Half-angle formulas
The half-angle formulas allow us to determine the values of trigonometric functions for half an angle, α/2, in terms of the full angle, α. $$ \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1-\cos \alpha}{2}} $$ $$ \cos \frac{\alpha}{2} = \pm \sqrt{\frac{1+\cos \alpha}{2}} $$ $$ \tan \frac{\alpha}{2} = \pm \frac{\sqrt{1-\cos \alpha}}{ \sqrt{1+\cos \alpha} } = \begin{cases} \frac{ \sin \alpha }{ 1 + \cos \alpha } \\ \\ \frac{ 1 - \cos \alpha }{ \sin \alpha } \end{cases} $$ $$ \cot \frac{\alpha}{2} = \pm \frac{\sqrt{1+\cos \alpha}}{ \sqrt{1-\cos \alpha} } $$
How do you choose the sign?
In the half-angle formulas, the plus-minus sign (±) appears, but both signs do not apply simultaneously.
The correct sign is determined by the sign of the trigonometric function for the angle α/2.
Example: If the sine of α/2 is negative because the terminal side is in the 3rd or 4th quadrant, the sine in the half-angle formula will also be negative. Conversely, if it’s in the 1st or 2nd quadrant, the sine in the formula will be positive. The same logic applies to cosine, tangent, and cotangent. You must always consider the sign of the relevant trigonometric function at the terminal side of angle α/2.
A practical example
Let’s consider the angle α = 120°:
$$ \alpha = 120° = \frac{2 \pi}{3} \ rad $$
The cosine of α = 120° is:
$$ \cos 120° = - \frac{1}{2} $$
Using this angle, we can find the sine, cosine, and tangent values for half the angle, α/2 = 60°, by applying the half-angle formulas.
First, apply the cosine half-angle formula:
$$ \cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}} $$
Since &cos(120°) = -1/2,
$$ \cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + (- \frac{1}{2})}{2}} $$
$$ \cos \frac{\alpha}{2} = \pm \sqrt{\frac{\frac{2-1}{2}}{2}} $$
$$ \cos \frac{\alpha}{2} = \pm \sqrt{\frac{\frac{1}{2}}{2}} $$
$$ \cos \frac{\alpha}{2} = \pm \sqrt{\frac{1}{2} \cdot \frac{1}{2}} $$
$$ \cos \frac{\alpha}{2} = \pm \sqrt{\frac{1}{4}} $$
$$ \cos \frac{\alpha}{2} = \pm \frac{1}{2} $$
We choose the positive sign because the cosine of α/2 = 60° lies in the 1st quadrant, making it positive.
$$ \cos \frac{\alpha}{2} = \frac{1}{2} $$
This result is correct: the cosine of 60° is indeed +1/2.
Now, apply the sine half-angle formula for α/2 = 60°:
$$ \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}} $$
Since &cos(120°) = -1/2,
$$ \sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - (-\frac{1}{2})}{2}} $$
$$ \sin \frac{\alpha}{2} = \pm \sqrt{\frac{\frac{2+1}{2}}{2}} $$
$$ \sin \frac{\alpha}{2} = \pm \sqrt{\frac{\frac{3}{2}}{2}} $$
$$ \sin \frac{\alpha}{2} = \pm \sqrt{\frac{3}{2} \cdot \frac{1}{2}} $$
$$ \sin \frac{\alpha}{2} = \pm \sqrt{\frac{3}{4}} $$
$$ \sin \frac{\alpha}{2} = \pm \frac{\sqrt{3}}{2} $$
We choose the positive sign because the sine of α/2 = 60° is positive in the 1st quadrant.
$$ \sin \frac{\alpha}{2} = + \frac{\sqrt{3}}{2} $$
This result is also correct: the sine of 60° is indeed √3/2.
