Question

find the equations of the tangent lines to the curve y = cos x at x = -\frac{\pi}{2},\frac{\pi}{2}, and \pi. graph the curve over the interval -\frac{3\pi}{2},2\pi 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 = -\frac{\pi}{2}? y=x+\frac{\pi}{2} what is the equation of the tangent line (ii) to the curve at x = \frac{\pi}{2}? y=-x+\frac{\pi}{2} what is the equation of the tangent line (iii) to the curve at x = \pi?

Explanation:

Step1: Recall the formula for the tangent - line

The equation of a 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 $y = \cos x$. The derivative $y^{\prime}=-\sin x$.

Step2: Find the tangent - line at $x =-\frac{\pi}{2}$

When $x =-\frac{\pi}{2}$, $y=\cos(-\frac{\pi}{2}) = 0$. And $y^{\prime}=-\sin(-\frac{\pi}{2})=1$. Using the point - slope form $y - y_0=f^{\prime}(x_0)(x - x_0)$, with $x_0 =-\frac{\pi}{2}$, $y_0 = 0$ and $f^{\prime}(x_0)=1$, we have $y-0 = 1\times(x+\frac{\pi}{2})$, so $y=x+\frac{\pi}{2}$.

Step3: Find the tangent - line at $x=\frac{\pi}{2}$

When $x=\frac{\pi}{2}$, $y=\cos(\frac{\pi}{2}) = 0$. And $y^{\prime}=-\sin(\frac{\pi}{2})=-1$. Using the point - slope form $y - y_0=f^{\prime}(x_0)(x - x_0)$, with $x_0=\frac{\pi}{2}$, $y_0 = 0$ and $f^{\prime}(x_0)=-1$, we get $y-0=-1\times(x - \frac{\pi}{2})$, so $y=-x+\frac{\pi}{2}$.

Step4: Find the tangent - line at $x = \pi$

When $x=\pi$, $y=\cos(\pi)=-1$. And $y^{\prime}=-\sin(\pi)=0$. Using the point - slope form $y - y_0=f^{\prime}(x_0)(x - x_0)$, with $x_0=\pi$, $y_0=-1$ and $f^{\prime}(x_0)=0$, we have $y+1=0\times(x - \pi)$, so $y=-1$.

Answer:

The equation of the tangent line (III) to the curve at $x = \pi$ is $y=-1$.