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20.4 Half-Angle Formulas

Formula for $\sin (α / 2)$ $\sin (α / 2)$ • Formula for $\cos (α / 2)$ $\cos (α / 2)$

If we let $θ = α / 2$ $θ = α / 2$ in the identity $\cos 2 θ = 1 - 2 \sin^{2} θ$ $\cos 2 θ = 1 - 2 \sin^{2} θ$ and then solve for $\sin (α / 2),$ $\sin (α / 2),$

\sin \frac{α}{2} = \pm \sqrt{\frac{1 - \cos α}{2}}

$\sin \frac{α}{2} = \pm \sqrt{\frac{1 - \cos α}{2}}$ (20.26)

Also, with the same substitution in the identity $\cos 2 θ = 2 \cos^{2} θ - 1,$ $\cos 2 θ = 2 \cos^{2} θ - 1,$ which is then solved for $\cos (α / 2),$ $\cos (α / 2),$ we have

\cos \frac{α}{2} = \pm \sqrt{\frac{1 + \cos α}{2}}

$\cos \frac{α}{2} = \pm \sqrt{\frac{1 + \cos α}{2}}$ (20.27)

CAUTION

In each of Eqs. (20.26) and (20.27), the sign chosen depends on the quadrant in which $\frac{α}{2}$ $\frac{α}{2}$ lies.

EXAMPLE 1 Evaluation using cos(α / 2) formula

We can find $\cos 165 °$ $\cos 165 °$ by using the relation

\begin{array}{rcll} \cos 165 ° & = & - \sqrt{\frac{1 + \cos 330 °}{2}} & using Eq . (20.27) \\ = & - \sqrt{\frac{1 + 0.8660}{2}} & = - 0.9659 \end{array}

$\begin{array}{rcll} \cos 165 ° & = & - \sqrt{\frac{1 + \cos 330 °}{2}} & using Eq . (20.27) \\ = & - \sqrt{\frac{1 + 0.8660}{2}} & = - 0.9659 \end{array}$

Here, the minus sign is used, since $165 °$ $165 °$ is in the second quadrant, and the cosine of a second-quadrant angle is negative.

EXAMPLE 2 Evaluation using sin(α / 2) formula

Simplify $\sqrt{\frac{1 - \cos 114 °}{2}}$ $\sqrt{\frac{1 - \cos 114 °}{2}}$ by expressing the result in terms of one-half the given angle. Then, using a calculator, show that the values are equal.

We note that the given expression fits the form of the right side of Eq. (20.26), which means that

\sqrt{\frac{1 - \cos 114 °}{2}} = \sin \frac{1}{2} (114 °) = \sin 57 °

$\sqrt{\frac{1 - \cos 114 °}{2}} = \sin \frac{1}{2} (114 °) = \sin 57 °$

Using a calculator shows that

\sqrt{\frac{1 - \cos 114 °}{2}} = 0.8386705679 and \sin 57 ° = 0.8386705679

$\sqrt{\frac{1 - \cos 114 °}{2}} = 0.8386705679 and \sin 57 ° = 0.8386705679$

which verifies the equation for these values.

EXAMPLE 3 Simplification using cos(α / 2) formula

Simplify the expression $\sqrt{\frac{9 + 9 \cos 6 x}{2}} .$ $\sqrt{\frac{9 + 9 \cos 6 x}{2}} .$

\begin{array}{rcll} \sqrt{\frac{9 + 9 \cos 6 x}{2}} & = & \sqrt{\frac{9 (1 + \cos 6 x)}{2}} & = 3 \sqrt{\frac{1 + \cos 6 x}{2}} \\ = & 3 \cos \frac{1}{2} (6 x) & using Eq . (20.27) with α = 6 x \\ = & 3 \cos 3 x \end{array}

$\begin{array}{rcll} \sqrt{\frac{9 + 9 \cos 6 x}{2}} & = & \sqrt{\frac{9 (1 + \cos 6 x)}{2}} & = 3 \sqrt{\frac{1 + \cos 6 x}{2}} \\ = & 3 \cos \frac{1}{2} (6 x) & using Eq . (20.27) with α = 6 x \\ = & 3 \cos 3 x \end{array}$

Noting the original expression, we see that cos 3x cannot be negative.

