How to solve this math problem: Two simultaneous trig equations

[barryj]

Homework Statement

I worked a physics problem and the answer came down to solving these two equations and I am stuck.

Homework Equations

[/B]
2 equations and two unknowns.

sin(x) = (2/3)sin(y)
1 + cos(x) = (2/3) cos(y)

The Attempt at a Solution

I tried this and it went nowhere..
z =sin^-1[(2/3)sin(y)]- cos^-1[(2/3)cos(y) -1]
and set this to zero to solve for y.
This did not seem to work so there must be a better way.
Suggestions??

 

[SammyS]

Homework Statement

I worked a physics problem and the answer came down to solving these two equations and I am stuck.

Homework Equations

[/B]
2 equations and two unknowns.

sin(x) = (2/3)sin(y)
1 + cos(x) = (2/3) cos(y)

The Attempt at a Solution

I tried this and it went nowhere..
z =sin^-1[(2/3)sin(y)]- cos^-1[(2/3)cos(y) -1]
and set this to zero to solve for y.
This did not seem to work so there must be a better way.
Suggestions??

Square both sides of each equation and add the resulting equations.

 

[barryj]

I thought I made a reply but I guess I didn't so here it is again.

I squared both equations and added them and used the trig identity to make sin^2 + cos^2 = 1.
I got an answer but it was not what I expected to see.
I will continue to work.

 

[SammyS]

I thought I made a reply but I guess I didn't so here it is again.

I squared both equations and added them and used the trig identity to make sin^2 + cos^2 = 1.
I got an answer but it was not what I expected to see.
I will continue to work.

What answer did you get?

 

[barryj]

I got cos x = -.777
I expected a positive answer.
Maybe i have my equations set up incorrectly.
I will continue to work and report back

 

[RPinPA]

That means an angle of greater than 90 degrees. What's wrong with that? Check the answer, work out what the sine is and plug back into the original equations. If the math is telling you the answer is more than 90 degrees, then it's more than 90 degrees, despite your expectations.

 

[barryj]

Actually, the answer is correct. The angle is and should be 140 deg. The squaring of both sides of the equation made the solution easier.

 

[Ray Vickson]

Actually, the answer is correct. The angle is and should be 140 deg. The squaring of both sides of the equation made the solution easier.

You could have lessened your anxiety by looking a bit more closely at your equations before setting out to solve them. From $1+\cos x = (2/3) \cos y$ it follows that we need $\cos x < 0$, because otherwise the left-hand-side would be $> 1,$ while the right-hand-side is $\leq 2/3.$ So, having $\cos x$ coming out negative is exactly what you need.

Anyway, the squaring and adding method yields a unique value for $\cos x$, so absent of algebraic errors, must be correct.

 

[barryj]

Actually, the equations were the result of a momentum problem in physics that I posted on the physics site. Had I performed the physics problem a bit smarted I would not have had the two equations at all.

Yes, I was glad my equations were correct however. the answer of 140 degrees is correct.