3.3 Double-angle, half-angle, and reduction formulas (Page 3/8)

Page 3 / 8

Using the power-reducing formulas to prove an identity

Use the power-reducing formulas to prove

\sin^{3} (2 x) = [\frac{1}{2} \sin (2 x)] [1 - \cos (4 x)]

We will work on simplifying the left side of the equation:

\begin{array}{l} \sin^{3} (2 x) = [\sin (2 x)] [\sin^{2} (2 x)] \\ = \sin (2 x) [\frac{1 - \cos (4 x)}{2}] & Substitute the power-reduction formula . \\ = \sin (2 x) (\frac{1}{2}) [1 - \cos (4 x)] \\ = \frac{1}{2} [\sin (2 x)] [1 - \cos (4 x)] \end{array}

Try It

Use the power-reducing formulas to prove that $10 \cos^{4} x = \frac{15}{4} + 5 \cos (2 x) + \frac{5}{4} \cos (4 x) .$

$\begin{array}{l} 10 \cos^{4} x = 10 \cos^{4} x = 10 {(\cos^{2} x)}^{2} \\ = 10 {[\frac{1 + \cos (2 x)}{2}]}^{2} & {Substitute reduction formula for cos}^{2} x . \\ = \frac{10}{4} [1 + 2 \cos (2 x) + \cos^{2} (2 x)] \\ = \frac{10}{4} + \frac{10}{2} \cos (2 x) + \frac{10}{4} (\frac{1 + \cos 2 (2 x)}{2}) & {Substitute reduction formula for cos}^{2} x . \\ = \frac{10}{4} + \frac{10}{2} \cos (2 x) + \frac{10}{8} + \frac{10}{8} \cos (4 x) \\ = \frac{30}{8} + 5 \cos (2 x) + \frac{10}{8} \cos (4 x) \\ = \frac{15}{4} + 5 \cos (2 x) + \frac{5}{4} \cos (4 x) \end{array}$

Using half-angle formulas to find exact values

The next set of identities is the set of half-angle formulas , which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace $θ$ with $\frac{α}{2},$ the half-angle formula for sine is found by simplifying the equation and solving for $\sin (\frac{α}{2}) .$ Note that the half-angle formulas are preceded by a $\pm$ sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which $\frac{α}{2}$ terminates.

The half-angle formula for sine is derived as follows:

\begin{array}{l} \sin^{2} θ = \frac{1 - \cos (2 θ)}{2} \\ \sin^{2} (\frac{α}{2}) = \frac{1 - (\cos 2 \cdot \frac{α}{2})}{2} \\ = \frac{1 - \cos α}{2} \\ \sin (\frac{α}{2}) = \pm \sqrt{\frac{1 - \cos α}{2}} \end{array}

To derive the half-angle formula for cosine, we have

\begin{array}{l} \cos^{2} θ = \frac{1 + \cos (2 θ)}{2} \\ \cos^{2} (\frac{α}{2}) = \frac{1 + \cos (2 \cdot \frac{α}{2})}{2} \\ = \frac{1 + \cos α}{2} \\ \cos (\frac{α}{2}) = \pm \sqrt{\frac{1 + \cos α}{2}} \end{array}

For the tangent identity, we have

\begin{array}{l} \tan^{2} θ = \frac{1 - \cos (2 θ)}{1 + \cos (2 θ)} \\ \tan^{2} (\frac{α}{2}) = \frac{1 - \cos (2 \cdot \frac{α}{2})}{1 + \cos (2 \cdot \frac{α}{2})} \\ = \frac{1 - \cos α}{1 + \cos α} \\ \tan (\frac{α}{2}) = \pm \sqrt{\frac{1 - \cos α}{1 + \cos α}} \end{array}

A General Note

Half-angle formulas

The half-angle formulas are as follows:

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

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

\begin{array}{l} \tan (\frac{α}{2}) = \pm \sqrt{\frac{1 - \cos α}{1 + \cos α}} \\ = \frac{\sin α}{1 + \cos α} \\ = \frac{1 - \cos α}{\sin α} \end{array}

Using a half-angle formula to find the exact value of a sine function

Find $\sin (15^{\circ})$ using a half-angle formula.

Since $15^{\circ} = \frac{30^{\circ}}{2},$ we use the half-angle formula for sine:

\begin{array}{l} \begin{array}{l} \sin \frac{30^{\circ}}{2} = \sqrt{\frac{1 - \cos 30^{\circ}}{2}} \end{array} \\ = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} \\ = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} \\ = \sqrt{\frac{2 - \sqrt{3}}{4}} \\ = \frac{\sqrt{2 - \sqrt{3}}}{2} \end{array}

How To

Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.

