1. ## A clarification on adding "rounds"

Okay I have been doing trigo sums over the past few days...

And I have also learnt that if solving for angle X gives me an answer of lets say, $\displaystyle -10^\circ$, I would need to add "1 round" to $\displaystyle -10^\circ$to get my answer since I can't have a negative angle.

I would like to clarify if the value of "1 round" is related to the domain of the question.

If it's true, two example would be this.

Example one

Angle $\displaystyle X$ found to be $\displaystyle -10^\circ$. Question domain is $\displaystyle 0^\circ < X < 360^\circ$
So adding 1 round = adding $\displaystyle 360^\circ$

Therefore, answer = $\displaystyle 350^\circ$.

Example two

Angle $\displaystyle X$ found to be $\displaystyle -10^\circ$. Question domain is $\displaystyle 0^\circ< X < 180^\circ$

So adding 1 round = adding $\displaystyle 180^\circ$
Therefore, answer = $\displaystyle 170^\circ$.

2. Originally Posted by Punch
Okay I have been doing trigo sums over the past few days...

And I have also learnt that if solving for angle X gives me an answer of lets say, $\displaystyle -10^\circ$, I would need to add "1 round" to $\displaystyle -10^\circ$to get my answer since I can't have a negative angle.

I would like to clarify if the value of "1 round" is related to the domain of the question.

If it's true, two example would be this.

Example one

Angle $\displaystyle X$ found to be $\displaystyle -10^\circ$. Question domain is $\displaystyle 0^\circ < X < 360^\circ$
So adding 1 round = adding $\displaystyle 360^\circ$

Therefore, answer = $\displaystyle 350^\circ$.

Example two

Angle $\displaystyle X$ found to be $\displaystyle -10^\circ$. Question domain is $\displaystyle 0^\circ< X < 180^\circ$

So adding 1 round = adding $\displaystyle 180^\circ$
Therefore, answer = $\displaystyle 170^\circ$.
First part correct.

Second part incorrect.

An angle of $\displaystyle -10^\circ$ is the same as an angle of $\displaystyle 350^\circ$.

You are not adding the domain, you are realising that if you go around the circle in the other direction, then you have to add $\displaystyle 360^\circ$.

3. Originally Posted by Prove It
First part correct.

Second part incorrect.

An angle of $\displaystyle -10^\circ$ is the same as an angle of $\displaystyle 350^\circ$.

You are not adding the domain, you are realising that if you go around the circle in the other direction, then you have to add $\displaystyle 360^\circ$.
Does this means that for every negative angle I get, I have to add $\displaystyle 360^\circ$?

However, I have just completed a question with domain $\displaystyle 0 < X < \pi$ and found that the answer was incorrect when I added $\displaystyle 2\pi$ but correct when I added $\displaystyle \pi$ instead. ( answer was provided by book )

4. Post the entire question. I have a feeling you've left out some important information...

5. Originally Posted by Prove It
Post the entire question. I have a feeling you've left out some important information...

It's the second part of the question where you equate it to zero and solve

$\displaystyle cos\theta sin(3\theta+\frac{\pi}{4})=0$

$\displaystyle cos\theta=0, sin(3\theta+\frac{\pi}{4})=0$

Answers are: $\displaystyle \frac{\pi}{4}, \frac{\pi}{2}, \frac{7\pi}{12}, \frac{11\pi}{12}$

What I attempted

$\displaystyle cos\theta=0$

basic angle =$\displaystyle \frac{\pi}{2}$

$\displaystyle \theta= \frac{\pi}{2} , \frac{3\pi}{4} (reject)$

__________________________________________

$\displaystyle sin(3\theta+\frac{\pi}{4})=0$

basic angle=0

$\displaystyle 3\theta+\frac{\pi}{4}=0, \pi, 2\pi$

$\displaystyle 3\theta=\frac{-\pi}{4}, \pi-\frac{\pi}{4}, 2\pi-\frac{\pi}{4}$

$\displaystyle 3\theta= \frac{-\pi}{4}. \frac{3\pi}{4}, \frac{7\pi}{4}$

$\displaystyle \theta=\frac{-\pi}{12}, \frac{1\pi}{4},\frac{7\pi}{12}$

$\displaystyle =\frac{-\pi}{12}+\pi, \frac{1\pi}{4},\frac{7\pi}{12}$

$\displaystyle =\frac{11\pi}{12}, \frac{\pi}{4},\frac{7\pi}{12}$

$\displaystyle \theta=\frac{\pi}{4}, \frac{\pi}{2}, \frac{7\pi}{12}, \frac{11\pi}{12}$

6. Originally Posted by Punch

It's the second part of the question where you equate it to zero and solve

$\displaystyle cos\theta sin(3\theta+\frac{\pi}{4})=0$

$\displaystyle cos\theta=0, sin(3\theta+\frac{\pi}{4})=0$

Answers are: $\displaystyle \frac{\pi}{4}, \frac{\pi}{2}, \frac{7\pi}{12}, \frac{11\pi}{12}$
When you bring in multiple angles (like $\displaystyle 2\theta$) the period changes.

