Question

find the equations of the tangent lines to the curve y = sin x at x = -π, -\frac{π}{2}, and 0. graph the curve over the interval -\frac{3π}{2},2π together with its tangent lines. label the curve and each tangent line.
what is the equation of the tangent line (i) to the curve at x = -π?
y = -x - π
what is the equation of the tangent line (ii) to the curve at x = -\frac{π}{2}?

Explanation:

Step1: Recall the formula for the tangent - line

The equation of the tangent line to the curve $y = f(x)$ at the point $(x_0,y_0)$ is given by $y - y_0=f^{\prime}(x_0)(x - x_0)$, where $f^{\prime}(x)$ is the derivative of $f(x)$. First, find the derivative of $y = \sin x$. The derivative of $y=\sin x$ is $y^{\prime}=\cos x$.

Step2: Find $y_0$ and $f^{\prime}(x_0)$ for $x_0=-\frac{\pi}{2}$

When $x_0 =-\frac{\pi}{2}$, $y_0=\sin(-\frac{\pi}{2})=-1$. And $f^{\prime}(x)=\cos x$, so $f^{\prime}(-\frac{\pi}{2})=\cos(-\frac{\pi}{2}) = 0$.

Step3: Substitute into the tangent - line formula

Substitute $x_0 =-\frac{\pi}{2}$, $y_0=-1$, and $f^{\prime}(-\frac{\pi}{2}) = 0$ into the equation $y - y_0=f^{\prime}(x_0)(x - x_0)$. We get $y-(-1)=0(x+\frac{\pi}{2})$, which simplifies to $y=-1$.

Answer:

$y = - 1$