Answer:
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.
Take the derivative of both sides with respect to x:
![\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5By%5D%3D%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Csin%28x%29%5Ccos%28x%29%5D)
We need to use the product rule:
So, differentiate:
![y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]](https://tex.z-dn.net/?f=y%27%3D%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Csin%28x%29%5D%5Ccos%28x%29%2B%5Csin%28x%29%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Ccos%28x%29%5D)
Evaluate:
Simplify:
Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':
Now, let's solve for x. First, we can use the difference of two squares to obtain:
Zero Product Property:
Solve for each case.
Case 1:
Add sin(x) to both sides:
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:
Subtract sine from both sides:
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:
And we're done!
Edit: Small Mistake :)
