The graphs of sine and cosine functions inter | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 ............

Similar Questions

The area of the region bounded by the curve $y=4x^{3}-6x^{2}+4x+1$ and the lines $x=1, x=5$ and the $x$-axis is

Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$.

The area bounded by the curve $y = x \sin x$ and the $x$-axis between $x = 0$ and $x = 2\pi$ is:

If the area bounded by the curve $x^2 = by$ and the lines $y = 1, y = 4$ in the first quadrant is $28$ sq. units, then the value of $b$ is...

Find the area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cut off by the line $x=\frac{a}{\sqrt{2}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module