EXAMPLE 4 Trig identity with sin(α / 2)—kinetic theory of gases

In the kinetic theory of gases, the expression $\sqrt{{(1 - \cos α)}^{2} + \sin^{2} α}$ $\sqrt{{(1 - \cos α)}^{2} + \sin^{2} α}$ is found. Show that this expression equals $2 \sin \frac{1}{2} α :$ $2 \sin \frac{1}{2} α :$

\begin{array}{rcll} \sqrt{{(1 - \cos α)}^{2} + \sin^{2} α} & = & \sqrt{1 - 2 \cos α + \cos^{2} α + \sin^{2} α} & expanding \\ = & \sqrt{1 - 2 \cos α + 1} & using Eq . (20.26) \\ = & \sqrt{2 - 2 \cos α} \\ = & \sqrt{2 (1 - \cos α)} & factoring \end{array}

$\begin{array}{rcll} \sqrt{{(1 - \cos α)}^{2} + \sin^{2} α} & = & \sqrt{1 - 2 \cos α + \cos^{2} α + \sin^{2} α} & expanding \\ = & \sqrt{1 - 2 \cos α + 1} & using Eq . (20.26) \\ = & \sqrt{2 - 2 \cos α} \\ = & \sqrt{2 (1 - \cos α)} & factoring \end{array}$

This last expression is very similar to that for $\sin \frac{1}{2} α,$ $\sin \frac{1}{2} α,$ except that no 2 appears in the denominator. Therefore, multiplying the numerator and the denominator under the radical by 2 leads to the solution:

\begin{array}{rcl} \sqrt{2 (1 - \cos α)} & = & \sqrt{\frac{4 (1 - \cos α)}{2}} = 2 \sqrt{\frac{1 - \cos α}{2}} \\ = & 2 \sin \frac{1}{2} α & \begin{array}{l} using Eq . (20.26) \end{array} \end{array}

$\begin{array}{rcl} \sqrt{2 (1 - \cos α)} & = & \sqrt{\frac{4 (1 - \cos α)}{2}} = 2 \sqrt{\frac{1 - \cos α}{2}} \\ = & 2 \sin \frac{1}{2} α & \begin{array}{l} using Eq . (20.26) \end{array} \end{array}$

Noting the original expression, we see that $\sin \frac{1}{2} α$ $\sin \frac{1}{2} α$ cannot be negative.

EXAMPLE 5 Evaluation using cos(α / 2) formula

Given that $\tan α = \frac{8}{15} (180 ° < α < 270 °),$ $\tan α = \frac{8}{15} (180 ° < α < 270 °),$ find $\cos (\frac{α}{2}) .$ $\cos (\frac{α}{2}) .$

Knowing that $\tan α = \frac{8}{15}$ $\tan α = \frac{8}{15}$ for a third-quadrant angle, we determine from Fig. 20.20 that $\cos α = - \frac{15}{17} .$ $\cos α = - \frac{15}{17} .$ This means

The terminal side of a position angle passes through (negative 15, negative 8) with length r = 17, and at counterclockwise angle alpha in quadrant 3. — Fig. 20.20

\begin{array}{rcl} \cos \frac{α}{2} & = & - \sqrt{\frac{1 + (- 15/17)}{2}} = - \sqrt{\frac{2}{34}} & using Eq . (20.27) \\ = & - \frac{1}{17} \sqrt{17} = - 0.2425 \end{array}

$\begin{array}{rcl} \cos \frac{α}{2} & = & - \sqrt{\frac{1 + (- 15/17)}{2}} = - \sqrt{\frac{2}{34}} & using Eq . (20.27) \\ = & - \frac{1}{17} \sqrt{17} = - 0.2425 \end{array}$

Because $180 ° < α < 270 °,$ $180 ° < α < 270 °,$ we know that $90 ° < \frac{α}{2} < 135 °$ $90 ° < \frac{α}{2} < 135 °$ and therefore $\frac{α}{2}$ $\frac{α}{2}$ is in the second quadrant. Because the cosine is negative for second-quadrant angles, we use the negative value of the radical.

