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, , I would need to add "1 round" to 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 found to be . Question domain is
So adding 1 round = adding
Therefore, answer = .
Example two
Angle found to be . Question domain is
So adding 1 round = adding
Therefore, answer = .
When you bring in multiple angles (like ) the period changes.
So in your case:
Case 1:
Here I think you can see straight away that the only that satisfies this equation in the given domain is .
Case 2:
Since when , where is the period of the function and is an integer representing how many times you have gone around the unit circle.
Notice how I have written . This doesn't change yet because I have assumed is not a multiple angle.
But in your case you have
So
.
Now, I have to divide both sides of the equation by . This INCLUDES the period . So do you see here how the period changes?
So
.
Here you have listed an infinity of solutions (since can take on any integer). So you now need to check which solutions lie in your given domain .
Let
Clearly is out of the domain.
is ok.
Let
is ok.
is ok.
Let
, which is out of the domain.
So now you have all your solutions in the domain :
.
So does it make sense that when you have a multiple angle (like in this case), the period will change accordingly?
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 , thus 1 cycle/round=
So by adding 1 round, I will be adding .
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.
Read my response properly.
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.
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.
"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.