theta = \frac{\sqrt{2}}{2}[/tex] when [tex]\theta = \frac{\pi}{4}[/tex] (aka 45 degrees)
Beyond this point, cosine is smaller than sine. This means that anything from 0 to pi/4 will have sine be smaller than cosine. It might help to graph y = sin(x) and y = cos(x) on the interval from x = 0 to x = pi.
The two curves y = sin(x) and y = cos(x) intersect at the point [tex]\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)[/tex]
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Here's a more detailed picture of whats going on.
[tex]\sin \theta < \cos \theta\\\\\sin \theta < \sqrt{1-\sin^2\theta}\\\\\sin^2 \theta < 1-\sin^2\theta \\\\2\sin^2 \theta < 1\\\\\sin^2 \theta < \frac{1}{2}\\\\\sin \theta < \sqrt{\frac{1}{2}}\\\\\sin \theta < \frac{1}{\sqrt{2}}\\\\\sin \theta < \frac{\sqrt{2}}{2}\\\\\theta < \arcsin\left(\frac{\sqrt{2}}{2}\right)\\\\\theta < \frac{\pi}{4}\\\\[/tex]
Intersect the intervals [tex]0 \le \theta < \pi[/tex] and [tex]\theta < \frac{\pi}{4}[/tex] and you'll end up with the final answer [tex]0 \le \theta < \frac{\pi}{4}[/tex]
