# 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

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

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.

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.

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

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^@.

sin 75^text(o)=+-sqrt((1-cos 150^(text(o)))/2)

=+-sqrt(((1+0.866))/2)

=0.9659

2. Find the value of sin(alpha/2) if cos alpha=12/13 where 0° < α < 90°.

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

"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

"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"

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