The sine and cosine curves intersect infinite | vedclass

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(B) The points of intersection of $y = \sin x$ and $y = \cos x$ are given by $\sin x = \cos x$,which implies $	an x = 1$. Thus,$x = \frac{\pi}{4} + n

Vedclass · Accepted Answer

$2\sqrt{2}$

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(B) The points of intersection of $y = \sin x$ and $y = \cos x$ are given by $\sin x = \cos x$,which implies $	an x = 1$. Thus,$x = \frac{\pi}{4} + n\pi$ for any integer $n$.Consider the interval between two consecutive intersection points,such as $\left[ \frac{\pi}{4}, \frac{5\pi}{4} ight]$.In this interval,$\sin x \ge \cos x$.The area of one such region is given by the integral:$A = \int_{\pi/4}^{5\pi/4} (\sin x - \cos x) \, dx$Evaluating the integral:$A = [-\cos x - \sin x]_{\pi/4}^{5\pi/4}$$A = \left( -\cos\left(\frac{5\pi}{4}ight) - \sin\left(\frac{5\pi}{4}ight) ight) - \left( -\cos\left(\frac{\pi}{4}ight) - \sin\left(\frac{\pi}{4}ight) ight)$$A = \left( -(-\frac{1}{\sqrt{2}}) - (-\frac{1}{\sqrt{2}}) ight) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} ight)$$A = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} ight) - \left( -\frac{2}{\sqrt{2}} ight)$$A = \frac{2}{\sqrt{2}} + \frac{2}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} 	ext{ sq. units}$

The sine and cosine curves intersect infinitely many times,creating bounded regions of equal areas. The area of one such region is

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