For which value of x is cos x sin x,where x | vedclass

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(C) We are given the inequality $\cos x > \sin x$ for $x \in \left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$.Dividing by $\cos x$ requires caution becau

Vedclass · Accepted Answer

$\left( \frac{5\pi}{4}, \frac{3\pi}{2} \right)$

Vedclass · Answer

(C) We are given the inequality $\cos x > \sin x$ for $x \in \left( \frac{\pi}{2}, \frac{3\pi}{2} ight)$.Dividing by $\cos x$ requires caution because $\cos x$ changes sign in this interval.Alternatively,consider the intersection of the graphs of $y = \cos x$ and $y = \sin x$.In the interval $\left( \frac{\pi}{2}, \frac{3\pi}{2} ight)$,$\cos x = \sin x$ when $	an x = 1$,which occurs at $x = \frac{5\pi}{4}$.For $x \in \left( \frac{\pi}{2}, \frac{5\pi}{4} ight)$,$\cos x$ is negative and $\sin x$ is positive (or $\cos x$ is less than $\sin x$ in the second quadrant).Specifically,at $x = \pi$,$\cos \pi = -1$ and $\sin \pi = 0$,so $-1 For $x \in \left( \frac{5\pi}{4}, \frac{3\pi}{2} ight)$,both $\sin x$ and $\cos x$ are negative.In this interval,$\cos x$ is closer to $0$ than $\sin x$ (e.g.,at $x = \frac{4\pi}{3}$,$\cos x = -0.5$ and $\sin x = -0.866$,so $-0.5 > -0.866$).Thus,$\cos x > \sin x$ holds for $x \in \left( \frac{5\pi}{4}, \frac{3\pi}{2} ight)$.

For which value of $x$ is $\cos x > \sin x$,where $x \in \left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$?

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