Answer
The equation of the tangent line is $y=-2x+\dfrac{2\pi+3\sqrt{3}}{3}$
Work Step by Step
$y=4\sin x\cos x$ $;$ $x=\dfrac{\pi}{3}$
$a)$
Evaluate the derivative of the function by using the product rule:
$y'=4[\sin x(\cos x)'+\cos x(\sin x)']=...$
Evaluate the indicated derivatives and simplify:
$...=4[\sin x(-\sin x)+\cos x(\cos x)]=4(\cos^{2}x-\sin^{2}x)$
Substitute $x=\dfrac{\pi}{3}$ in the derivative found and simplify to obtain the slope of the tangent line at the given point:
$m=4\Big[\cos^{2}\Big(\dfrac{\pi}{3}\Big)-\sin^{2}\Big(\dfrac{\pi}{3}\Big)\Big]=4\Big[\Big(\dfrac{1}{2}\Big)^{2}-\Big(\dfrac{\sqrt{3}}{2}\Big)^{2}\Big]=...$
$...=4\Big(\dfrac{1}{4}-\dfrac{3}{4}\Big)=1-3=-2$
Substitute $x=\dfrac{\pi}{3}$ in the original expression to obtain the $y$-coordinate of the point given:
$y$-coordinate $=4\sin\dfrac{\pi}{3}\cos\dfrac{\pi}{3}=4\Big(\dfrac{\sqrt{3}}{2}\Big)\Big(\dfrac{1}{2}\Big)=\sqrt{3}$
The point is $\Big(\dfrac{\pi}{3},\sqrt{3}\Big)$
The slope of the tangent line and a point through which it passes are now known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point:
$y-\sqrt{3}=-2\Big(x-\dfrac{\pi}{3}\Big)$
$y-\sqrt{3}=-2x+\dfrac{2\pi}{3}$
$y=-2x+\dfrac{2\pi}{3}+\sqrt{3}$
$y=-2x+\dfrac{2\pi+3\sqrt{3}}{3}$
The graph of the function and the tangent line is shown in the answer section.
