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 the curves $y=\sin x$ and $y=\cos x$. To find the intersection point, set $\sin x = \cos x$, which implies $	an x = 1$. Thus, $x = \frac{\p

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the curves $y=\sin x$ and $y=\cos x$. To find the intersection point, set $\sin x = \cos x$, which implies $	an x = 1$. Thus, $x = \frac{\pi}{4}$. Now, find the slopes of the tangents at $x = \frac{\pi}{4}$. For $y = \sin x$, the slope $m_1 = \frac{dy}{dx} = \cos x$. At $x = \frac{\pi}{4}$, $m_1 = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. For $y = \cos x$, the slope $m_2 = \frac{dy}{dx} = -\sin x$. 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 the tangent to the curve $y=x^{3}-x^{2}+x$ at the point $(a, b)$ is also tangent to the curve $y=5x^{2}+2x-25$ at the point $(2, -1)$,then $|2a+9b|$ is equal to $........$

If the angle between the curves $y^2=4x$ and $y=e^{-x/2}$ is $\theta$,then $\operatorname{cosec}^2(\theta/2)=$

If the tangent to the curve $y = f(x) = x \log_{e} x$ $(x > 0)$ at a point $(c, f(c))$ is parallel to the line segment joining the points $(1, 0)$ and $(e, e)$,then $c$ is equal to:

The angle of intersection of the curves $y^2 = \frac{2x}{\pi}$ and $y = \sin x$ is

If a variable tangent to the curve $x^2y = c^3$ makes intercepts $a$ and $b$ on the $x$-axis and $y$-axis respectively,then the value of $a^2b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module