# Trigonometric derivation question

1. Oct 29, 2009

### iamsmooth

1. The problem statement, all variables and given/known data
Determine all points on the function $y=\sin2x-2sinx$ where the tangent is parallel to the x-axis.

2. Relevant equations
f'(sin(x)) = cos(x)

chain rule:
f'(f(g(x)) = g'f + f'g

3. The attempt at a solution
$$y=sin2x-2sinx$$

$$y\prime=cos(2x)(2)-[2cosx+0sinx]$$

$$y\prime=cos(2x)(2)-2cosx$$

If I multiply 2 and cos(2x), do I get 2cos(2x) or do I get 2cos(4x)?

Is there a way to simplify it further by subtracting the -2cosx?

Do I use the sum and difference identities? My trig is still pretty rusty :(

In a similar question without trig, we'd derive the equation, and factor it to find out the points of x; but I'm not quite sure how to do this :(

Thanks for the help.

Last edited: Oct 30, 2009
2. Oct 29, 2009

### lanedance

2*cos(2x) = 2cos(2x)

what is y'(x) when y' (the tangent) is parallel to the x axis?

3. Oct 29, 2009

### iamsmooth

y'(x) is 0, so

0 = 2cos(2x) - 2cos(x)

Messing with the equation, I can get:

2cos(2x)=2cos(x)

cos(2x) = 2cos(x)/2

cos(2x) = cos(x)

And I'm stumped :(

Basically all I know is that:

y' = 2cos(2x) - 2cos(x) is the slope of the tangent at the point where y' = 0.

4. Oct 29, 2009

### lanedance

when does cos(2x) = cos(x) happen, think about the peridocity of the cos function (2pi)

5. Oct 30, 2009

### iamsmooth

I guess x would have to be 0, for them to be the same. Which means anywhere on the unit circle where x is 0? There's 3 points that satisfy this condition:

$$(2k\pi+\frac{2\pi}{3},-\sqrt{3}), (2k\pi+\frac{4\pi}{3},\sqrt{3}),(2k\pi,0)$$

I don't get how you can figure out any of these points.

I'm looking at the unit circle and none of these coordinates seem obvious as a solution to me.

Somehow I get those x's as mentioned, and when plugging them back in to the original formula y = sin2x-2sinx, you'll get the y coordinates. None of the x's in the answers mentioned are 0s so I'm even more confused now.

6. Oct 30, 2009

### Staff: Mentor

This isn't help for this particular problem, but some guidance on using the notation correctly. Writing f'(sin(x)) doesn't mean what you think it means. IOW, it does not mean "take the derivative of sin(x)." Instead it means evaluate the derivative of some unspecified function f at sin(x).

Better would be d/dx(sin(x)) = cos(x). Not quite as good, IMO, would be (sin(x))' = cos(x).

For the chain rule, you have actually written the product rule, not the chain rule. Again, the left-most f' does not mean "take the derivative of ..." The operator d/dx does mean take the derivative of whatever is to its right. f' already is the derivative of a function f. You should write the product rule as d/dx(fg) = g'f + f'g.

The chain rule is like this: d/dx(f(g(x))) = f'(g(x))*g'(x)

7. Oct 30, 2009

### iamsmooth

Thanks Mark. Appreciate all the help you (and everyone else) has given me in my short time on these forums. For once, I feel like I'm starting to understand the material--and I feel like I'm pretty prepared for my midterm tomorrow :D

Also, what's great is you guys don't just show how to do something, but explain in depth why we do something. Most of the other students in class just seem to do the problems like robots and not understand why they're doing something--which I must confess I still do--but nevertheless, things are starting to click!

Once again, thanks!