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

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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{matrix} \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{matrix}

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Try It

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

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

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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{matrix} \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{matrix}

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

\begin{matrix} \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{matrix}

For the tangent identity, we have

\begin{matrix} \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{matrix}

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 $\sin (15°)$ using a half-angle formula.

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

\begin{matrix} \sin \frac{30°}{2} & = & \sqrt{\frac{1 - \cos 30°}{2}} \\ = & \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{matrix}

Remember that we can check the answer with a graphing calculator.

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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.

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

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.
$\begin{matrix} \cos \frac{α}{2} & = & \pm \sqrt{\frac{1 + \cos α}{2}} \\ = & \pm \sqrt{\frac{1 + (- \frac{15}{17})}{2}} \\ = & \pm \sqrt{\frac{\frac{2}{17}}{2}} \\ = & \pm \sqrt{\frac{2}{17} \cdot \frac{1}{2}} \\ = & \pm \sqrt{\frac{1}{17}} \\ = & - \frac{\sqrt{17}}{17} \end{matrix}$

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.
$\begin{matrix} \tan \frac{α}{2} & = & \pm \sqrt{\frac{1 - \cos α}{1 + \cos α}} \\ = & \pm \sqrt{\frac{1 - (- \frac{15}{17})}{1 + (- \frac{15}{17})}} \\ = & \pm \sqrt{\frac{\frac{32}{17}}{\frac{2}{17}}} \\ = & \pm \sqrt{\frac{32}{2}} \\ = & - \sqrt{16} \\ = & −4 \end{matrix}$

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.

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Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

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