Respuesta :
A trig identity is asinucosu=a/2sin(2u)So you can write your equation asy=sin(x)cos(x)=1/2sin(2x)Use the crain rule herey′=d/dx1/2sin(2x)=1/2cos(2x)d/dx2x=cos(2x)The curve will have horizontal tangents when y' = 0.y′=0=cos(2x)On the interval [-pi, pi], solution to that isx=±π4,±3π4
Answer:
[tex]x=\dfrac{\pi}{4},-\dfrac{\pi}{4},\dfrac{3\pi}{4},-\dfrac{\pi}{4}[/tex]
Step-by-step explanation:
Function: y= sin(x) cos(x)
To find the value of x where tangent line is horizontal on the interval [−π, π]
Slope = y'
[tex]y=\sin x\cos x[/tex]
derivative of y
[tex]y'=\cos^2x-\sin^2x[/tex]
For horizontal tangent line, slope must be 0
[tex]\cos^2x-\sin^2x=0[/tex]
[tex]\tan^2x=1[/tex]
[tex]\tan x=\pm 1[/tex]
[tex]x=\dfrac{\pi}{4},-\dfrac{\pi}{4},\dfrac{3\pi}{4},-\dfrac{\pi}{4}[/tex]
Horizontal tangents are,
[tex]y=\dfrac{1}{2}\ and \ y=-\dfrac{1}{2}[/tex]