Verified Answer :

y = sin x + cos x 

the tangent will be horizontal when the derivative is 0 

y ' = cos x - sin x 
0 = cos x - sin x 
sin x = cos x 
tan x = 1 
this happens when x = pi/4 or 5pi/4 

y(pi/4) = sqrt(2) / 2 + sqrt(2) / 2 = sqrt(2) 
y(5pi/4) = -sqrt(2) / 2 + -sqrt(2) / 2 = -sqrt(2) 

the two points are: (pi/4 , srqt(2)) and (5pi/4 , -srqt(2))

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The required values are,

\(x=\frac{\pi}{4},\frac{3\pi}{4} ,\frac{5\pi}{4}\)

Given function is,

\(y=sinx cosx\)

Tangent to the graph:

The tangent to the graph  \(y=f(x)\) is horizontal at \(x=a\) if \(\frac{dy}{dx} =0\) at \(x=a\)

We can write the given equation as,

\(y=sinxcos\\y=\frac{1}{2}(sinxcosx)\\ y=\frac{1}{2}sin2x\)

Now, differentiating the given function with respect to \(x\).

\(\frac{dy}{dx} =\frac{1}{2} (2cos2x)\\\frac{dy}{dx} =cos2x\\\frac{dy}{dx} =0\\cos2x=0\\x=\frac{\pi}{4} ,\frac{3\pi}{4} ,\frac{5\pi}{4} ,\frac{7\pi}{4} ,\frac{9\pi}{4} \\y=\frac{\pi}{4} ,\frac{3\pi}{4} ,\frac{5\pi}{4}\)

Learn more about a tangent to the graph:

https://brainly.com/question/26453506

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