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^2alpha/2-1`
Reverse the equation:
`2\ cos^2alpha/2-1=cos\ alpha`
Add 1 to both sides:
`2cos^2alpha/2=1+cos\ alpha`
Divide both sides by `2`
`cos^2alpha/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)`
We can also write the tangent of a half angle as follows:
`tan\ alpha/2=(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.
Example 2
Find the value of `cos\ 165^@` using the cosine half-angle relationship given above.
Example 3
Show that `2\ cos^2x/2-cos\ x=1`
Exercises: Evaluating and Proving Half-Angle Identities
1. Use the half angle formula to evaluate `sin\ 75^@`.
2. Find the value of `sin(alpha/2)` if `cos\ alpha=12/13` where 0° < α < 90°.
3. Prove the identity: `2\ sin^2x/2+cos\ x=1`
4. Prove the identity: `2\ cos^2theta/2sec\ theta=sec\ theta+1`
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