4. Half-Angle Formulas
by M. Bourne
We will develop formulas for the sine, cosine and tangent of a half angle.
Half Angle Formula - Sine
We start with the formula for the cosine of a double angle that we met in the last section.
cos 2θ = 1− 2sin2 θ
Formula Summary
We derive the following formulas on this page:
`sin (alpha/2)=+-sqrt((1-cos alpha)/2`
`cos (alpha/2)=+-sqrt((1+cos alpha)/2`
`tan (alpha/2)=(1-cos alpha)/(sin alpha`
Now, if we let
`theta=alpha/2`
then 2θ = α and our formula becomes:
`cos α = 1 − 2\ sin^2(α/2)`
We now solve for
`sin(alpha/2)`
(That is, we get `sin(alpha/2)` on the left of the equation and everything else on the right):
`2\ sin^2(α/2) = 1 − cos α`
`sin^2(α/2) = (1 − cos α)/2`
Solving gives us the following sine of a half-angle identity:
`sin (alpha/2)=+-sqrt((1-cos alpha)/2`
The sign (positive or negative) of `sin(alpha/2)` depends on the quadrant in which `α/2` lies.
If `α/2` is in the first or second quadrants, the formula uses the positive case:
`sin (alpha/2)=sqrt(1-cos alpha)/2`
If `α/2` is in the third or fourth quadrants, the formula uses the negative case:
`sin (alpha/2)=-sqrt(1-cos alpha)/2`
Half Angle Formula - Cosine
Using a similar process, with the same substitution of `theta=alpha/2` (so 2θ = α) we subsitute into the identity
cos 2θ = 2cos2 θ − 1 (see cosine of a double angle)
We obtain
`cos alpha=2\ cos^2(alpha/2)-1`
Reverse the equation:
`2\ cos^2(alpha/2)-1=cos alpha`
Add 1 to both sides:
`2\ cos^2(alpha/2)=1+cos alpha`
Divide both sides by `2`
`cos^2(alpha/2)=(1+cos alpha)/2`
Solving for `cos(α/2)`, we obtain:
`cos (alpha/2)=+-sqrt((1+cos alpha)/2`
As before, the sign we need depends on the quadrant.
If `α/2` is in the first or fourth quadrants, the formula uses the positive case:
`cos (alpha/2)=sqrt((1+cos alpha)/2`
If `α/2` is in the second or third quadrants, the formula uses the negative case:
`cos (alpha/2)=-sqrt((1+cos alpha)/2`
Half Angle Formula - Tangent
The tangent of a half angle is given by:
`tan (alpha/2)=(1-cos alpha)/(sin alpha)`
Proof
First, we recall `tan x = (sin x) / (cos x)`.
`tan a/2=(sin a/2)/(cos a/2)`
Then we use the sine and cosine of a half angle, as given above:
`=sqrt((1-cos a)/2)/sqrt((1+cos a)/2)`
Next line is the result of multiplying top and bottom by `sqrt 2`.
`=sqrt((1-cos a)/(1+cos a))`
We then multiply top and bottom (under the square root) by `(1 − cos α)`
`=sqrt(((1-cos a)^2)/((1+cos a)(1-cos a)))`
Next is a difference of 2 squares.
`=sqrt((1-cos a)^2/(1-cos^2a))`
We then make use of the identity `sin^2theta+cos^2theta=1`
`=sqrt((1-cos a)^2/(sin^2a))`
We then find the square root:
`=(1-cos a)/(sin a)`
Of course, we would need to make allowance for positive and negative signs, depending on the quadrant in question.
We can also write the tangent of a half angle as follows:
`tan (alpha/2)=(sin alpha)/(1+cos alpha)`
Proof
We multiply numerator (top) and denominator (bottom) of the right hand side of our first result by `1+cos alpha`, and obtain:
`(1-cos alpha)/(sin alpha) xx (1+cos alpha)/(1+cos alpha)`
Next, we use the difference of 2 squares.
`=(1-cos^2alpha)/(sin alpha(1+cos alpha))`
We recall `sin^2θ + cos^2θ = 1`, and use it to obtain:
`=(sin^2alpha)/(sin alpha(1+cos alpha))`
Finally, we cancel out the sin α.
`=(sin alpha)/(1+cos alpha`
Summary of Tan of a Half Angle
`tan (alpha/2)=(1-cos alpha)/(sin alpha)=(sin alpha)/(1+cos alpha`
Using t
It is sometimes useful to define t as the tan of a half angle:
`t=tan (alpha/2)`
This gives us the results:
`sin a=(2t)/(1+t^2)`
`cos alpha=(1-t^2)/(1+t^2)`
`tan\ alpha=(2t)/(1-t^2)`
Tan of the Average of 2 Angles
With some algebraic manipulation, we can obtain:
`tan\ (alpha+beta)/2=(sin alpha+sin beta)/(cos alpha+cos beta)`
Example 1
Find the value of `sin 15^@` using the sine half-angle relationship given above.
Answer
With α = 30° and the formula
`(sin alpha)/2=+-sqrt((1-cos a)/2`
we obtain:
`sin 15^text(o)=+-sqrt((1-cos 30^text(o))/2)` `=+-sqrt((1-0.866...)/2)` `=0.2588`
Here the positive value is chosen because 15° is in the first quadrant.
[Why bother doing this when we can use a calculator? This is just to illustrate that the formula works.]
Example 2
Find the value of `cos 165^@` using the cosine half-angle relationship given above.
Answer
We use α = 330°, and so `α/2 = 165^@`.
`cos 165^text(o)=+-sqrt((1+cos 330^text(o))/2)`
`=+-sqrt((1+0.866...)/2)`
`=-0.9659`
Here the minus sign is used because 165° is in the second quadrant.
Example 3
Show that `2\ cos^2(x/2)-cos x=1`
Answer
Using the above formula to substitute for `(cos alpha)/2`, we get:
`"LHS"=2 cos^2(x/2)-cos x`
`=2(sqrt((1+cos x)/2))^2-cos x`
`=2((1+cos x)/2)-cos x`
`=1+cos x-cos x`
`=1`
`="RHS"`
Exercises: Evaluating and Proving Half-Angle Identities
1. Use the half angle formula to evaluate `sin 75^@`.
Answer
`sin 75^text(o)=+-sqrt((1-cos 150^(text(o)))/2)`
`=+-sqrt(((1+0.866))/2)`
`=0.9659`
First Quadrant, so it's positive.
2. Find the value of `sin(alpha/2)` if `cos alpha=12/13` where 0° < α < 90°.
Answer
`sin (alpha/2)=+-sqrt((1-cos alpha)/2)`
`=sqrt((1-12/13)/2)`
`=sqrt((1/13)/2)`
`=sqrt(1/26)`
`=0.1961`
We choose positive because we are in the first quadrant.
3. Prove the identity: `2\ sin^2(x/2)+cos x=1`
Answer
`"LHS"=2 sin^2(x/2)+cos x`
`=2(sqrt((1-cos x)/(2)))^2+cos x`
`=2((1-cos x)/2)+cos x`
`=1-cos x+cos x`
`=1`
`="RHS"`
4. Prove the identity: `2\ cos^2(theta/2)sec theta=sec theta+1`
Answer
`"LHS"=2\ cos^2(theta/2)sec theta`
`=2(sqrt((1+cos theta)/(2)))^2sec\ theta`
`=(1+cos theta)sec\ theta`
`=(1+cos theta)1/(cos theta)`
`=sec\ theta+1`
`="RHS"`