The graphs of sine and cosine functions inter | vedclass

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(A) 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}, \f

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(A) 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}, \frac{5\pi}{4}, \dots$ The area $A$ between two consecutive points of intersection $\frac{\pi}{4}$ and $\frac{5\pi}{4}$ is given by: $A = \int_{\pi/4}^{5\pi/4} |\sin x - \cos x| \, dx$ In the interval $[\frac{\pi}{4}, \frac{5\pi}{4}]$,$\sin x \geq \cos x$. $A = \int_{\pi/4}^{5\pi/4} (\sin x - \cos x) \, dx$ $A = [-\cos x - \sin x]_{\pi/4}^{5\pi/4}$ $A = \left( -\cos\left(\frac{5\pi}{4}ight) - \sin\left(\frac{5\pi}{4}ight) ight) - \left( -\cos\left(\frac{\pi}{4}ight) - \sin\left(\frac{\pi}{4}ight) ight)$ $A = \left( -(-\frac{1}{\sqrt{2}}) - (-\frac{1}{\sqrt{2}}) ight) - \left( -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} ight)$ $A = \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} ight) - \left( -\frac{2}{\sqrt{2}} ight) = \frac{2}{\sqrt{2}} + \frac{2}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$ Now,$A^{4} = (2\sqrt{2})^{4} = 2^{4} 	imes (\sqrt{2})^{4} = 16 	imes 4 = 64$.

The graphs of sine and cosine functions intersect each other at a number of points,and between two consecutive points of intersection,the two graphs enclose the same area $A$. Then $A^{4}$ is equal to ............

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