Find the equation of the tangent line to the graph of the function $f(x)=x(cosx)^2$ when $x=\pi$

Question

$f(x)=x(cosx)^2$

$x=\pi$

Need help solving the question, don't know where to start.

Answer 1

Hint:

you want the line that passes thorough the point $P=(\pi,f(\pi))$, so that it has equation $y-f(\pi)=m(x-\pi)$,

and the slope must be $m=f'(\pi)$.

Answer 2

First $f(\pi) = \pi (\cos \pi)^2 = \pi$ Now $f'(x) = \cos^2 x - 2x \cos x \sin x = \cos^2 x -x \sin 2x$ So at $f'(\pi) = 1$ So we obtain: $y=x+b$, now we plug in the value at the point $(\pi ,\pi)$ to obtain: $\pi = \pi +b$. Hence $b=0$ So we get the line: $y=x$