Quick trigonometric function problem

[Elpinetos]

Homework Statement

Sketch the curve of y=sin(x)+sin(2x)

I got all the derivatives, and the roots N(k*∏|0), but now I'm stuck with the maxima/minima

Homework Equations

First derivative y'=cos(x)+2cos(2x)

The Attempt at a Solution

I set cos(x)+2cos(2x)=0 but now I'm stuck. I just don't know how to solve this oO
Graph-plotters and the solution book spit out -2,57 -0,94 and so on, but how do I get this solution?

 

[Dick]

Homework Statement

Sketch the curve of y=sin(x)+sin(2x)

I got all the derivatives, and the roots N(k*∏|0), but now I'm stuck with the maxima/minima

Homework Equations

First derivative y'=cos(x)+2cos(2x)

The Attempt at a Solution

I set cos(x)+2cos(2x)=0 but now I'm stuck. I just don't know how to solve this oO
Graph-plotters and the solution book spit out -2,57 -0,94 and so on, but how do I get this solution?

Use a double angle formula to express cos(2x) in terms of cos(x).

 

[Elpinetos]

So cos(2x) = 2cos^2(x)-1?

When I calculate that through I get cos(x)+4cos(x) = 1

What now? Sorry, I'm really bad at trigonometric identities :(

 

[Mark44]

I got all the derivatives, and the roots N(k*∏|0)

What does ∏|0 mean?

 

[Dick]

So cos(2x) = 2cos^2(x)-1?

When I calculate that through I get cos(x)+4cos(x) = 1

What now? Sorry, I'm really bad at trigonometric identities :(

What happened to the cos(x)^2? You should get a quadratic equation in cos(x). Use that to find the possible values of cos(x).

 

[Student100]

So cos(2x) = 2cos^2(x)-1?

When I calculate that through I get cos(x)+4cos(x) = 1

What now? Sorry, I'm really bad at trigonometric identities :(

You made a mistake with your multiplication here, look at what the 2 should be multiplying, you also lost a power.

 

[Elpinetos]

Sorry, I made a typo, I meant I get cos(x)+4*cos^2(x) = 1
I don't know what to do with this equation :(

@Mark: It means x=k*pi and y=0

 

[Student100]

Sorry, I made a typo, I meant I get cos(x)+4*cos^2(x) = 1
I don't know what to do with this equation :(

@Mark: It means x=k*pi and y=0

Still wrong. 2[cos(2x)]

 

[Elpinetos]

How can I change cos(x)+2cos(2x) to 2[cos(2x)]? oO
I don't get it :(

 

[Student100]

How can I change cos(x)+2cos(2x) to 2[cos(2x)]? oO
I don't get it :(

No no, take a realllll close look at your algebra after you switch the 2[cos(2x)] with 2[2cos²(x)-1].

Since you want to find the roots, what kind of equation does this now produce? What can you do with this equation to make your life easier? What has to be done to the equation to find the roots? Hint: Don't move the constant like you've been doing and maybe it'll all make more sense.

 

[Elpinetos]

Oh okay, so I get cos(x)+4cos^2(x)-2 = 0

And now? Oo

 

[Student100]

Oh okay, so I get cos(x)+4cos^2(x)-2 = 0

And now? Oo

Haha good job.

Now what type of equation is that?

 

[Elpinetos]

It's a quadratic equation, but can I use the formula to solve those for cos(x) too?
Sorry, I'm so rusty in algebra and math, haven't been doing any for 3 years now *sigh*

 

[Student100]

It's a quadratic equation, but can I use the formula to solve those for cos(x) too?
Sorry, I'm so rusty in algebra and math, haven't been doing any for 3 years now *sigh*

Don't feel bad, I'm still pretty bad at math.

You certainly can. You could use u substitution to put it in terms of u if it's confusing for you:
u = cos(x)
4u² + u -2.

Or you could also just plug it into the quadratic formula as is.

You just have to remember that cos(x) oscillates, so you have to put the anwser in terms of 2\pin. Unless your question states a closed interval. Also, you have to use the inverse trig function.

You can brush up on algebra here: http://www.purplemath.com/modules/index.htm

 

[Elpinetos]

So the answer would be cos(x) = -1(+-)sqrt(33)/8 ?

This gives me the solutions from above :) Thank you :)

How do I convert this into 2pi*n though?

 

[Student100]

So the answer would be cos(x) = -1(+-)sqrt(33)/8 ?

This gives me the solutions from above :) Thank you :)

How do I convert this into 2pi*n though?

Does the problem state an interval to draw the original function on? It's probably assumed you'd only have to draw from (-2\pi , 2\pi) So it isn't important for this problem most likely, but it's nice to know:

2\pi n \pm \arccos( \frac{-1 \pm \sqrt(33)} {8})

This would give you the relative extrema for as long as you'd want to look at the original function.

 

[Elpinetos]

Thank you :)

 

[Mark44]

So the answer would be cos(x) = -1(+-)sqrt(33)/8 ?

What you wrote is something different from what you mean.
This is what you wrote:
$$cos(x) = - 1 \pm \frac{\sqrt{33}}{8}$$

What you meant was this:
$$cos(x) = \frac{-1 \pm \sqrt{33}}{8}$$

If you don't use LaTeX to write this, you need parentheses around the entire numerator, like so:
cos(x) = (-1 ± √33)/8

Apparently what you wrote confused Student100, who is not working with the correct value.

This gives me the solutions from above :) Thank you :)

How do I convert this into 2pi*n though?

Does the problem state an interval to draw the original function on? It's probably assumed you'd only have to draw from [-2\pi , 2\pi] So it isn't important for this problem most likely, but it's nice to know:

2\pi n \pm \arccos( \frac{-1 \pm \sqrt(33)} {8})

Student100, I don't believe this is correct. It looks like you are saying cos^-1(A + B) = cos^-1(A) + cos^-1(B). That is NOT true.

 

[Student100]

Student100, I don't believe this is correct. It looks like you are saying cos^-1(A + B) = cos^-1(A) + cos^-1(B). That is NOT true.

Your right Mark, I was trying to save time by putting everything in one expression. I could see how that might be confusing.

Quick question:

Which root is outside the domain? I thought I did a quick check on the calculator and both were valid. I may be wrong.

 

[Mark44]

Your right Mark, I was trying to save time by putting everything in one expression. I could see how that might be confusing.

Not just confusing, but wrong, if my understanding of what you did is correct. Is this what you did? cos^-1(A + B) = cos^-1(A) + cos^-1(B)?

Quick question:

Which root is outside the domain? I thought I did a quick check on the calculator and both were valid. I may be wrong.

Neither is outside the domain. I got fixated on the OP's solution of -1 ± (√33)/8 and this threw me off. The correctly written values, (-1 ± √33)/8, are both within the interval [-1, 1].

 

[Student100]

Not just confusing, but wrong, if my understanding of what you did is correct. Is this what you did? cos^-1(A + B) = cos^-1(A) + cos^-1(B)?

2\pi n + \arccos( \frac{-1 - \sqrt{33}} {8})

2\pi n - \arccos( \frac{-1 - \sqrt{33}} {8})

2\pi n + \arccos( \frac{\sqrt{33}-1} {8})

2\pi n - \arccos( \frac{\sqrt{33}-1} {8})

Is what I meant, which can be obtained from the first expression I posted. Please correct me if I'm wrong.