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

 

[IRstuff]

I tried 50 values from WA's solution in Mathcad, and they all successfully came back with equalities, out to 40 decimal places.

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

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

faq731-376 forum1529

 

[hacksaw]

Dear MrsSpock,

The following offers a bit more insight into the question you've raised

presumbly the file has uploaded

nonrational equations are always interesting

 

[electricpete]

Bad timing for me. I just tried to solve and was proud tome come up with something. Then realized similar thing was just posted by hacksaw. Oh well.
sin(x) = sin(2x+s)+1
cos(x) = cos(2x+s)

since sin^2 + cos^2 = 1 (lhs), we know we can do the same to the RHS
[sin(2x+s)+1]^2 + cos(2x+s)^2 = 1
[sin^2(2x+s) +2*sin(2x+s)+1] + cos^2(2x+s) = 1

subtract 1 each side
sin^2(2x+s) +2*sin(2x+s) + cos^2(2x+s) = 0

Combine sin^2 and cos^2 into 1
1 + 2*sin(2x+s) = 0
sin(2x+s) = -1/2


=====================================
(2B)+(2B)' ?

 

[MRSSPOCK]

@ hacksaw

surely line 1 of your solution, (as shown below), doesn't reflect what is stated in post #1

You have stated "2Cos" instead of just "Cos"

Sin(x) = 1 + Sin(2x+s)
Cos(x) = 2Cos(2x+s)

 

[electricpete]

I don't think he was intending to solve the identical problem you posted, maybe just a similar problem. It points out the use of the simplification sin^2+cos^2=1. That makes the solution pretty simple. I think the way I showed above is the solution unless I made a silly algebra error.

Just to clarify my terminology

since sin^2 + cos^2 = 1 (lhs), we know we can do the same to the RHS

could maybe have been written more succinctly as
LHS(eq1)^2 + LHS(eq2)^2 = 1 [⇒] RHS(eq1)^2 + RHS(eq2)^2 = 1

The answer given above was
sin(2x+s) = -1/2
for any given value of x you can generate corresponding values of s as follows:
s = asin(-1/2) - 2x
where
asin(-1/2) = [7*pi/6 + n*(2*pi)] or [11*pi/6 + n*(2*pi)]
and n= any integer positive, negative or zero



=====================================
(2B)+(2B)' ?

 

[IRstuff]

Which were indeed the solutions generated by Wolfram Alpha.

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

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

faq731-376 forum1529

 

[hacksaw]

typical for me, I stand corrected,

 

[electricpete]

sin(x) = sin(2x+s)+1 [eq 1]

cos(x) = cos(2x+s) [eq 2]

Find x and s to satisfy the two above statements.

That's the part I solved above. Call my solution [equation 3]

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

Is that is an additional requirement that needs to be solved?
If so its an additional constraint on the solution... odd to tack it on at the very end and maybe explains why hacksaw was looking at derivatives.
It would still be a somewhat murky requirement (which two out of the four listed functions LHS(1),RHS(1),LHS(2),RHS(2) are supposed to be tangential and in which plane (y,x) or (y,s).

=====================================
(2B)+(2B)' ?

 

[MRSSPOCK]

@electricpete

I don't know what you "mean tack it on at the end"

The fourth line of post #1 provides the derivatives of the two equations in the third line of the first post, so the requirement for the tangential solution was there from the very beginning. See below

[highlight #4E9A06]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)[/highlight] these are the derivatives from the previous line, hence the common tangent definitions..

 

[MRSSPOCK]

So, after all the posts above, I still don't know how to calculate the phase shift.

Possibly it is because I didn't lay out my question correctly in the first instance.

For that reason I decided to reformulate the question and create a new post.

Maybe that will help clarify any confusion.

Thanks

The new post is here

http://www.eng-tips.com/viewthread.cfm?qid=412137