The angle between the curves y=sin x and y=co | vedclass

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

Question

(A) Given curves are $y=\sin x$ and $y=\cos x$. To find the intersection point,set $\sin x = \cos x$,which gives $	an x = 1$,so $x = \frac{\pi}{4}$.

Vedclass · Accepted Answer

$	an ^{-1}(2 \sqrt{2})$

Vedclass · Answer

(A) Given curves are $y=\sin x$ and $y=\cos x$. To find the intersection point,set $\sin x = \cos x$,which gives $	an x = 1$,so $x = \frac{\pi}{4}$. Now,find the slopes of the tangents at $x = \frac{\pi}{4}$. For $y = \sin x$,$m_1 = \frac{dy}{dx} = \cos x$. At $x = \frac{\pi}{4}$,$m_1 = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. For $y = \cos x$,$m_2 = \frac{dy}{dx} = -\sin x$. At $x = \frac{\pi}{4}$,$m_2 = -\sin(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$. The angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. Substituting the values,$	an 	heta = |\frac{\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})}{1 + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})}| = |\frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}}| = |\frac{\sqrt{2}}{\frac{1}{2}}| = 2\sqrt{2}$. Therefore,$	heta = 	an^{-1}(2\sqrt{2})$.

The angle between the curves $y=\sin x$ and $y=\cos x$ is

Similar Questions

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=2x+3$ and $g(x)=x^2+7$,then the values of $x$ such that $g(f(x))=8$ are

Which of the following are not electromagnetic waves?

$A$ diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from $V$ to $32V$,the efficiency of the engine is

Assertion $(A):$ Above Curie temperature, a ferromagnetic material becomes paramagnetic.

Reason $(R):$ When a magnetic material is heated to very high temperature, it loses its magnetic properties.

If the roots of the equation $x^2 - 8x + (a^2 - 6a) = 0$ are real,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module