# Trig Equation

1. Nov 18, 2008

### Draggu

1. The problem statement, all variables and given/known data
Solve the following equation, where 0 $$\leq$$u $$\leq$$ 2pi [360 degrees]

2sinucosu = 0

2. Relevant equations

3. The attempt at a solution

(2sinucosu/2) = (0/2)

sinucosu = 0

2. Nov 18, 2008

### CrystalEyes

3. Nov 19, 2008

### Draggu

sin2u?

tried that as well, and not sure what to do.

4. Nov 19, 2008

### Staff: Mentor

If 2 sin(u)cos(u) = 0, then sin(2u) = 0. What can you say about 2u? You should several values for 2u, and several for u.

5. Nov 19, 2008

### icystrike

You gotta read this step by step alright no peeping into the next step.

1st step
Refer to the rule
Sin 2A=2 SinA CosA

0 $$\leq$$u $$\leq$$ 2pi
0 $$\leq$$2u $$\leq$$ 4pi

6. Nov 19, 2008

### CrystalEyes

going off of what he was saying, imagine for a second that the 2 wasn't there, so that you have sin(u)=0. For confusions sake, let's call it n now. So you've got sin(n)=0 Essentially saying the sign of some angle "n" equals 0 right? You should be able to figure out what that angle n is. If you think of it like that and can solve for n, then all you have to do now is think of n=2u, and that i think you can figure out =). Hope that wasn't too revealing =X

7. Nov 19, 2008

### HallsofIvy

Staff Emeritus
So either sin u= 0 or cos u= 0. What values if y satisfy those?

8. Nov 19, 2008

### icystrike

haha.. tats true we nid consider tat factor..

Therefore u = 2pi

9. Nov 20, 2008

### Draggu

Well.... I got as far as sinu = 0 . I honestly have no idea how to continue it further, since no special triangle can get you to equal 0. On a sin graph with a period of 2pi though, sin touches 0 at pi and 2pi

10. Nov 20, 2008

### HallsofIvy

Staff Emeritus
Apparently you didn't consider it. sin(u)= 0 for more values than 2pi, cos(u)= 0 for other values.

11. Nov 20, 2008

### HallsofIvy

Staff Emeritus
Yes, exactly. And since your problem says "$0\le u\le 2\pi$, those are two solutions. Now, what values of u make cos(u)= 0?

12. Nov 20, 2008

### Draggu

pi/2 and 3pi/2

I have a question though, how would I show my work for it? It's worth 6 marks...would I sketch the graph for each of them or? :)

13. Nov 20, 2008

### Draggu

Also, is using cosu necessary for the equation too? If so, changing it to sin2u = 0 is irrelevant. Since sin2u = 0 and diving sin2u by 2 and 0 by 2, giving us sinu = 0 , then would cos not be apart of this?

14. Nov 21, 2008

### HallsofIvy

Staff Emeritus
No sin(au)/a is NOT equal to sin(u).

If xy= 0, then either x= 0 or y= 0. Surely you know that?

15. Nov 21, 2008

### Staff: Mentor

You should get the same solutions either way:
By solving sin(2u) = 0
By solving sin(u)cos(u) = 0

16. Nov 21, 2008

### HallsofIvy

Staff Emeritus
But NOT by "dividing sin 2u= 0 by 2" to get sin u= 0!

17. Nov 24, 2008

### Draggu

u=theta

sin2u=0 , is the double angle identity. Sorry but cos has nothing to do with the answer.

2pi/2 = pi pi is the period.

18. Nov 24, 2008

### Draggu

u=theta

sin2u=0 , is the double angle identity. Sorry but cos has nothing to do with the answer.

(2pi/2) = pi ---->pi is the period.

19. Nov 24, 2008

### Chaos2009

Mark44 was right

$$sin(2u) = 0$$ is the same thing as $$sin(u)cos(u) = 0$$. You will get the same answers either way, it is just a matter of preference.

20. Nov 24, 2008

### Draggu

Hmm, well my answers were (pi/2), (3pi/2), 2pi, pi