EXAMPLE 6 Simplification—calculator verification

Show that $2 \cos^{2} \frac{x}{2} - \cos x = 1 .$ $2 \cos^{2} \frac{x}{2} - \cos x = 1 .$

The first step is to substitute for $\cos \frac{x}{2},$ $\cos \frac{x}{2},$ which will result in each term on the left being in terms of x and no $\frac{x}{2}$ $\frac{x}{2}$ terms will exist. This might allow us to combine terms. So we perform this operation, and we have for the left side

\begin{array}{rcll} 2 \cos^{2} \frac{x}{2} - \cos x & = & 2 (\frac{1 + \cos x}{2}) - \cos x & \begin{array}{l} using Eq . (20.27) with \\ both sides squared \end{array} \\ = & 1 + \cos x - \cos x = 1 \end{array}

$\begin{array}{rcll} 2 \cos^{2} \frac{x}{2} - \cos x & = & 2 (\frac{1 + \cos x}{2}) - \cos x & \begin{array}{l} using Eq . (20.27) with \\ both sides squared \end{array} \\ = & 1 + \cos x - \cos x = 1 \end{array}$

From Fig. 20.21, we verify that the graph of $y_{1} = 2 {[\cos (x / 2)]}^{2} - \cos x$ $y_{1} = 2 {[\cos (x / 2)]}^{2} - \cos x$ is the same as the graph of $y_{2} = 1 .$ $y_{2} = 1 .$

A horizontal line passes through (0, 1). Part of the line is red and part is blue. — Fig. 20.21

EXAMPLE 7 Formulas for other functions of α / 2

We can find relations for other functions of $\frac{α}{2}$ $\frac{α}{2}$ by expressing these functions in terms of $\sin (\frac{α}{2})$ $\sin (\frac{α}{2})$ and $\cos (\frac{α}{2}) .$ $\cos (\frac{α}{2}) .$ For example,

\begin{array}{rcl} \sec \frac{α}{2} & = & \frac{1}{\cos \frac{α}{2}} = \pm \frac{1}{\sqrt{\frac{1 + \cos α}{2}}} & using Eq . (20.27) \\ = & \pm \sqrt{\frac{2}{1 + \cos α}} \end{array}

$\begin{array}{rcl} \sec \frac{α}{2} & = & \frac{1}{\cos \frac{α}{2}} = \pm \frac{1}{\sqrt{\frac{1 + \cos α}{2}}} & using Eq . (20.27) \\ = & \pm \sqrt{\frac{2}{1 + \cos α}} \end{array}$

EXERCISES 20.4

In Exercises 1 and 2, make the given changes in the indicated examples of this section and then solve the given problem.

In Example 2, change the $-$ $-$ sign in the numerator to $+ .$ $+ .$
In Example 5, change $\frac{8}{15} (180 ° < α < 270 °)$ $\frac{8}{15} (180 ° < α < 270 °)$ to $- \frac{8}{15} (270 ° < α < 360 °) .$ $- \frac{8}{15} (270 ° < α < 360 °) .$

In Exercises 3–8, use the half-angle formulas to evaluate the given functions.

cos 15°
sin 22.5°
sin 105°
cos 112.5°
$\cos \frac{3 π}{8}$ $\cos \frac{3 π}{8}$
$\sin \frac{11 π}{12}$ $\sin \frac{11 π}{12}$

In Exercises 9–12, simplify the given expressions by giving the results in terms of one-half the given angle. Then use a calculator to verify the result.

$\sqrt{\frac{1 - \cos 236 °}{2}}$ $\sqrt{\frac{1 - \cos 236 °}{2}}$
$\sqrt{\frac{1 + \cos 98 °}{2}}$ $\sqrt{\frac{1 + \cos 98 °}{2}}$
$\sqrt{\frac{1 + \cos 96 °}{2}}$ $\sqrt{\frac{1 + \cos 96 °}{2}}$
$\sqrt{\frac{1 - \cos 328 °}{2}}$ $\sqrt{\frac{1 - \cos 328 °}{2}}$

In Exercises 13–20, simplify the given expressions.

