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

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(C) The curves $y = \sin x$ and $y = \cos x$ intersect where $\sin x = \cos x$,which implies $	an x = 1$. Since $0 < x < \frac{\pi}{2}$,the intersect

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

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(C) The curves $y = \sin x$ and $y = \cos x$ intersect where $\sin x = \cos x$,which implies $	an x = 1$. Since $0 Let $m_1$ and $m_2$ be the slopes of the tangents to the curves at $x = \frac{\pi}{4}$.For $y = \sin x$,$\frac{dy}{dx} = \cos x$. Thus,$m_1 = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.For $y = \cos x$,$\frac{dy}{dx} = -\sin x$. Thus,$m_2 = -\sin(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.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 $y = \cos x$,$0 < x < \frac{\pi}{2}$,is

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