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. what is the equation of the tangent line (l) to the curve at x = -\frac{\pi}{2}?

Explanation:

Step1: Find the derivative of $y = \cos x$

The derivative of $y=\cos x$ is $y'=-\sin x$.

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

Substitute $x =-\frac{\pi}{2}$ into $y'$. So, $m =-\sin(-\frac{\pi}{2})= 1$.

Step3: Find the y - value at $x =-\frac{\pi}{2}$

Substitute $x =-\frac{\pi}{2}$ into $y=\cos x$. We get $y=\cos(-\frac{\pi}{2}) = 0$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-\frac{\pi}{2},0)$ and $m = 1$.
$y-0 = 1\times(x+\frac{\pi}{2})$.

Answer:

$y=x+\frac{\pi}{2}$