The angle between the curves y=sin 2x and y=c | vedclass

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(B) Given equations of the curves are: $y = \sin 2x$ $(i)$ $y = \cos 2x$ (ii) To find the point of intersection,we set the equations equal: $\sin 2x =

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

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(B) Given equations of the curves are: $y = \sin 2x$ $(i)$ $y = \cos 2x$ (ii) To find the point of intersection,we set the equations equal: $\sin 2x = \cos 2x$ $	an 2x = 1 = 	an \frac{\pi}{4}$ $2x = \frac{\pi}{4} \Rightarrow x = \frac{\pi}{8}$ When $x = \frac{\pi}{8}$,$y = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$. So,the point of intersection is $(\frac{\pi}{8}, \frac{1}{\sqrt{2}})$. Now,find the slopes $m_1$ and $m_2$ at this point: $m_1 = \frac{dy}{dx} (\sin 2x) = 2 \cos 2x$. At $x = \frac{\pi}{8}$,$m_1 = 2 \cos \frac{\pi}{4} = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}$. $m_2 = \frac{dy}{dx} (\cos 2x) = -2 \sin 2x$. At $x = \frac{\pi}{8}$,$m_2 = -2 \sin \frac{\pi}{4} = -2 \cdot \frac{1}{\sqrt{2}} = -\sqrt{2}$. The angle $	heta$ between the curves is given by: $	an 	heta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$ $	an 	heta = |\frac{-\sqrt{2} - \sqrt{2}}{1 + (\sqrt{2})(-\sqrt{2})}| = |\frac{-2\sqrt{2}}{1 - 2}| = |\frac{-2\sqrt{2}}{-1}| = 2\sqrt{2}$. Therefore,$	heta = 	an^{-1}(2\sqrt{2})$.

The angle between the curves $y=\sin 2x$ and $y=\cos 2x$ is

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