Simultaneous trigonometry equation ? 3

[MRSSPOCK]

I don't even know what this type of equation is really called, so I'm having difficulty even searching for a solution.

I wonder can anyone show a step by step solution to this, or is it purely a case of solving it by iteration?

sin(x) = sin(2x+s)+1

cos(x) = cos(2x+s)

Find x and s to satisfy the two above statements.

I do believe there are at least two solutions to this. (+ve s and -ve s)

(I suppose in this case there are infinite solutions after each period of sin x, since 2x is a direct multiple of x, but in reality if we suppose 2x was replaced by 1.9278x or some other random value, then the solution is no longer going to be every period of sin x).

The solution has a real life application, but I didn't want to include my application lest it just caused more confusion.

Below is an illustration of a graphically guessed solution. (guessed value of s=1.3)

I'm trying to solve where the two waves are tangential.

Thanks

 

[snarkysparky]

try making these substitutions

sin(A+B) = sin(A)*cos(B)+sin(B)*cos(A)

cos(A+B) = cos(A)*cos(B)-sin(A)*sin(B)

http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html

 

[MRSSPOCK]

Thanks, but what's next?


sin(x) = (sin(2x)*cos(s)+sin(s)*cos(2x))+1


cos(x) = cos(2x)*cos(s)-sin(2x)*sin(s)

 

[MacGyverS2000]

If you're looking for tangent points, it seems like going back to basics is the best path.

If the first derivative of a function gives you the slope of the tangent, you want to determine where the first derivative of one equals the first derivative of the other. The first derivative of each is straightforward, so after that, it's plug and play.

Dan - Owner

http://www.Hi-TecDesigns.com

 

[MRSSPOCK]

@MacGyver

Thanks, but that is exactly what I have already done.

The first equation defines possible intersection points, and the second line is the derivatives of each half of the first statement, hence the two tangents are coincident.

 

[IRstuff]

Since you're using what appears to be Mathcad, why not use use a solve block?

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529

 

[MRSSPOCK]

I could also do it iteratively myself, but I want to learn how to do it mathematically rather than just use software, that is if there is a way other than iteration.

 

[MacGyverS2000]

Start off with something more simple, like s=0. If that's the case, then you have cos(x)=cos(2x). This can only be true when x=0°, right (and 360° increments, if my rusty memory serves)?

So what happens when s!=0?


Dan - Owner

http://www.Hi-TecDesigns.com

 

[MacGyverS2000]

Essentially, they are only equivalent once around the entire unit circle. That point is when x=2x+s, right? So, any time you have s=-x, the equations are equivalent. Solve for any angle x, which gives you a valid value of s. Bob's your uncle.

Dan - Owner

http://www.Hi-TecDesigns.com

 

[MRSSPOCK]

Sorry, you totally lost me with that one.

Can you please show it will real numbers to clarify.

Thanks

 

[MacGyverS2000]

Ignore s for the moment... you're left with cos(x)=cos(2x). The only time that is true over the unit circle is when x=0°, so we only have to worry about one solution, the direct one (no versions of x +/- π/2, etc.).

Bringing s back in, the only way for cos(x)=cos(2x+s) to work is for the "innards" to be equivalent, as well, i.e., x=2x+s. Solving for s, you get s=-x. No matter what value x is, setting s=-x will provide you with an equivalent value on both sides.

Let x=12°. Therefore, s must equal -12° if you want the two sides to be equivalent. When equivalent, the slopes of the signal at that point (x=12°) are also equivalent. Pick another value for x, same thing. If you plot x and s, they are merely negatives of each other(or out of phase by 180°, pick your viewpoint)...

Dan - Owner

http://www.Hi-TecDesigns.com

 

[MRSSPOCK]

Well, here are two manually adjusted graphs.

First thing, I was wrong to assume -ve s and +ve s would give the same results.

I had to do a fudge of s = -ve 1.3 or s = +ve 4.5

@MacGuyver You can see in both these solutions that x and s are not a common value of opposite signs.

In the upper solution, when x is about 3, s is 4.5

and in the lower solution, when x is about 0.1, s is -1.3

 

[MRSSPOCK]

This is how I have broken it into separate stages, but I still don't know how to calculate the true value of s.

The first stage shows the two superimposed sine waves.

Stage two shifts the higher frequency wave upwards by a value of 1, hence the lowest point is tangential with the horizontal axis.


Stage three then slides the higher frequency wave horizontally until it becomes tangent with the lower frequency wave.



Common in the three plots is sin (x)

Then the superimposed wave is:

Plot one has sin(2x+0)
Plot two has sin(2x+0)+1
Plot three has sin(2x-1.3)+1

 

[IRstuff]

I assume you tried the symbolic solver in Mathcad? I've tried it a couple of ways and there's symbolic answer from Mathcad, which means that it's more than likely that there is no symbolic solution.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529

 

[VEBill]

https://www.wolframalpha.com/

Enter: Find x and s where sin(x) = sin(2x+s)+1, cos(x) = cos(2x+s)

Click Here

Disclaimer: I'm not sure if it's interpreted it the way you want or not. Perhaps the input needs adjustment.

 

[IRstuff]

So that's how to do that.; thanks

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529

 

[MRSSPOCK]

hmmmmm ...

I still can't see any numbers that show s around -1.3 or 4.5 on the Wolfram page



Is there a fundamental flaw in my equation?

Am I correct in defining my first derivatives of

sin(x) and sin(2x+s)+1

as

cos(x) and cos(2x+s) ???

 

[MacGyverS2000]

I'm not seeing any mistakes in my math (though it's possible they exist)...

In your graph, however, I see crossings, not identical slopes (like at x=3)... not the same thing. It does not appear as if your "solution" is actually one. You cannot look at a graph like that and pick a solution by eye.

You're also graphing the original equations, not their derivatives. If you graph the derivatives, I think you'll find things are a bit different looking...

Dan - Owner

http://www.Hi-TecDesigns.com

 

[MRSSPOCK]

@MacGyverS2000 -- But the derivatives are the tangent points.

There's no need to plot them.

They're shown below by the estimated green line, representing the common tangent point.



Thanks for everyone's input, but it appears not such a simple solution after all.

I had hoped someone might be fit to illustrate it if is was very simple, but it seems not to be the case.

Thanks nevertheless.

 

[VEBill]

The Wolfram Alpha link that I provided above offers an 'Implicit Plot'.

Notice that the Implicit Plot has red dots. If you hover your mouse pointer over the red dots, then numbers will magically appear...

{s= -1.5708, x= 0.523599}
{s= 4.71239, x= 0.523599}
...etc.

Perhaps I'm confused, but those values (presumably truncated) appear to be simultaneous solutions to the OP's original formula (the two formula at the very top of this thread). I checked the first example with the first formula, and it worked out as approximately 0.5 = 0.5. But I can't be bothered to check them all.

The Implicit Plot hints that there are an infinite number of solutions. If I understand the WA output correctly, then the 'Results' formulas provides the machinery to generate the solutions. Crank in the c1 and c2 from 'The Set of all Integers'.

Disclaimer: I'm not very good at math, so hopefully this is a correct interpretation.