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

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(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

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

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(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

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