How do I solve this trigonometry equation involving sine and cosine?

[Draggu]

Homework Statement

Solve the following equation, where 0 \lequ \leq 2pi [360 degrees]

2sinucosu = 0

Homework Equations

The Attempt at a Solution

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

sinucosu = 0

 

[CrystalEyes]

Try seeing if there's a specific trig identity to help you

 

[Draggu]

sin2u?

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

 

[Mark44]

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.

 

[icystrike]

Homework Statement

Solve the following equation, where 0 \lequ \leq 2pi [360 degrees]

2sinucosu = 0

Homework Equations

The Attempt at a Solution

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

sinucosu = 0

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

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

Your range of 2U
0 \lequ \leq 2pi
0 \leq2u \leq 4pi

 

[CrystalEyes]

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.

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

 

[HallsofIvy]

Homework Statement

Solve the following equation, where 0 \lequ \leq 2pi [360 degrees]

2sinucosu = 0

Homework Equations

The Attempt at a Solution

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

sinucosu = 0

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

 

[icystrike]

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

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

Therefore u = 2pi

 

[Draggu]

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

Therefore u = 2pi

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

 

[HallsofIvy]

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

Therefore u = 2pi

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

 

[HallsofIvy]

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

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?

 

[Draggu]

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?

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? :)

 

[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? :)

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?

 

[HallsofIvy]

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

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

 

[Mark44]

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?

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

 

[HallsofIvy]

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

 

[Draggu]

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

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

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.

 

[Draggu]

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

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

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.

 

[Chaos2009]

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.

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.

 

[Draggu]

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