Trig question

Original post by Plato's Trousers

...

One could note that 3x = 1/2 (x + 5x), and that might suggest something.

Does seem a bit fiddly, so may not be the best method (or perhaps that's just how I'm seeing it later on)

Reply 2

SecondHand

Express cos(3x) and cos(5x) in terms of cos(x) so you'll have a quintic in cos(x) and show from there? It's a long way but is a possibility.

Reply 3

skibur

12

could you not use complex numbers? may be quicker than expressing cos3x and cos5x in terms of cosx

Reply 4

TheGrinningSkull

19

Original post by Plato's Trousers

How would I show that this:

\cos (x) + \cos(3x) + \cos (5x) = -\frac{1}{2}

has exactly two solutions in the interval

[ 0,\frac{\pi}{2}]

?

So, I can see that the function is continuous on the interval and also that it has a min value of -3 and a max value of +3, but how do I show it has two zeros? (and it's not a graph sketching question)

I know that

\cos(3x) = 4\cos^3(x)-3\cos(x)

if that helps :L

Other than that, trial and error?

Reply 5

Maybe use sum-to-product?

\cos(x)+\cos(5x)

Spoiler

2\cos(3x)\cos(2x)

Anyone know what to do with the random -1/2? Move it to the LHS and convert it into cos?

(edited 11 years ago)

Reply 6

BabyMaths

The quintic can be factorised into a quadratic and a cubic.

Eventually, with some assistance, I arrived at x=pi/5 and x = 2pi/7.

Reply 7

Plato's Trousers

OP

hmm... I think it's something to do with the fact that the LHS sum has a min of -3 and a max of +3, and because it's continuous it must therefore go through the value -1/2. But I need to show it does that twice...

Reply 8

BabyMaths

Original post by Plato's Trousers

I think it's something to do with the fact that the LHS sum has a min of -3 and a max of +3,

Not on that interval.

Reply 9

Are you sure it's not a sketch/algebra question? Showing it crosses y=-1/2 at least twice should be simple, but wouldn't showing it crosses less than thrice require looking at every single point?

Maybe use calculus to show that it only has 1 minimum below -1/2?

(edited 11 years ago)

Rolle's theorem?

Original post by Plato's Trousers

hmm... I think it's something to do with the fact that the LHS sum has a min of -3 and a max of +3, and because it's continuous it must therefore go through the value -1/2. But I need to show it does that twice...

Well, it certainly obtains +3, but does it obtain -3 on that somewhat restricted domain?

Reply 12

Original post by ghostwalker

Well, it certainly obtains +3, but does it obtain -3 on that somewhat restricted domain?

You're right, the minimum in the domain is

-\sqrt{2}

. Nice catch!

EDIT: Is there a way to reason that there's a local minimum at pi/4 without calculus? Does it have to do with matching up the periods?

(edited 11 years ago)

Reply 13

Intriguing Alias

16

Original post by aznkid66

You're right, the minimum in the domain is

-\sqrt{2}

. Nice catch!

EDIT: Is there a way to reason that there's a local minimum at pi/4 without calculus? Does it have to do with matching up the periods?

I don't think it's as low as -root(2)

Can you give the value of x that would give that?

Reply 14

raheem94

13

Original post by hassi94

I don't think it's as low as -root(2)

Can you give the value of x that would give that?

You are right.

See the graph.

Graph

Reply 15

Original post by hassi94

I don't think it's as low as -root(2)

Can you give the value of x that would give that?

Ugh, forgot to divide by 2. Sorry 'bout that.
Does -1/sqrt(2) work?

EDIT: Hmm, is there no way to factorise -sinx-3sin3x-5sin5x?

(edited 11 years ago)

Reply 16

Intriguing Alias

16

Original post by aznkid66

Ugh, forgot to divide by 2. Sorry 'bout that.
Does -1/sqrt(2) work?

EDIT: Hmm, is there no way to factorise -sinx-3sin3x-5sin5x?

According to the graph, the minimum is pretty damn close to -1/root(2) but it isn't that. It's 0.01 off or so.

Reply 17

ghostwalker

Original post by aznkid66

You're right, the minimum in the domain is

-\sqrt{2}

. Nice catch!

EDIT: Is there a way to reason that there's a local minimum at pi/4 without calculus? Does it have to do with matching up the periods?

If you can show there is a local minimum at x=pi/4, you've made an error. It's close, but not quite.

What I was originally hinting at, and you spelled out in your first post is the way to go I think; assuming you're not using calculus or other fancy techniques.

What ever the value of the minimum, it's only important that it's < -1/2.

Reply 18

Plato's Trousers

OP

Original post by ghostwalker

If you can show there is a local minimum at x=pi/4, you've made an error. It's close, but not quite.

What I was originally hinting at, and you spelled out in your first post is the way to go I think; assuming you're not using calculus or other fancy techniques.

What ever the value of the minimum, it's only important that it's < -1/2.

I don't need to find the solutions. I need to show that there are two

ghostwalker - yes, without using calculus, just by using analysis methods

Reply 19

ghostwalker