Question

Give any values in the itnerval [0,2pi] for which the given function has a horizontal tangent line.

y= cosx / cscx

Answer 1

There are lots of hard ways and one relatively easy way. How's your trigonometry?

Answer 2

Well what I did was I first found the derivative of the function, ended up being -tanx+cotx / csc^2(x) and set that equal to zero

Answer 3

So basically I set -tanx+cotx = 0 but then that's where I have trouble.

Answer 4

Yeah, that's horrible. Let's do something else.

csc(x) = 1/sin(x). Substitute that.

Answer 5

So then we get y = cosxsinx , the derivative being y= -sin^2x + cos^2x

Answer 6

That is another way to go.

\(\sin(2x) = 2\sin(x)\cos(x)\) - Substitute that.

Answer 7

Where are you getting the sin2x from?

Answer 8

This is why as asked about your trig skills. This is an important identity.

\(\sin(x)\cos(x) = ½(2\sin(x)\cos(x)) = ½\sin(2x)\)

Answer 9

I see. I'm not that advanced into trig yet. The process I'm supposed to be using is finding the derivative of the function and solving the problem by setting the derivative equal to zero.

Answer 10

What advanced. Trigonometry is a prerequisite to calculus. You should have these identities will in hand before you ever hear the word "derivative".

You also had a chance to see it back here...

With \(y = \sin(x)\cos(x)\), you correctly found the derivative \(y' = -\sin^{2}(x) + \cos^{2}(x)\). You absolutely should have enough trig to recognize this as \(y' = \cos(2x)\)

Anyway, what do you get for the derivative from \(y = ½\sin(2x)\)?

Answer 11

y'=cos2x

Answer 12

And then the points would be pi/4, 3pi/4, 5pi/4, 7pi/4

Answer 13

Isn't it awesome that we got there two different ways?!

Note: You could have gotten there the way you started out. Always be on the lookout for ways to improve (simplify) you life. This reduces errors if nothing else.

Brush up on your trigonometry!