$\sqrt{\frac{1 - \cos 8 x}{2}}$ $\sqrt{\frac{1 - \cos 8 x}{2}}$
$\sqrt{\frac{4 + 4 \cos 8 β}{8}}$ $\sqrt{\frac{4 + 4 \cos 8 β}{8}}$
$\sqrt{\frac{1 + \cos 6 x}{2}}$ $\sqrt{\frac{1 + \cos 6 x}{2}}$
$\sqrt{2 - 2 \cos 64 x}$ $\sqrt{2 - 2 \cos 64 x}$
$\sqrt{4 - 4 \cos 10 θ}$ $\sqrt{4 - 4 \cos 10 θ}$
$\sqrt{18 + 18 \cos 1.4 x}$ $\sqrt{18 + 18 \cos 1.4 x}$
$2 \sin^{2} \frac{x}{2} + \cos x$ $2 \sin^{2} \frac{x}{2} + \cos x$
$2 \cos^{2} \frac{θ}{2} \sec θ$ $2 \cos^{2} \frac{θ}{2} \sec θ$

In Exercises 21–24, evaluate the indicated functions.

Find the value of $\sin (\frac{α}{2})$ $\sin (\frac{α}{2})$ if $\cos α = \frac{12}{13} (0 ° < α < 90 °) .$ $\cos α = \frac{12}{13} (0 ° < α < 90 °) .$
Find the value of $\cos (\frac{α}{2})$ $\cos (\frac{α}{2})$ if $\sin α = - \frac{4}{5} (180 ° < α < 270 °) .$ $\sin α = - \frac{4}{5} (180 ° < α < 270 °) .$
Find the value of $\cos (\frac{α}{2})$ $\cos (\frac{α}{2})$ if $\tan α = - 0.2917 (90 ° < α < 180 °) .$ $\tan α = - 0.2917 (90 ° < α < 180 °) .$
Find the value of $\sin (\frac{α}{2})$ $\sin (\frac{α}{2})$ if $\cos α = 0.4706 (270 ° < α < 360 °) .$ $\cos α = 0.4706 (270 ° < α < 360 °) .$

In Exercises 25–28, derive the required expressions.

Derive an expression for $\csc (\frac{α}{2})$ $\csc (\frac{α}{2})$ in terms of $\sec α$ $\sec α$
Derive an expression for $\sec (\frac{α}{2})$ $\sec (\frac{α}{2})$ in terms of $\sec α .$ $\sec α .$
Derive an expression for $\tan (\frac{α}{2})$ $\tan (\frac{α}{2})$ in terms of $\sin α$ $\sin α$ and $\cos α .$ $\cos α .$
Derive an expression for $\cot (\frac{α}{2})$ $\cot (\frac{α}{2})$ in terms of $\sin α$ $\sin α$ and $\cos α .$ $\cos α .$

In Exercises 29–32, prove the given identities.

$\sin \frac{α}{2} = \frac{1 - \cos α}{2 \sin \frac{α}{2}}$ $\sin \frac{α}{2} = \frac{1 - \cos α}{2 \sin \frac{α}{2}}$
$\cos \frac{θ}{2} = \frac{\sin θ}{2 \sin \frac{θ}{2}}$ $\cos \frac{θ}{2} = \frac{\sin θ}{2 \sin \frac{θ}{2}}$
$2 \cos \frac{x}{2} = (1 + \cos x) \sec \frac{x}{2}$ $2 \cos \frac{x}{2} = (1 + \cos x) \sec \frac{x}{2}$
$\cos^{2} \frac{x}{2} [1 + {(\frac{\sin x}{1 + \cos x})}^{2}] = 1$ $\cos^{2} \frac{x}{2} [1 + {(\frac{\sin x}{1 + \cos x})}^{2}] = 1$

In Exercises 33–36, verify each identity by comparing the graph of the left side with the graph of the right side on a calculator.

