sweatingbear said:
Forum, do you have any idea how to solve the trigonometric inequality $$\cos (x) < \sin (x)$$ strictly algebraically?
The conventional(?) approach is to first solve $$\cos(x) = \sin(x)$$ 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 $$\sin (90^\circ - x) < \sin (x)$$ which yields $$x > 45^\circ + n \cdot 180^\circ$$ but I am unable to find an upper bound for $$x$$.
When \displaystyle \begin{align*} -\frac{\pi}{2} + 2\pi n < x < \frac{\pi}{2} + 2\pi n \end{align*} where \displaystyle \begin{align*} n \in \mathbf{Z} \end{align*}, we have \displaystyle \begin{align*} \cos{(x)} > 0 \end{align*}, so in this region we can say
\displaystyle \begin{align*} \sin{(x)} &> \cos{(x)} \\ \frac{\sin{(x)}}{\cos{(x)}} &> 1 \\ \tan{(x)} &> 1 \end{align*}
We know that \displaystyle \begin{align*} \tan{(x)} > 1 \end{align*} when \displaystyle \begin{align*} \frac{ \pi}{ 4} + 2\pi n < x < \frac{\pi}{2} + 2\pi n \end{align*} and when \displaystyle \begin{align*} -\frac{3\pi}{4} + 2\pi n < x < -\frac{\pi}{2} + 2\pi n \end{align*}
So putting these together, that means we have one solution set for \displaystyle \begin{align*} \sin{(x)} > \cos{(x)} \end{align*} as \displaystyle \begin{align*} \frac{\pi}{4} + 2\pi n < x < \frac{\pi}{2} + 2\pi n \end{align*}.Now when \displaystyle \begin{align*} -\pi + 2\pi n < x < -\frac{\pi}{2} + 2\pi n \end{align*} or \displaystyle \begin{align*} \frac{\pi}{2} + 2\pi n < x \leq \pi + 2\pi n \end{align*} we have \displaystyle \begin{align*} \cos{(x)} < 0 \end{align*}, so that would mean in that region we have
\displaystyle \begin{align*} \sin{(x)} &> \cos{(x)} \\ \frac{\sin{(x)}}{\cos{(x)}} &< 1 \\ \tan{(x)} &< 1 \end{align*}
We know that \displaystyle \begin{align*} \tan{(x)} < 1 \end{align*} when \displaystyle \begin{align*} -\frac{\pi}{2} + 2\pi n < x < \frac{\pi}{4} + 2\pi n \end{align*} or \displaystyle \begin{align*} -\pi + 2\pi n < x < -\frac{3\pi}{4} + 2\pi n \end{align*} or \displaystyle \begin{align*} \frac{\pi}{2} + 2\pi n < x \leq \pi + 2\pi n \end{align*}.
So putting these together, that means another possible solution set for \displaystyle \begin{align*} \sin{(x)} > \cos{(x)} \end{align*} is \displaystyle \begin{align*} -\pi + 2\pi n \leq x < -\frac{3\pi}{4} + 2\pi n \end{align*} or \displaystyle \begin{align*} \frac{\pi}{2} + 2\pi n < x \leq \pi + 2\pi n \end{align*}.Therefore the solution to \displaystyle \begin{align*} \sin{(x)} > \cos{(x)} \end{align*} is \displaystyle \begin{align*} -\pi + 2\pi n \leq x \leq -\frac{3\pi}{4} + 2\pi n \end{align*} or \displaystyle \begin{align*} \frac{\pi}{4} + 2\pi n < x < \frac{\pi}{2} + 2\pi n \end{align*} or \displaystyle \begin{align*} \frac{\pi}{2} + 2\pi n < x \leq \pi + 2\pi n \end{align*}.
