Half angle formula , what is happening in these formulas

AI Thread Summary
The discussion centers on the half angle formulas in trigonometry, particularly their application in determining values like Cos A/2 when given Sin A and the quadrant of angle A. It highlights the confusion around why these formulas are called "half angle" when they don't simply divide an angle in half, as the resulting angle's quadrant affects the sign of the answer. Participants clarify that understanding the quadrant is crucial for determining the correct square root when using the half angle formulas. The conversation emphasizes the importance of grasping the underlying concepts rather than just memorizing formulas. Ultimately, the half angle formulas are adaptations of the double angle formulas, and knowing the quadrant helps in selecting the appropriate value.
Jurrasic
Messages
98
Reaction score
0
The half angle formulas used when you solve problems that are like the following I mean: Ok, they usually look something like (I may have left something out, but I am putting down the general kind of idea of how they look and work:)
Something like, Find Cos A/2 if Sin A =-1/2 and SinA is in Quadrant 3
Then you have to put Cos A between these things <cosA< and then 180 goes on one side of that and 270 goes on the other side, or what have you, and then you divide 180 and 270 in half, but aren't we just dividing an angle in half, so why does that change what quadrant it's in? That is so weird? lol
So what are these formulas having you do when you use them? Apparently they aren't dividing an actual angle in half. So why do they even call them half angle formulas.
 
Physics news on Phys.org
Jurrasic said:
The half angle formulas used when you solve problems that are like the following I mean: Ok, they usually look something like (I may have left something out, but I am putting down the general kind of idea of how they look and work:)
Something like, Find Cos A/2 if Sin A =-1/2 and SinA is in Quadrant 3
Then you have to put Cos A between these things <cosA< and then 180 goes on one side of that and 270 goes on the other side, or what have you, and then you divide 180 and 270 in half, but aren't we just dividing an angle in half, so why does that change what quadrant it's in? That is so weird? lol
So what are these formulas having you do when you use them? Apparently they aren't dividing an actual angle in half. So why do they even call them half angle formulas.

Your example was a little off:

Something like, Find Cos A/2 if Sin A =-1/2 and SinA is in Quadrant 3

should read

Something like, Find Cos A/2 if Sin A =-1/2 and A is in Quadrant 3.

Also, if 180<A<270, then 90 < A/2 < 135 which is in the second quadrant, and we thus know what sign should be in the answer to Cos A/2. That is useful if your calculations present you with two possible answers, and one of them is positive and the other negative.

I believe the half angle formulae are just adaptations of the Double angle formulae - such as Sin2A = 2.SinA.CosA
 
You should first try to find the value of A based on the information and then find cos A/2. You only need to worry about the limits for sin A.
 
Jurrasic said:
The half angle formulas used when you solve problems that are like the following I mean: Ok, they usually look something like (I may have left something out, but I am putting down the general kind of idea of how they look and work:)
Something like, Find Cos A/2 if Sin A =-1/2 and SinA is in Quadrant 3
Then you have to put Cos A between these things <cosA< and then 180 goes on one side of that and 270 goes on the other side, or what have you, and then you divide 180 and 270 in half, but aren't we just dividing an angle in half, so why does that change what quadrant it's in? That is so weird? lol
So what are these formulas having you do when you use them? Apparently they aren't dividing an actual angle in half. So why do they even call them half angle formulas.

you could use a half angle formula directly: like CosA = 2 Cos^2(A/2) - 1

By transposing this Cos(A/2) would be the square root of (CosA +1)/2

But is is the positive square root, or the negative square root?

By telling you which quadrant you are working with, you know which root to accept.
 
Jurrasic said:
The half angle formulas used when you solve problems that are like the following I mean: Ok, they usually look something like (I may have left something out, but I am putting down the general kind of idea of how they look and work:)
Something like, Find Cos A/2 if Sin A =-1/2 and SinA is in Quadrant 3
Then you have to put Cos A between these things <cosA< and then 180 goes on one side of that and 270 goes on the other side, or what have you, and then you divide 180 and 270 in half, but aren't we just dividing an angle in half, so why does that change what quadrant it's in? That is so weird? lol
So what are these formulas having you do when you use them? Apparently they aren't dividing an actual angle in half. So why do they even call them half angle formulas.
I strongly recommend that you learn what formulas mean instead of just memorizing symbols!

"Cos A between these things <cosA< and then 180 goes on one side of that and 270 goes on the other side, or what have you", is mostly nonsense. What you should have is 180< A< 270, not "cos A". That means that the angle, A, lies between 180 degrees and 270 degrees, exactly what is meant by " A is in the third quadrant". If 180< A then, dividing both sides by the positive number 2, 90< A/2. Similarly, if A< 270, then A/2< 135. Putting those together, 90< A/2< 135 so A/2 is in the first quadrant. That tells you that all trig functions of A/2 will have positive values and, as others have told you, that will tell you which sign to use for the square root.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top