Finally, apply the tangent half-angle formula for α/2 = 60°:
$$ \tan \frac{\alpha}{2} = \pm \frac{\sqrt{1-\cos \alpha}}{ \sqrt{1+\cos \alpha} } $$
Since &cos(120°) = -1/2,
$$ \tan \frac{\alpha}{2} = \pm \frac{\sqrt{1-(-\frac{1}{2}) }}{ \sqrt{1 + (-\frac{1}{2}) } } $$
$$ \tan \frac{\alpha}{2} = \pm \frac{\sqrt{\frac{2+1}{2}}}{ \sqrt{\frac{2-1}{2}} } $$
$$ \tan \frac{\alpha}{2} = \pm \frac{\sqrt{\frac{3}{2}}}{ \sqrt{\frac{1}{2}} } $$
$$ \tan \frac{\alpha}{2} = \pm \sqrt{\frac{3}{2}} \cdot \sqrt{\frac{2}{1}} $$
$$ \tan \frac{\alpha}{2} = \pm \sqrt{3} $$
We choose the positive sign because the tangent of α/2 = 60° is positive in the 1st quadrant.
$$ \tan \frac{\alpha}{2} = + \sqrt{3} $$
This result is correct: the tangent of 60° is indeed √3.
Thus, we have obtained the sine, cosine, and tangent values for α/2 = 60° starting from the cosine of its double angle, α = 120°.
The proof
Half-angle formula for cosine
We can express the cosine of angle α as:
$$ \cos \alpha = \cos(2 \cdot \frac{\alpha}{2}) $$
Using the cosine double-angle formula:
$$ \cos \alpha = 2 \cos^2 (\frac{\alpha}{2}) - 1 $$
Solving for &cos(α/2), we get:
$$ \cos^2 (\frac{\alpha}{2}) = \frac{1 + \cos \alpha}{2} $$
Taking the square root on both sides:
$$ \sqrt{\cos^2 (\frac{\alpha}{2})} = \sqrt{\frac{1 + \cos \alpha}{2}} $$
This gives us the cosine half-angle formula:
$$ \cos (\frac{\alpha}{2}) = \pm \sqrt{\frac{1 + \cos \alpha}{2}} $$
Half-angle formula for sine
Similarly, we can express the cosine of angle α as:
$$ \cos \alpha = \cos(2 \cdot \frac{\alpha}{2}) $$
Using the cosine double-angle formula in terms of sine:
$$ \cos \alpha = 1 - 2 \cdot \sin^2 (\frac{\alpha}{2}) $$
Solving for &sin(α/2), we get:
$$ - \sin^2 (\frac{\alpha}{2}) = \frac{\cos \alpha - 1}{2} $$
Multiplying both sides by -1:
$$ \sin^2 (\frac{\alpha}{2}) = \frac{1 - \cos \alpha}{2} $$
Taking the square root on both sides:
$$ \sqrt{\sin^2 (\frac{\alpha}{2})} = \sqrt{\frac{1 - \cos \alpha}{2}} $$
This gives us the sine half-angle formula:
$$ \sin (\frac{\alpha}{2}) = \pm \sqrt{\frac{1 - \cos \alpha}{2}} $$
Half-angle formula for tangent
The tangent of any angle α is the ratio of its sine to cosine:
$$ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} $$
For the angle α/2, we have:
$$ \tan \frac{\alpha}{2} = \frac{\sin (\frac{\alpha}{2})}{\cos (\frac{\alpha}{2})} $$
Substituting the half-angle formulas for sine and cosine:
$$ \tan \frac{\alpha}{2} = \frac{\sqrt{\frac{1 - \cos \alpha}{2}}}{\sqrt{\frac{1 + \cos \alpha}{2}}} $$
Simplifying, we get:
$$ \tan \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2} \cdot \frac{2}{1 + \cos \alpha}} $$
And this gives us the tangent half-angle formula:
$$ \tan \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} $$
Half-angle formula for cotangent
The cotangent is the reciprocal of the tangent:
$$ \cot \frac{\alpha}{2} = \frac{1}{\tan \frac{\alpha}{2}} $$
Substituting the tangent half-angle formula:
$$ \cot \frac{\alpha}{2} = \frac{1}{\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}} $$
Simplifying, we get the cotangent half-angle formula:
$$ \cot \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} $$
Other half-angle formulas for tangent
Other forms of the tangent half-angle formulas include:
$$ \tan \frac{\alpha}{2} = \begin{cases} \frac{\sin \alpha}{1 + \cos \alpha} \\ \\ \frac{1 - \cos \alpha}{\sin \alpha} \end{cases} $$
These two formulas are yet to be proven.
