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

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(A) To find the angle between the curves $y = \sin x$ and $y = \cos x$,we first find their point of intersection by setting $\sin x = \cos x$.This giv

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

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(A) To find the angle between the curves $y = \sin x$ and $y = \cos x$,we first find their point of intersection by setting $\sin x = \cos x$.This gives $	an x = 1$,so $x = \pi/4$.Let $m_1$ be the slope of the first curve $y = \sin x$. Then $m_1 = \frac{dy}{dx} = \cos x$. At $x = \pi/4$,$m_1 = \cos(\pi/4) = \frac{1}{\sqrt{2}}$.Let $m_2$ be the slope of the second curve $y = \cos x$. Then $m_2 = \frac{dy}{dx} = -\sin x$. At $x = \pi/4$,$m_2 = -\sin(\pi/4) = -\frac{1}{\sqrt{2}}$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values: $	an 	heta = \left| \frac{\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})}{1 + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})} ight| = \left| \frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}} ight| = \left| \frac{\sqrt{2}}{\frac{1}{2}} ight| = 2\sqrt{2}$.Thus,$	heta = 	an^{-1}(2\sqrt{2})$.

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

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