Click here to see part two of this four week series.
Last week, we figured out a way to figure out trigonometric functions for twice a given angle. This week, we will do the same, but for determining the trigonometric functions for half a given angle. First, let's discuss what half of an angle means.
Remember that there are 360° in a circle, or 360° in a full revolution. This means that a number like 370° can also be expressed as 10°. Though they are different measurements, plugging 370° into a trigonometric function will yield the same answer as 10°. It is also equivalent in most other situations in mathematics.
For double angle formulas, we did not need to discuss this. This is because when doing double angle formula calculations with these measurements, there would be no issue.
sin(2 • 10°) = sin(20°)
sin(2 • 370°) = sin(740°) = sin(740° - 720°) = sin(20°)
Note that 720° is two full revolutions around a circle, and thus, it can be subtracted off when performing a trigonometric operation.
But performing a half angle calculation will create more of an issue. Let's use 10° and 370° again.
sin(1/2 • 10°) = sin(5°)
sin(1/2 • 370°) = sin(185°)
These answers are not the same. Since they are both in the 0° - 360° interval, we cannot make any assumptions. We do know that the sine and cosine of 185° are the negative sine and negative cosine of 5° respectively, but this proves that they are not equal. If one were go up to 730°, they would be back to normal, however.
sin(1/2 • 730°) = sin(365°) = sin(365° - 360°) = sin(5°)
This means that for every angle, the half sine and half cosine function should yield two answers. The two answers should have the same absolute value, but different signs (they are the same number, but one is negative and one is positive). You may already have a function in your head that can create this type of situation, but we will be able to derive it as well.
Take a variation of the cosine double angle formula that we derived last week:
cos(2α) = 1 – 2sin2α
Let's try to isolate sinα. We would first add that to the left hand side and subtract the cosine of 2α over to the right hand side.
2sin2α = 1 – cos(2α)
Divide through by 2 to get:
sin2α = (1 – cos(2α))/2
And square root both sides to get:
sinα = ±√((1 - cos(2α))/2)
Notice how there is a ± sign in front of the square root. This is because when one squares a positive or negative value, it becomes positive. For instance, the equation x2 = 25 would be solved as x = ±5 because (5)(5) = 25 and (–5)(–5) = 25. The same thing happened here. But also remember what we found before. We proved through logic that the half sine and half cosine of an angle has two answers, one negative and one positive. With that in mind, we can see that this is the accurate way to write the square root (some derivations call for just a positive answer such as the Distance Formula).
Let's replace angle α with α/2 to keep the half angle definition. This gives a formula of:
sin(α/2) = ±√((1 - cosα)/2)
Let's derive a cosine half angle formula. We can take another variation on the cosine double angle formula and go forward.
cos(2α) = 2cos2α – 1
This time, we will only need to add one to both sides to isolate the cosα term. Let's also flip the equation around to make it simpler.
2cos2α = 1 + cos(2α)
Divide through by 2 to get:
cos2α = (1 + cos(2α))/2
And square rooting both sides yields:
cosα = ±√((1 + cos(2α))/2)
Again, we end up with a ± sign in the formula. This means that we probably did everything correctly, as logic shows we will need this sort of sign to create two answers. Rewriting α as α/2 gives a final formula of:
cos(α/2) = ±√((1 + cosα)/2)
It is tough to see what these formulas actually look like in this formatting, so I have written them out in LaTeX so you can see what is going on more conveniently.
These formulas themselves are pretty cool, but the logic involved in finding them is also very interesting. The fact that we could predict the nature of the function before we even found it is really helpful. This can be huge in trying to figure out the right way to go about solving a problem.