Draw a triangle to represent the given information.
Determine the correct half-angle formula.
Substitute values into the formula based on the triangle.
Simplify.

Finding exact values using half-angle identities

Given that $\tan α = \frac{8}{15}$ and $α$ lies in quadrant III, find the exact value of the following:

$\sin (\frac{α}{2})$
$\cos (\frac{α}{2})$
$\tan (\frac{α}{2})$

Using the given information, we can draw the triangle shown in [link] . Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate $\sin α = - \frac{8}{17}$ and $\cos α = - \frac{15}{17} .$

Diagram of a triangle in the x,y-plane. The vertices are at the origin, (-15,0), and (-15,-8). The angle at the origin is alpha. The angle formed by the side (-15,-8) to (-15,0) forms a right angle with the x axis. The hypotenuse across from the right angle is length 17.

Before we start, we must remember that, if $α$ is in quadrant III, then $180° < α < 270°,$ so $\frac{180°}{2} < \frac{α}{2} < \frac{270°}{2} .$ This means that the terminal side of $\frac{α}{2}$ is in quadrant II, since $90° < \frac{α}{2} < 135° .$
To find $\sin \frac{α}{2},$ we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in [link] and simplify.

$\begin{array}{l} \begin{array}{l} \sin \frac{α}{2} = \pm \sqrt{\frac{1 - \cos α}{2}} \end{array} \\ = \pm \sqrt{\frac{1 - (- \frac{15}{17})}{2}} \\ = \pm \sqrt{\frac{\frac{32}{17}}{2}} \\ = \pm \sqrt{\frac{32}{17} \cdot \frac{1}{2}} \\ = \pm \sqrt{\frac{16}{17}} \\ = \pm \frac{4}{\sqrt{17}} \\ = \frac{4 \sqrt{17}}{17} \end{array}$

We choose the positive value of $\sin \frac{α}{2}$ because the angle terminates in quadrant II and sine is positive in quadrant II.
To find $\cos \frac{α}{2},$ we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in [link] , and simplify.
$\begin{array}{l} \begin{array}{l} \cos \frac{α}{2} = \pm \sqrt{\frac{1 + \cos α}{2}} \end{array} \\ = \pm \sqrt{\frac{1 + (- \frac{15}{17})}{2}} \\ = \pm \sqrt{\frac{\frac{2}{17}}{2}} \\ = \pm \sqrt{\frac{2}{17} \cdot \frac{1}{2}} \\ = \pm \sqrt{\frac{1}{17}} \\ = - \frac{\sqrt{17}}{17} \end{array}$

We choose the negative value of $\cos \frac{α}{2}$ because the angle is in quadrant II because cosine is negative in quadrant II.
To find $\tan \frac{α}{2},$ we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in [link] and simplify.
$\begin{array}{l} \begin{array}{l} \tan \frac{α}{2} = \pm \sqrt{\frac{1 - \cos α}{1 + \cos α}} \end{array} \\ = \pm \sqrt{\frac{1 - (- \frac{15}{17})}{1 + (- \frac{15}{17})}} \\ = \pm \sqrt{\frac{\frac{32}{17}}{\frac{2}{17}}} \\ = \pm \sqrt{\frac{32}{2}} \\ = - \sqrt{16} \\ = - 4 \end{array}$

We choose the negative value of $\tan \frac{α}{2}$ because $\frac{α}{2}$ lies in quadrant II, and tangent is negative in quadrant II.

<< Chapter < Page Page > Chapter >>

Practice Key Terms 3

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Essential precalculus, part 2. OpenStax CNX. Aug 20, 2015 Download for free at http://legacy.cnx.org/content/col11845/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Essential precalculus, part 2' conversation and receive update notifications?

Ask

	Macroeconomics MCQ By Candice Butts Start Quiz
	1 AP Key Terms 01 Human Body Anatomy Physiology By OpenStax Start Key Terms
	Immunology Practice Test By Sandhills MLT Start Test
	11 AP 11 Muscular System MCQ By OpenStax Start Quiz
	13 AP Key Terms 13 Anatomy of the Nervous System By OpenStax Start Key Terms
	13 Lec:13 Hypothesis Testing P-values By Janet Forrester Start Quiz
©flickr: quinn	Neuroanatomy By George Turner Start Quiz
	Deciduous Forest By Hope Percle Start Quiz
	22 AP 22 Respiratory System Essay By OpenStax Start Flashcards
	26 AP Key Terms 26 Fluid, Electrolyte, Acid-Base By OpenStax Start Key Terms