The equation of the tangent to the curve y = | vedclass

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(B) Given the curve $y = \sin x$. First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = \cos x$. Now,evaluate the slope at

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$x + y = \pi$

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(B) Given the curve $y = \sin x$. First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = \cos x$. Now,evaluate the slope at the point $(\pi, 0)$: $m = \left( \frac{dy}{dx} \right)_{(\pi, 0)} = \cos(\pi) = -1$. The equation of the tangent line passing through $(x_1, y_1) = (\pi, 0)$ with slope $m = -1$ is given by $y - y_1 = m(x - x_1)$. Substituting the values: $y - 0 = -1(x - \pi)$. This simplifies to $y = -x + \pi$,which can be rewritten as $x + y = \pi$.

The equation of the tangent to the curve $y = \sin x$ at the point $(\pi, 0)$ is $......$.

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