$2 \sin^{2} \frac{α}{2} - \cos^{2} \frac{α}{2} = \frac{1 - 3 \cos α}{2}$ $2 \sin^{2} \frac{α}{2} - \cos^{2} \frac{α}{2} = \frac{1 - 3 \cos α}{2}$
$\cos^{2} \frac{A}{2} - \sin^{2} \frac{A}{2} = \frac{\sin 2 A}{2 \sin A}$ $\cos^{2} \frac{A}{2} - \sin^{2} \frac{A}{2} = \frac{\sin 2 A}{2 \sin A}$
$2 \sin^{2} \frac{θ}{2} = \frac{\sin^{2} θ}{1 + \cos θ}$ $2 \sin^{2} \frac{θ}{2} = \frac{\sin^{2} θ}{1 + \cos θ}$
$\tan \frac{α}{2} = \frac{\sin α}{1 + \cos α}$ $\tan \frac{α}{2} = \frac{\sin α}{1 + \cos α}$

In Exercises 37–48, use the half-angle formulas to solve the given problems.

Find $\tan θ$ $\tan θ$ if $\sin (θ / 2) = 3/5 .$ $\sin (θ / 2) = 3/5 .$
In a right triangle with sides and angles as shown in Fig. 20.22, show that $\sin^{2} \frac{A}{2} = \frac{c - b}{2 c} .$ $\sin^{2} \frac{A}{2} = \frac{c - b}{2 c} .$

Fig. 20.22
Find the exact value of $\tan 22.5 °$ $\tan 22.5 °$ using half-angle formulas.
If $180 ° < θ < 270 °$ $180 ° < θ < 270 °$ and $\tan (θ / 2) = - π / 3,$ $\tan (θ / 2) = - π / 3,$ find $\sin θ .$ $\sin θ .$
If $90 ° < θ < 180 °$ $90 ° < θ < 180 °$ and $\sin θ = 4/5,$ $\sin θ = 4/5,$ find $\cos (θ / 2) .$ $\cos (θ / 2) .$
Find the area of the segment of the circle in Fig. 20.23, expressing the result in terms of $θ / 2 .$ $θ / 2 .$

Fig. 20.23
In finding the path of a sliding particle, the expression $\sqrt{8 - 8 \cos θ}$ $\sqrt{8 - 8 \cos θ}$ is used. Simplify this expression.
In designing track for a railway system, the equation $d = 4 r \sin^{2} \frac{A}{2}$ $d = 4 r \sin^{2} \frac{A}{2}$ is used. Solve for d in terms of cos A.
In electronics, in order to find the root-mean-square current in a circuit, it is necessary to express $\sin^{2} ω t$ $\sin^{2} ω t$ in terms of $\cos 2 ω t .$ $\cos 2 ω t .$ Show how this is done.
In studying interference patterns of radio signals, the expression $2 E^{2} - 2 E^{2} \cos (π - θ)$ $2 E^{2} - 2 E^{2} \cos (π - θ)$ arises. Show that this can be written as $4 E^{2} \cos^{2} (θ / 2) .$ $4 E^{2} \cos^{2} (θ / 2) .$
The index of refraction n, the angle A of a prism, and the minimum angle of deflection $ϕ$ $ϕ$ are related by $n = \frac{\sin \frac{1}{2} (A + ϕ)}{\sin \frac{1}{2} A} .$ $n = \frac{\sin \frac{1}{2} (A + ϕ)}{\sin \frac{1}{2} A} .$

See Fig. 20.24. Show that an equivalent expression is $n = \sqrt{\frac{1 - \cos A \cos ϕ + \sin A \sin ϕ}{1 - \cos A}}$ $n = \sqrt{\frac{1 - \cos A \cos ϕ + \sin A \sin ϕ}{1 - \cos A}}$

$A prism with angle Ay. A light ray passes through the prism opposite angle Ay, refracting at angle phi so that it exits at a different angle than it entered.$
Fig. 20.24
For the structure shown in Fig. 20.25, show that $x = 2 l \sin^{2} \frac{1}{2} θ .$ $x = 2 l \sin^{2} \frac{1}{2} θ .$

Fig. 20.25

Answers to Practice Exercises

5 sin 2x
$\pm \sqrt{\frac{2}{1 - \cos x}}$ $\pm \sqrt{\frac{2}{1 - \cos x}}$

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Table of Contents for 20.4 Half-Angle Formulas

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20.4 Half-Angle Formulas