Question

CALCULUS - Find the values of in the interval (0,2pi) where the tangent line to the graph of y = sinxcosx is

horizontal. Please use the derivative. Thank you!

Answer 1

`\{(\pi)/(4), (3\pi)/(4),(5\pi)/(4),(7\pi)/(4)\}`

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.

`y=\sin(x)\cos(x)`

Take the derivative of both sides with respect to x:

`(d)/(dx)[y]=(d)/(dx)[\sin(x)\cos(x)]`

We need to use the product rule:

`(uv)'=u'v+uv'`

So, differentiate:

`y'=(d)/(dx)[\sin(x)]\cos(x)+\sin(x)(d)/(dx)[\cos(x)]`

Evaluate:

`y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))`

Simplify:

`y'=\cos^2(x)-\sin^2(x)`

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

`0=\cos^2(x)-\sin^2(x)`

Now, let's solve for x. First, we can use the difference of two squares to obtain:

`0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))`

Zero Product Property:

`0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)`

Solve for each case.

Case 1:

`0=\cos(x)-\sin(x)`

Add sin(x) to both sides:

`\cos(x)=\sin(x)`

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:

`0=\cos(x)+\sin(x)`

Subtract sine from both sides:

`\cos(x)=-\sin(x)`

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:

`\{(\pi)/(4), (3\pi)/(4),(5\pi)/(4),(7\pi)/(4)\}`

And we're done!

Edit: Small Mistake :)