The first formula
The tangent of α/2 can be written as:
$$ \tan \frac{\alpha}{2} = \frac{\sin (\frac{\alpha}{2})}{\cos (\frac{\alpha}{2})} $$
Multiplying both the numerator and denominator by 2&cos(α/2):
$$ \tan \frac{\alpha}{2} = \frac{\sin (\frac{\alpha}{2})}{\cos (\frac{\alpha}{2})} \cdot \frac{2 \cos (\frac{\alpha}{2})}{2 \cos (\frac{\alpha}{2})} $$
$$ \tan \frac{\alpha}{2} = \frac{2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2})}{2 \cos^2 (\frac{\alpha}{2})} $$
Now apply the sine double-angle formula:
$$ \tan \frac{\alpha}{2} = \frac{\sin \alpha}{2 \cos^2 (\frac{\alpha}{2})} $$
Note: According to the sine double-angle formula, $$ \sin 2a = 2 \sin a \cos a $$ Here, the angle is a = α/2, so $$ \sin \alpha = 2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2}) $$
Rewriting the denominator:
$$ \tan \frac{\alpha}{2} = \frac{\sin \alpha}{2 \cdot [\cos (\frac{\alpha}{2}) \cdot \cos (\frac{\alpha}{2})]} $$
Now, using the cosine half-angle formula:
$$ \tan \frac{\alpha}{2} = \frac{\sin \alpha}{2 \cdot \left(\sqrt{\frac{1 + \cos \alpha}{2}}\right)^2} $$
$$ \tan \frac{\alpha}{2} = \frac{\sin \alpha}{2 \cdot \frac{(1 + \cos \alpha)^2}{4}} $$
$$ \tan \frac{\alpha}{2} = \frac{\sin \alpha}{\frac{(1 + \cos \alpha)}{2}} $$
And this gives the first tangent half-angle formula:
$$ \tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha} $$
The second formula
The tangent of α/2 can also be expressed as:
$$ \tan \frac{\alpha}{2} = \frac{\sin (\frac{\alpha}{2})}{\cos (\frac{\alpha}{2})} $$
Multiplying both numerator and denominator by 2&sin(α/2):
$$ \tan \frac{\alpha}{2} = \frac{\sin (\frac{\alpha}{2})}{\cos (\frac{\alpha}{2})} \cdot \frac{2 \sin (\frac{\alpha}{2})}{2 \sin (\frac{\alpha}{2})} $$
$$ \tan \frac{\alpha}{2} = \frac{2 \sin^2 (\frac{\alpha}{2})}{2 \cos (\frac{\alpha}{2}) \cdot \sin (\frac{\alpha}{2})} $$
Apply the sine double-angle formula to the denominator:
$$ \tan \frac{\alpha}{2} = \frac{2 \sin^2 (\frac{\alpha}{2})}{\sin \alpha} $$
Note: According to the sine double-angle formula, $$ \sin 2a = 2 \sin a \cos a $$ In this case, a = α/2, so $$ \sin \alpha = 2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2}) $$
Now rewrite the denominator:
$$ \tan \frac{\alpha}{2} = \frac{2 (\sin (\frac{\alpha}{2}))^2}{\sin \alpha} $$
Substituting the sine half-angle formula:
$$ \tan \frac{\alpha}{2} = \frac{2 \cdot \left(\sqrt{\frac{1 - \cos \alpha}{2}}\right)^2}{\sin \alpha} $$
$$ \tan \frac{\alpha}{2} = \frac{2 \cdot \frac{(1 - \cos \alpha)^2}{4}}{\sin \alpha} $$
$$ \tan \frac{\alpha}{2} = \frac{1 - \cos \alpha}{\sin \alpha} $$
And this gives the second tangent half-angle formula.
