# Trigonometry Help

1. Sep 9, 2007

### Champ07

I've been really stumped by this one from my math class. It's a two variable trig. question with two equations. I believe there simultaneous but I'm not sure

3 = 1.2(cosx) + v(cos30)
0 = 1.2(sinx) - v(sin30)

Solve for the values of x & v

2. Sep 9, 2007

### arildno

Well, solve for v first from, say, the 2nd eq.
Then you get a trig equation; use the summation formula to reshape it so that you can solve for relevant x's

3. Sep 9, 2007

### Champ07

Sorry I'm not familar with the summation formula. I solved for V in the second equation and got V = .6sinx.

I substituted that into the second equation and got sin = 0, 180 or 360 but then the 2nd equation doesn't check out. I know I'm doing it wrong somehow just not sure how

4. Sep 9, 2007

### HallsofIvy

Staff Emeritus
Actually, I suspect the problem asked YOU to explain how you got your answer or shiow work. I don't have to! Let x= cos(x) and y= sin(x). then you have the two equations
3= 1.2x+ vy and 0 = 1.2x+ vy as well as the obvious equation $x^2+ y^2= 1$. What happens if you add the first two equations?

5. Sep 9, 2007

### nicktacik

HallsofIvy how do you get those equations?

Seems it should be
$$3 = 1.2x + v\frac{\sqrt{3}}{2}$$
$$0 = 1.2y - 0.5v$$
$$x^2 + y^2 = 1$$

6. Sep 9, 2007

### HallsofIvy

Staff Emeritus
How did I get those two equations? I copied them from the first post in this thread!
3 = 1.2(cosx) + v(cos30)
0 = 1.2(sinx) - v(sin30)
I now see, looking more closely, that it might be good idea to multiply the first equation by sin(30) and the second equation by cos(30). Although it wasn't said, I assume that is "30 degrees" so that cos(30)= $\sqrt{2}/3$ and sin(30)= 1/2.

7. Sep 9, 2007

### AlephZero

That's the right idea, but the solution to 0 = 1.2(sinx) - v(sin30) = 1.2(sinx) -v(0.5) isn't v = 0.6 sin x.

8. Sep 9, 2007

### rocomath

i just solved it, long prob