# 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

${\mathrm{sin}}^{3}\left(2x\right)=\left[\frac{1}{2}\text{\hspace{0.17em}}\mathrm{sin}\left(2x\right)\right]\text{\hspace{0.17em}}\left[1-\mathrm{cos}\left(4x\right)\right]$

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

Use the power-reducing formulas to prove that $\text{\hspace{0.17em}}10\text{\hspace{0.17em}}{\mathrm{cos}}^{4}x=\frac{15}{4}+5\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)+\frac{5}{4}\text{\hspace{0.17em}}\mathrm{cos}\left(4x\right).$

## 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 $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}\frac{\alpha }{2},$ the half-angle formula for sine is found by simplifying the equation and solving for $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\alpha }{2}\right).\text{\hspace{0.17em}}$ Note that the half-angle formulas are preceded by a $\text{\hspace{0.17em}}±\text{\hspace{0.17em}}$ sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which $\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ terminates.

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

$\begin{array}{ccc}\hfill {\mathrm{sin}}^{2}\theta & =& \frac{1-\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ \hfill {\mathrm{sin}}^{2}\left(\frac{\alpha }{2}\right)& =& \frac{1-\left(\mathrm{cos}2\cdot \frac{\alpha }{2}\right)}{2}\hfill \\ & =& \frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}\hfill \\ \hfill \mathrm{sin}\left(\frac{\alpha }{2}\right)& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \end{array}$

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

$\begin{array}{ccc}\hfill {\mathrm{cos}}^{2}\theta & =& \frac{1+\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ \hfill {\mathrm{cos}}^{2}\left(\frac{\alpha }{2}\right)& =& \frac{1+\mathrm{cos}\left(2\cdot \frac{\alpha }{2}\right)}{2}\hfill \\ & =& \frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}\hfill \\ \hfill \mathrm{cos}\left(\frac{\alpha }{2}\right)& =& ±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \end{array}$

For the tangent identity, we have

$\begin{array}{ccc}\hfill {\mathrm{tan}}^{2}\theta & =& \frac{1-\mathrm{cos}\left(2\theta \right)}{1+\mathrm{cos}\left(2\theta \right)}\hfill \\ \hfill {\mathrm{tan}}^{2}\left(\frac{\alpha }{2}\right)& =& \frac{1-\mathrm{cos}\left(2\cdot \frac{\alpha }{2}\right)}{1+\mathrm{cos}\left(2\cdot \frac{\alpha }{2}\right)}\hfill \\ & =& \frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }\hfill \\ \hfill \mathrm{tan}\left(\frac{\alpha }{2}\right)& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \end{array}$

## Half-angle formulas

The half-angle formulas    are as follows:

$\mathrm{sin}\left(\frac{\alpha }{2}\right)=±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}$
$\mathrm{cos}\left(\frac{\alpha }{2}\right)=±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}$
$\begin{array}{l}\mathrm{tan}\left(\frac{\alpha }{2}\right)=±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{\mathrm{sin}\text{\hspace{0.17em}}\alpha }\hfill \end{array}$

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

Find $\text{\hspace{0.17em}}\mathrm{sin}\left(15°\right)\text{\hspace{0.17em}}$ using a half-angle formula.

Since $\text{\hspace{0.17em}}15°=\frac{30°}{2},$ we use the half-angle formula for sine:

$\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\frac{30°}{2}& =& \sqrt{\frac{1-\mathrm{cos}30°}{2}}\hfill \\ & =& \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\hfill \\ & =& \sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}}\hfill \\ & =& \sqrt{\frac{2-\sqrt{3}}{4}}\hfill \\ & =& \frac{\sqrt{2-\sqrt{3}}}{2}\hfill \end{array}$

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

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.

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

## Finding exact values using half-angle identities

Given that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha =\frac{8}{15}$ and $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ lies in quadrant III, find the exact value of the following:

1. $\mathrm{sin}\left(\frac{\alpha }{2}\right)$
2. $\mathrm{cos}\left(\frac{\alpha }{2}\right)$
3. $\mathrm{tan}\left(\frac{\alpha }{2}\right)$

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 $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =-\frac{8}{17}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha =-\frac{15}{17}.$

1. Before we start, we must remember that if $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ is in quadrant III, then $\text{\hspace{0.17em}}180°<\alpha <270°,$ so $\text{\hspace{0.17em}}\frac{180°}{2}<\frac{\alpha }{2}<\frac{270°}{2}.\text{\hspace{0.17em}}$ This means that the terminal side of $\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ is in quadrant II, since $\text{\hspace{0.17em}}90°<\frac{\alpha }{2}<135°.$

To find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2},$ 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.

$\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ & =& ±\sqrt{\frac{1-\left(-\frac{15}{17}\right)}{2}}\hfill \\ & =& ±\sqrt{\frac{\frac{32}{17}}{2}}\hfill \\ & =& ±\sqrt{\frac{32}{17}\cdot \frac{1}{2}}\hfill \\ & =& ±\sqrt{\frac{16}{17}}\hfill \\ & =& ±\frac{4}{\sqrt{17}}\hfill \\ & =& \frac{4\sqrt{17}}{17}\hfill \end{array}$

We choose the positive value of $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ because the angle terminates in quadrant II and sine is positive in quadrant II.

2. To find $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2},$ we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in [link] , and simplify.
$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ & =& ±\sqrt{\frac{1+\left(-\frac{15}{17}\right)}{2}}\hfill \\ & =& ±\sqrt{\frac{\frac{2}{17}}{2}}\hfill \\ & =& ±\sqrt{\frac{2}{17}\cdot \frac{1}{2}}\hfill \\ & =& ±\sqrt{\frac{1}{17}}\hfill \\ & =& -\frac{\sqrt{17}}{17}\hfill \end{array}$

We choose the negative value of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ because the angle is in quadrant II because cosine is negative in quadrant II.

3. To find $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2},$ 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.
$\begin{array}{ccc}\hfill \mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \\ & =& ±\sqrt{\frac{1-\left(-\frac{15}{17}\right)}{1+\left(-\frac{15}{17}\right)}}\hfill \\ & =& ±\sqrt{\frac{\frac{32}{17}}{\frac{2}{17}}}\hfill \\ & =& ±\sqrt{\frac{32}{2}}\hfill \\ & =& -\sqrt{16}\hfill \\ & =& -4\hfill \end{array}$

We choose the negative value of $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ because $\text{\hspace{0.17em}}\frac{\alpha }{2}\text{\hspace{0.17em}}$ lies in quadrant II, and tangent is negative in quadrant II.

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