# 9.3 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

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

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.

what is the function of sine with respect of cosine , graphically
tangent bruh
Steve
cosx.cos2x.cos4x.cos8x
sinx sin2x is linearly dependent
what is a reciprocal
The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1
Shemmy
Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1
Jeza
each term in a sequence below is five times the previous term what is the eighth term in the sequence
I don't understand how radicals works pls
How look for the general solution of a trig function
stock therom F=(x2+y2) i-2xy J jaha x=a y=o y=b
sinx sin2x is linearly dependent
cr
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
Wrong question
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined? By By By OpenStax By Anh Dao By OpenStax By Brooke Delaney By Jessica Collett By OpenStax By Stephanie Redfern By Richley Crapo By JavaChamp Team By Madison Christian