Respuesta :
Answer:
[tex]\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}[/tex]
Step-by-step explanation:
We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.
Since the tangent line is horizontal, this means that our derivative at those points are 0.
So, first, let's find the derivative of our function.
[tex]y=\sin(x)\cos(x)[/tex]
Take the derivative of both sides with respect to x:
[tex]\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)][/tex]
We need to use the product rule:
[tex](uv)'=u'v+uv'[/tex]
So, differentiate:
[tex]y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)][/tex]
Evaluate:
[tex]y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))[/tex]
Simplify:
[tex]y'=\cos^2(x)-\sin^2(x)[/tex]
Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':
[tex]0=\cos^2(x)-\sin^2(x)[/tex]
Now, let's solve for x. First, we can use the difference of two squares to obtain:
[tex]0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))[/tex]
Zero Product Property:
[tex]0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)[/tex]
Solve for each case.
Case 1:
[tex]0=\cos(x)-\sin(x)[/tex]
Add sin(x) to both sides:
[tex]\cos(x)=\sin(x)[/tex]
To solve this, we can use the unit circle.
Recall at what points cosine equals sine.
This only happens twice: at π/4 (45°) and at 5π/4 (225°).
At both of these points, both cosine and sine equals √2/2 and -√2/2.
And between the intervals 0 and 2π, these are the only two times that happens.
Case II:
We have:
[tex]0=\cos(x)+\sin(x)[/tex]
Subtract sine from both sides:
[tex]\cos(x)=-\sin(x)[/tex]
Again, we can use the unit circle. Recall when cosine is the opposite of sine.
Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.
At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).
At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.
Therefore, our solution set is:
[tex]\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}[/tex]
And we're done!
Edit: Small Mistake :)