What is the angle between the curves y = sin | vedclass

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(C) 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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(C) 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 $x = \pi /4$.At $x = \pi /4$,$y = \sin(\pi /4) = 1/\sqrt{2}$. The point of intersection is $(\pi /4, 1/\sqrt{2})$.Now,find the slopes of the tangents at this point:For $y = \sin x$,$dy/dx = \cos x$. At $x = \pi /4$,$m_1 = \cos(\pi /4) = 1/\sqrt{2}$.For $y = \cos x$,$dy/dx = -\sin x$. At $x = \pi /4$,$m_2 = -\sin(\pi /4) = -1/\sqrt{2}$.The angle $	heta$ between the curves is given by $	an 	heta = |(m_1 - m_2) / (1 + m_1 m_2)|$.$	an 	heta = |(1/\sqrt{2} - (-1/\sqrt{2})) / (1 + (1/\sqrt{2})(-1/\sqrt{2}))|$$	an 	heta = |(2/\sqrt{2}) / (1 - 1/2)| = |\sqrt{2} / (1/2)| = 2\sqrt{2}$.Therefore,$	heta = 	an^{-1}(2\sqrt{2})$.

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

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