$\displaystyle 2\sqrt{2}\cos{\theta}\sin{\left(3\theta + \frac{\pi}{4}\right)} = 0$

Case 1: $\displaystyle \cos{\theta} = 0$

Here I think you can see straight away that the only $\displaystyle \theta$ that satisfies this equation in the given domain is $\displaystyle \frac{\pi}{2}$.

Case 2: $\displaystyle \sin{\left(3\theta + \frac{\pi}{4}\right)} = 0$

Since $\displaystyle \sin{x} = 0$ when $\displaystyle x = \left\{0, \pi \right\} + 2\pi n$, where $\displaystyle 2\pi$ is the period of the function and $\displaystyle n$ is an integer representing how many times you have gone around the unit circle.

Notice how I have written $\displaystyle + 2\pi n$. This doesn't change yet because I have assumed $\displaystyle x$ is not a multiple angle.

But in your case you have

$\displaystyle \sin{\left(3\theta + \frac{\pi}{4}\right)} = 0$

So $\displaystyle 3\theta + \frac{\pi}{4} = \left\{0, \pi \right\} + 2\pi n$

$\displaystyle 3\theta = \left\{ -\frac{\pi}{4}, \frac{3\pi}{4} \right\} + 2\pi n$.

Now, I have to divide both sides of the equation by $\displaystyle 3$. This INCLUDES the period $\displaystyle 2\pi$. So do you see here how the period changes?

So $\displaystyle \theta = \left\{ -\frac{\pi}{12}, \frac{3\pi}{12} \right\} + \frac{2\pi}{3}n$

$\displaystyle \theta = \left\{ -\frac{\pi}{12}, \frac{\pi}{4} \right\} + \frac{2\pi}{3}n$.

Here you have listed an infinity of solutions (since $\displaystyle n$ can take on any integer). So you now need to check which solutions lie in your given domain $\displaystyle 0 \leq \theta \leq \pi$.

Let $\displaystyle n = 0$

Clearly $\displaystyle -\frac{\pi}{12}$ is out of the domain.

$\displaystyle \frac{\pi}{4}$ is ok.

Let $\displaystyle n = 1$

$\displaystyle -\frac{\pi}{12} + \frac{2\pi}{3} = \frac{7\pi}{12}$ is ok.

$\displaystyle \frac{\pi}{4} + \frac{2\pi}{3} = \frac{11\pi}{12}$ is ok.

Let $\displaystyle n = 2$

$\displaystyle -\frac{\pi}{12} + \frac{4\pi}{3} = \frac{5\pi}{4}$, which is out of the domain.

So now you have all your solutions in the domain $\displaystyle 0 \leq \theta \leq \pi$:

$\displaystyle \theta = \left\{\frac{\pi}{4}, \frac{\pi}{2}, \frac{7\pi}{12}, \frac{11\pi}{12} \right\}$.

So does it make sense that when you have a multiple angle (like in this case), the period will change accordingly?

7. I have included my workings in the previous post which proves that adding 1round=max of domain provides a solution for negative angles.

Also, am I right to say that when the period changes, so does the domain. So in this case domain is $\displaystyle \pi$, thus 1 cycle/round=$\displaystyle \pi$

So by adding 1 round, I will be adding $\displaystyle \pi$.

In conclusion, for any negative angles, adding 1 round/cycle to the angle will enable finding the positive angle. And the domain of the question provides me with the value of 1 round/cycle.

I have explained that you originally get a negative angle, but since it is not in your specified domain, you DISREGARD IT. However, you add however many multiples of the period to that negative angle that you need to to show which solutions DO lie within the domain.

9. Originally Posted by Prove It

I have explained that you originally get a negative angle, but since it is not in your specified domain, you DISREGARD IT. However, you add however many multiples of the period to that negative angle that you need to to show which solutions DO lie within the domain.
Sorry but I don't really understand the way it actually is being done, but using the method in my previous last post actually really does work for me.

10. You don't change the given domain at all. All that the given domain does is tell you which of an infinity of solutions you are allowed to accept.

However, the multiplicity of an angle changes the period - thus will change the number of acceptable solutions.

11. Originally Posted by Punch
Okay I have been doing trigo sums over the past few days...

And I have also learnt that if solving for angle X gives me an answer of lets say, $\displaystyle -10^\circ$, I would need to add "1 round" to $\displaystyle -10^\circ$to get my answer since I can't have a negative angle.

I would like to clarify if the value of "1 round" is related to the domain of the question.

If it's true, two example would be this.

Example one

Angle $\displaystyle X$ found to be $\displaystyle -10^\circ$. Question domain is $\displaystyle 0^\circ < X < 360^\circ$
So adding 1 round = adding $\displaystyle 360^\circ$

Therefore, answer = $\displaystyle 350^\circ$.

Example two

Angle $\displaystyle X$ found to be $\displaystyle -10^\circ$. Question domain is $\displaystyle 0^\circ< X < 180^\circ$

So adding 1 round = adding $\displaystyle 180^\circ$
Therefore, answer = $\displaystyle 170^\circ$.
"Domain" only makes sense for functions. What functions are you talking about? All the trig functions have 360 degrees as a period so adding 360 to a negative number will work. But what "reference" angle between 0 and 180 willl give the correct value for the function depends on the specific function.

12. Thanks everybody, I realised that for any domain, 1 round = 360 or 2Pi